Contents / Calculus / Multiple Integrals

Chapter 12

Multiple Integrals

Double and triple integrals, polar, cylindrical and spherical coordinates, and change of variables.

Introduction

A single integral adds up a quantity along an interval: chop $[a,b]$ into pieces, multiply the value of $f$ on each piece by its length, add, and take a limit. A double integral does the same over a region of the plane, a triple integral over a solid. The recipe — chop, multiply, add, take the limit — does not change at all. What changes is the geometry: a region of the plane is a far more varied object than an interval, so the central skill of this chapter is describing a region by inequalities in the right order, and the second skill is choosing coordinates in which that description is simple.

The pay-off is large. Volumes, masses, centres of mass, moments of inertia, surface areas and joint probabilities all become double or triple integrals, and each is computed by the same machine: reduce to iterated single integrals and use the Fundamental Theorem of Calculus one variable at a time. Along the way you will meet the two most useful coordinate changes in mathematics — polar and spherical — and the theorem that explains both of them, the change of variables formula with its Jacobian factor.

You will need comfortable single-variable integration (substitution, parts, the standard antiderivatives) and the partial derivatives of the previous chapter; nothing else.

12.1Double integrals over rectangles

Let $R = [a,b]\times[c,d] = \{(x,y) : a \le x \le b,\ c \le y \le d\}$ be a closed rectangle and let $f$ be a function of two variables defined on $R$ . Divide $[a,b]$ into $m$ subintervals of equal width $\Delta x = (b-a)/m$ and $[c,d]$ into $n$ subintervals of width $\Delta y = (d-c)/n$ . The lines through the division points cut $R$ into $mn$ subrectangles $R_{ij}$ , each of area $\Delta A = \Delta x\,\Delta y$ . In each subrectangle choose a sample point $(x_{ij}^*, y_{ij}^*)$ . If $f \ge 0$ , the box with base $R_{ij}$ and height $f(x_{ij}^*, y_{ij}^*)$ has volume $f(x_{ij}^*, y_{ij}^*)\,\Delta A$ , and adding these boxes gives the double Riemann sum

\sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A,

an approximation to the volume of the solid that lies under the graph of $f$ and above $R$ . Refining the grid improves the approximation; the limit is the double integral.

Definition 12.1 (Double integral over a rectangle). The double integral of $f$ over the rectangle $R$ is

\iint_R f(x,y)\,dA = \lim_{m,n\to\infty} \sum_{i=1}^{m}\sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A,

provided the limit exists and has the same value for every choice of sample points. A function for which the limit exists is called integrable on $R$ .

The definition has the same shape as the single-variable one, and the same theorem governs existence: every function continuous on $R$ is integrable, and so is every bounded function whose discontinuities lie on finitely many smooth curves. The proof belongs to real analysis and rests on the uniform continuity of a continuous function on a closed bounded set; for this chapter the practical rule is that every function you are asked to integrate is integrable.

When $f \ge 0$ the integral is the volume $V$ of the solid under the graph $z = f(x,y)$ and above $R$ . When $f$ takes both signs the integral is a signed volume: the part of the solid above the $xy$ -plane counts positively and the part below counts negatively, exactly as a single integral counts area below the axis as negative.

Intuition. A rectangular field receives rain, and the depth of water that fell is $f(x,y)$ centimetres at the point $(x,y)$ : heavier near a hill, lighter in its shadow. To find the total volume of rain, a surveyor cannot measure at every point, so she marks the field into a grid, reads the depth at the centre of each cell, multiplies by the cell's area, and adds. That sum is a double Riemann sum, and the finer her grid the closer she is to the exact volume $\iint_R f\,dA$ .

The surveyor's procedure — sample each cell at its centre — is the two-variable Midpoint Rule.

Method 12.2 (Midpoint Rule for double integrals). With $\bar x_i$ the midpoint of the $i$ -th subinterval of $[a,b]$ and $\bar y_j$ the midpoint of the $j$ -th subinterval of $[c,d]$ ,

\iint_R f(x,y)\,dA \approx \sum_{i=1}^{m}\sum_{j=1}^{n} f(\bar x_i, \bar y_j)\,\Delta A.

Example 12.3 (A midpoint estimate). Use the Midpoint Rule with $m = n = 2$ to estimate $\iint_R (x - 3y^{2})\,dA$ , where $R = [0,2]\times[1,2]$ .

Solution.

The subintervals are $[0,1],[1,2]$ in $x$ and $[1,1.5],[1.5,2]$ in $y$ , so $\Delta A = 1\cdot 0.5 = 0.5$ and the midpoints are $\bar x_1 = 0.5$ , $\bar x_2 = 1.5$ , $\bar y_1 = 1.25$ , $\bar y_2 = 1.75$ .
Evaluate $f(x,y) = x - 3y^{2}$ at the four centres: $f(0.5,1.25) = -4.1875$ , $f(0.5,1.75) = -8.6875$ , $f(1.5,1.25) = -3.1875$ , $f(1.5,1.75) = -7.6875$ .
The sum of the four values is $-23.75$ , so the estimate is $-23.75\times 0.5 = -11.875$ .

The exact value, computed later in this section, is $-12$ ; the estimate is within about one percent with only four sample points. The integral is negative because $x - 3y^{2} < 0$ on all of $R$ , so the whole solid lies below the $xy$ -plane.□

Double integrals inherit the algebraic properties of single integrals, and for the same reason: each property holds for every Riemann sum, so it survives the limit.

Proposition 12.4 (Properties of the double integral). Let $f$ and $g$ be integrable on $R$ and let $c$ be a constant. Then

\iint_R (f + g)\,dA = \iint_R f\,dA + \iint_R g\,dA, \qquad \iint_R cf\,dA = c\iint_R f\,dA.

If $f(x,y) \ge g(x,y)$ on $R$ then $\iint_R f\,dA \ge \iint_R g\,dA$ . If $m \le f(x,y) \le M$ on $R$ then

m\,A(R) \le \iint_R f\,dA \le M\,A(R),

where $A(R) = (b-a)(d-c)$ is the area of the rectangle; in particular $\iint_R 1\,dA = A(R)$ .

Proof. For the first two, the Riemann sum of $f + g$ is the sum of the Riemann sums of $f$ and $g$ with the same sample points, and the Riemann sum of $cf$ is $c$ times that of $f$ ; take limits. For monotonicity, every Riemann sum of $f$ is at least the corresponding sum of $g$ , and limits preserve non-strict inequalities. The last statement is monotonicity applied to the constant functions $m$ and $M$ , whose Riemann sums are $m\,A(R)$ and $M\,A(R)$ for every grid.∎

Nobody evaluates a double integral from the definition. The working method is to integrate one variable at a time, and the theorem that licenses it is the most important in the chapter.

Definition 12.5 (Partial integration and iterated integrals). For $f$ defined on $R = [a,b]\times[c,d]$ , the partial integral of $f$ with respect to $y$ is

A(x) = \int_c^d f(x,y)\,dy,

computed by holding $x$ fixed and integrating in $y$ as in one variable; the result is a function of $x$ alone. Integrating that in turn gives the iterated integral

\int_a^b \int_c^d f(x,y)\,dy\,dx = \int_a^b \left[\int_c^d f(x,y)\,dy\right] dx.

The iterated integral $\int_c^d\int_a^b f(x,y)\,dx\,dy$ is defined in the same way with the roles of the variables exchanged.

The rule for reading an iterated integral is: work from the inside out, and the innermost differential names the variable of the innermost integral. In $\int_a^b\int_c^d f\,dy\,dx$ the $dy$ is inside, so the $y$ -integration with limits $c$ and $d$ is done first, treating $x$ as a constant.

Example 12.6 (An iterated integral in both orders). Evaluate $\displaystyle\int_0^3\int_1^2 x^{2} y\,dy\,dx$ and $\displaystyle\int_1^2\int_0^3 x^{2} y\,dx\,dy$ .

Solution.

In the first, $y$ is the inner variable. Holding $x$ fixed,

\int_1^2 x^2 y\,dy = x^2\left[\frac{y^2}{2}\right]_{y=1}^{y=2} = x^2\cdot\frac{4-1}{2} = \frac{3x^2}{2}.

Now integrate the result in $x$ :

\int_0^3 \frac{3x^2}{2}\,dx = \left[\frac{x^3}{2}\right]_0^3 = \frac{27}{2}.

In the second, $x$ is inner: $\int_0^3 x^{2} y\,dx = y\left[x^{3}/3\right]_0^3 = 9y$ , and then $\int_1^2 9y\,dy = 9\cdot\frac{3}{2} = \frac{27}{2}$ .

Both orders give $\frac{27}{2}$ . That is not a coincidence.□

Theorem 12.7 (Fubini's Theorem). If $f$ is continuous on the rectangle $R = [a,b]\times[c,d]$ , then

\iint_R f(x,y)\,dA = \int_a^b\int_c^d f(x,y)\,dy\,dx = \int_c^d\int_a^b f(x,y)\,dx\,dy.

More generally the conclusion holds whenever $f$ is bounded on $R$ , is discontinuous only on finitely many smooth curves, and both iterated integrals exist.

Proof. Here is the geometric argument for $f \ge 0$ ; the general case is the same argument applied to the positive and negative parts of $f$ separately, and the fully rigorous version, which controls the interchange of two limits in the Riemann sum, is a theorem of real analysis.

The double integral is the volume $V$ of the solid $S$ under the graph. Slice $S$ by the plane $x = x_0$ . The cross-section is the plane region under the curve $z = f(x_0, y)$ for $c \le y \le d$ , so its area is

A(x_0) = \int_c^d f(x_0, y)\,dy.

The slicing principle for volumes — the one that gives volumes of revolution — says that a solid whose cross-sectional area perpendicular to the $x$ -axis is $A(x)$ has volume $\int_a^b A(x)\,dx$ . Hence

V = \int_a^b A(x)\,dx = \int_a^b\int_c^d f(x,y)\,dy\,dx.

Slicing by planes $y = y_0$ instead gives the other iterated integral. Both equal $V$ , so both equal $\iint_R f\,dA$ .∎

Fubini's Theorem is two statements in one. First, the double integral, defined as a limit of sums, can be computed by two ordinary integrations. Second, the two possible orders give the same answer, so you may choose whichever is easier — and as the next examples show, the difference can be between a two-line computation and two rounds of integration by parts.

Example 12.8 (The exact value behind the midpoint estimate). Evaluate $\iint_R (x - 3y^{2})\,dA$ for $R = [0,2]\times[1,2]$ .

Solution.

Integrate in $y$ first, holding $x$ fixed:

\int_1^2 (x - 3y^2)\,dy = \left[xy - y^3\right]_{y=1}^{y=2} = (2x - 8) - (x - 1) = x - 7.

Then integrate in $x$ :

\int_0^2 (x - 7)\,dx = \left[\frac{x^2}{2} - 7x\right]_0^2 = 2 - 14 = -12.

Sanity check: the other order gives $\int_0^2 (x - 3y^{2})\,dx = 2 - 6y^{2}$ , and $\int_1^2 (2 - 6y^{2})\,dy = 2 - 2(8-1) = -12$ . The midpoint estimate $-11.875$ was close.

□

Example 12.9 (Choosing the easier order). Evaluate $\iint_R y\sin(xy)\,dA$ , where $R = [1,2]\times[0,\pi]$ .

Solution.

Integrating in $y$ first means $\int_0^\pi y\sin(xy)\,dy$ , which needs integration by parts, and the result must then be integrated in $x$ by parts again. Try $x$ first instead.
With $y$ held fixed, $\sin(xy)$ has antiderivative $-\cos(xy)/y$ in $x$ , so

\int_1^2 y\sin(xy)\,dx = \Big[-\cos(xy)\Big]_{x=1}^{x=2} = \cos y - \cos 2y.

\int_0^\pi (\cos y - \cos 2y)\,dy = \left[\sin y - \frac{\sin 2y}{2}\right]_0^\pi = 0.

The integral is $0$ : over this rectangle the surface $z = y\sin(xy)$ encloses as much volume below the $xy$ -plane as above it.□

Example 12.10 (A volume). Find the volume of the solid bounded by the elliptic paraboloid $z = 16 - x^{2} - 2y^{2}$ , the planes $x = 2$ and $y = 2$ , and the three coordinate planes.

Solution.

The solid lies above the square $R = [0,2]\times[0,2]$ and under the graph of $f(x,y) = 16 - x^{2} - 2y^{2}$ , which is positive on $R$ (its smallest value there is $16 - 4 - 8 = 4$ ). So $V = \iint_R f\,dA$ .
Integrate in $x$ first:

\int_0^2 (16 - x^2 - 2y^2)\,dx = \left[16x - \frac{x^3}{3} - 2y^2 x\right]_0^2 = 32 - \frac{8}{3} - 4y^2 = \frac{88}{3} - 4y^2.

Then in $y$ :

V = \int_0^2 \left(\frac{88}{3} - 4y^2\right) dy = \frac{176}{3} - \frac{32}{3} = 48.

Sanity check: the height ranges from $4$ to $16$ over a base of area $4$ , so $V$ must lie between $16$ and $64$ . It does.

□

One special case is worth stating on its own, because it turns many double integrals into a product of two single integrals.

Corollary 12.11 (Product of single integrals). If $f(x,y) = g(x)\,h(y)$ on $R = [a,b]\times[c,d]$ , with $g$ and $h$ continuous, then

\iint_R g(x)\,h(y)\,dA = \left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right).

Proof. By Fubini's Theorem the integral is $\int_a^b\int_c^d g(x)h(y)\,dy\,dx$ . In the inner integral $g(x)$ is a constant, so it comes out: the inner integral is $g(x)\int_c^d h(y)\,dy$ . Now $\int_c^d h(y)\,dy$ is a number, so it comes out of the outer integral too.∎

Example 12.12 (Using the product form). Evaluate $\iint_R \sin x\cos y\,dA$ over $R = [0,\pi/2]\times[0,\pi/2]$ .

Solution.

The integrand is a product of a function of $x$ and a function of $y$ , and the region is a rectangle, so the corollary applies.
$\int_0^{\pi/2}\sin x\,dx = [-\cos x]_0^{\pi/2} = 1$ and $\int_0^{\pi/2}\cos y\,dy = [\sin y]_0^{\pi/2} = 1$ .
The double integral is $1\cdot 1 = 1$ .

□

The product form also makes averages quick. The average value of $f$ over a rectangle (or any region) $R$ is

f_{\text{avg}} = \frac{1}{A(R)}\iint_R f(x,y)\,dA,

the constant height of the box that would hold the same volume. For instance, the average of $x^{2} y$ over $[0,1]\times[0,2]$ is $\frac{1}{2}\int_0^1 x^{2}\,dx\int_0^2 y\,dy = \frac{1}{2}\cdot\frac{1}{3}\cdot 2 = \frac{1}{3}$ .

Pitfall. The limits belong to the differentials, not to the order in which you happen to see them. In $\int_0^3\int_1^2 f\,dy\,dx$ the inner integral is in $y$ from $1$ to $2$ and the outer in $x$ from $0$ to $3$ ; the numbers $0$ and $3$ never go with $y$ . Attaching the wrong limits to the wrong variable is the commonest way to lose a whole question, and it is invisible in the arithmetic afterwards. Say to yourself, every time: innermost differential, innermost limits.

Exercises for §12.1 →

12.2Double integrals over general regions

Most regions are not rectangles: the base of a solid is a triangle, a disk, the space between two parabolas. The definition extends by a simple trick. Enclose the region in a rectangle, extend the function by zero, and integrate over the rectangle.

Definition 12.13 (Double integral over a general region). Let $D$ be a bounded region of the plane and $R$ a rectangle containing $D$ . Define $F$ on $R$ by $F(x,y) = f(x,y)$ for $(x,y)$ in $D$ and $F(x,y) = 0$ otherwise. Then

\iint_D f(x,y)\,dA = \iint_R F(x,y)\,dA,

whenever the right-hand side exists.

The extended function $F$ may jump at the boundary of $D$ , but if the boundary consists of finitely many smooth curves and $f$ is continuous on $D$ , the discontinuities of $F$ live on those curves and $F$ is integrable. The choice of $R$ does not matter, because $F$ is zero on whatever part of $R$ lies outside $D$ . The volume interpretation survives unchanged: for $f \ge 0$ , $\iint_D f\,dA$ is the volume of the solid under the graph of $f$ and above $D$ , and the properties of the previous section — linearity, monotonicity, the bounds $m\,A(D) \le \iint_D f\,dA \le M\,A(D)$ — carry over word for word. One property is new. If $D$ is the union of two regions $D_1$ and $D_2$ that overlap only along a boundary curve, then

\iint_D f\,dA = \iint_{D_1} f\,dA + \iint_{D_2} f\,dA,

which lets you cut an awkward region into pieces that are each easy to describe.

Easy to describe means one of two shapes.

Definition 12.14 (Type I and type II regions). A region is of type I if it lies between the graphs of two continuous functions of $x$ :

D = \{(x,y) : a \le x \le b,\ g_1(x) \le y \le g_2(x)\}.

It is of type II if it lies between the graphs of two continuous functions of $y$ :

D = \{(x,y) : c \le y \le d,\ h_1(y) \le x \le h_2(y)\}.

A type I region is one that a vertical line sweeps out cleanly: for each $x$ in $[a,b]$ the vertical line at $x$ enters $D$ at height $g_1(x)$ and leaves at height $g_2(x)$ . A type II region is the same with horizontal lines. Many regions — disks, triangles — are both, and then you have a choice of order; some are only one, and some are neither and must be cut.

Theorem 12.15 (Iterated integrals over type I and type II regions). If $f$ is continuous on a type I region $D = \{a \le x \le b,\ g_1(x) \le y \le g_2(x)\}$ , then

\iint_D f(x,y)\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy\,dx.

If $f$ is continuous on a type II region $D = \{c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$ , then

\iint_D f(x,y)\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x,y)\,dx\,dy.

Proof. Take the type I case. Choose a rectangle $R = [a,b]\times[c,d]$ containing $D$ and let $F$ be $f$ extended by zero. By the definition and Fubini's Theorem,

\iint_D f\,dA = \iint_R F\,dA = \int_a^b\int_c^d F(x,y)\,dy\,dx.

Fix $x$ in $[a,b]$ . Along the vertical segment $c \le y \le d$ , the function $F(x,y)$ is zero for $y < g_1(x)$ and for $y > g_2(x)$ , and equals $f(x,y)$ in between, so

\int_c^d F(x,y)\,dy = \int_{g_1(x)}^{g_2(x)} f(x,y)\,dy.

Substituting this for the inner integral gives the formula. The type II case is identical with horizontal segments.∎

Notice the shape of the result: the inner limits are functions of the outer variable, and the outer limits are constants. That is always the case, and it is the quickest check that an iterated integral over a general region has been set up correctly.

Intuition. Think of a type I region as a piece of land surveyed by walking north along every line of longitude $x$ . On each walk the surveyor records where she crosses into the property, $y = g_1(x)$ , and where she leaves it, $y = g_2(x)$ . The inner integral is what she collects on one walk; the outer integral adds up all the walks from $x = a$ to $x = b$ . If the property is easier to survey walking east along lines of latitude, that is a type II description, and the order of integration swaps.

Example 12.16 (A region between two parabolas). Evaluate $\iint_D (x + 2y)\,dA$ , where $D$ is the region bounded by the parabolas $y = 2x^{2}$ and $y = 1 + x^{2}$ .

Solution.

The parabolas meet where $2x^{2} = 1 + x^{2}$ , that is $x = \pm 1$ . Between those values $1 + x^{2} \ge 2x^{2}$ , so the region is type I:

D = \{(x,y) : -1 \le x \le 1,\ 2x^2 \le y \le 1 + x^2\}.

Inner integral in $y$ :

\int_{2x^2}^{1+x^2} (x + 2y)\,dy = \Big[xy + y^2\Big]_{y=2x^2}^{y=1+x^2} = x(1+x^2) + (1+x^2)^2 - 2x^3 - 4x^4.

Expanding, $x + x^{3} + 1 + 2x^{2} + x^{4} - 2x^{3} - 4x^{4} = -3x^{4} - x^{3} + 2x^{2} + x + 1$ .
The odd powers integrate to zero over $[-1,1]$ , so

\iint_D (x+2y)\,dA = 2\int_0^1 (-3x^4 + 2x^2 + 1)\,dx = 2\left(-\frac{3}{5} + \frac{2}{3} + 1\right) = \frac{32}{15}.

Sanity check: the region has area $\int_{-1}^1 (1 - x^{2})\,dx = \frac{4}{3}$ , and the integrand lies between $-1$ and $5$ on $D$ ; the answer $\frac{32}{15} \approx 2.13$ is comfortably inside $[-\frac{4}{3}, \frac{20}{3}]$ .

□

Example 12.17 (Volume of a tetrahedron by a double integral). Find the volume of the tetrahedron bounded by the planes $x + 2y + z = 2$ , $x = 2y$ , $x = 0$ and $z = 0$ .

Solution.

The solid lies under the plane $z = 2 - x - 2y$ and above its shadow $D$ in the $xy$ -plane. That shadow is bounded by the lines $x = 2y$ , $x = 0$ and $x + 2y = 2$ (where the plane meets $z = 0$ ): a triangle with vertices $(0,0)$ , $(0,1)$ and $(1,\tfrac12)$ .
Sweeping vertically: for $0 \le x \le 1$ , the line at $x$ enters at $y = x/2$ and leaves at $y = 1 - x/2$ . So

V = \int_0^1\int_{x/2}^{1 - x/2} (2 - x - 2y)\,dy\,dx.

Inner integral:

\Big[(2-x)y - y^2\Big]_{y = x/2}^{y = 1 - x/2} = (2-x)\left(1 - \frac{x}{2}\right) - \left(1 - \frac{x}{2}\right)^2 - (2-x)\frac{x}{2} + \frac{x^2}{4}.

Since $(2-x)(1 - x/2) = (2-x)^2/2$ and $(1-x/2)^2 = (2-x)^2/4$ , the first two terms combine to $(2-x)^2/4 = 1 - x + x^{2}/4$ , and the last two give $-x + 3x^{2}/4$ . The total is $1 - 2x + x^{2} = (1-x)^2$ . 4. Then $V = \int_0^1 (1-x)^2\,dx = \frac{1}{3}$ . 5. Sanity check: a tetrahedron has volume $\frac{1}{3}(\text{base area})(\text{height})$ . The base triangle $D$ has area $\frac{1}{2}\cdot 1\cdot 1 = \frac{1}{2}$ (base along $x = 0$ from $y = 0$ to $y = 1$ , apex at $x = 1$ ), and the apex of the solid is $(0,0,2)$ at height $2$ , so $V = \frac{1}{3}\cdot\frac{1}{2}\cdot 2 = \frac{1}{3}$ .□

Example 12.18 (A region that is type II but not conveniently type I). Evaluate $\iint_D xy\,dA$ , where $D$ is bounded by the line $y = x - 1$ and the parabola $y^{2} = 2x + 6$ .

Solution.

Solve simultaneously: $(x-1)^2 = 2x + 6$ gives $x^{2} - 4x - 5 = 0$ , so $x = -1$ or $x = 5$ ; the intersection points are $(-1,-2)$ and $(5,4)$ .
As a type I region $D$ would need to be split at $x = -1$ , because its lower boundary is the parabola's lower branch $y = -\sqrt{2x+6}$ for $-3 \le x \le -1$ and the line $y = x - 1$ for $-1 \le x \le 5$ . Sweeping horizontally avoids this: for $-2 \le y \le 4$ the horizontal line at height $y$ enters at $x = y^{2}/2 - 3$ and leaves at $x = y + 1$ . So $D$ is type II:

D = \{(x,y) : -2 \le y \le 4,\ \tfrac{1}{2}y^2 - 3 \le x \le y + 1\}.

Inner integral in $x$ :

\int_{y^2/2 - 3}^{y+1} xy\,dx = \frac{y}{2}\left[(y+1)^2 - \left(\frac{y^2}{2} - 3\right)^2\right] = \frac{y}{2}\left(-\frac{y^4}{4} + 4y^2 + 2y - 8\right) = -\frac{y^5}{8} + 2y^3 + y^2 - 4y.

Outer integral:

\int_{-2}^{4}\left(-\frac{y^5}{8} + 2y^3 + y^2 - 4y\right) dy = \left[-\frac{y^6}{48} + \frac{y^4}{2} + \frac{y^3}{3} - 2y^2\right]_{-2}^{4} = 32 - (-4) = 36.

□

The freedom to choose the order is more than a convenience. Sometimes the inner antiderivative in one order does not exist in closed form — $\sin(y^{2})$ , $e^{x^{2}}$ , $\sqrt{1 + x^{3}}$ — while the other order is elementary.

Method 12.19 (Reversing the order of integration).

Read the region off the limits as inequalities: for $\int_a^b\int_{g_1(x)}^{g_2(x)} f\,dy\,dx$ they are $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$ .
Sketch $D$ from those inequalities, marking the corner points.
Describe the same region the other way: find the overall range of $y$ , and for each $y$ the entering and leaving values of $x$ as functions of $y$ .
Rewrite the iterated integral with the new limits and the differentials in the new order.

Example 12.20 (An order that cannot be done becomes one that can). Evaluate $\displaystyle\int_0^1\int_x^{1} \sin(y^{2})\,dy\,dx$ .

Solution.

As written, the inner integral $\int\sin(y^{2})\,dy$ has no elementary antiderivative, so this order is impossible.
The limits say $0 \le x \le 1$ and $x \le y \le 1$ : the triangle with vertices $(0,0)$ , $(0,1)$ , $(1,1)$ , above the line $y = x$ .
Sweeping horizontally: $y$ runs from $0$ to $1$ , and for each $y$ the horizontal line enters the triangle at $x = 0$ and leaves at $x = y$ . So

\int_0^1\int_x^1 \sin(y^2)\,dy\,dx = \int_0^1\int_0^y \sin(y^2)\,dx\,dy.

The inner integral is now trivial, since $\sin(y^{2})$ is constant in $x$ : it equals $y\sin(y^{2})$ . That factor of $y$ is exactly what makes the substitution $u = y^{2}$ work:

\int_0^1 y\sin(y^2)\,dy = \left[-\frac{\cos(y^2)}{2}\right]_0^1 = \frac{1 - \cos 1}{2} \approx 0.23.

Sanity check: on the triangle $\sin(y^{2})$ lies between $0$ and $\sin 1 \approx 0.84$ , and the triangle has area $\frac12$ , so the integral lies between $0$ and $0.42$ .

□

Setting $f = 1$ turns a double integral into an area.

Proposition 12.21 (Area as a double integral). For a bounded region $D$ with a piecewise-smooth boundary,

A(D) = \iint_D 1\,dA.

Proof. The integral of $1$ over $D$ is the volume of the solid cylinder with base $D$ and height $1$ , and a cylinder's volume is base area times height.∎

For a type I region this reads $A(D) = \int_a^b [g_2(x) - g_1(x)]\,dx$ , the familiar area-between-curves formula, so the proposition is not new information. Its value is conceptual: it says that area, mass and probability are all the same kind of object, and in polar coordinates and after a change of variables it produces area formulas that are not obvious at all.

Example 12.22 (Area between a parabola and a line). Find the area of the region bounded by $y = x^{2}$ and $y = 2x$ .

Solution.

The curves meet where $x^{2} = 2x$ , at $x = 0$ and $x = 2$ , and between them $2x \ge x^{2}$ .
As a double integral, $A = \int_0^2\int_{x^{2}}^{2x} 1\,dy\,dx = \int_0^2 (2x - x^{2})\,dx = 4 - \frac{8}{3} = \frac{4}{3}$ .

□

Pitfall. The outer limits of an iterated integral must be numbers. If you find yourself writing $\int_0^{y}\int_{0}^{1} f\,dx\,dy$ with a $y$ in the outermost limit, the region has been described backwards: the answer would be a function of $y$ , not a number. Equally, an inner limit that depends on the inner variable itself, such as $\int_0^1\int_0^{x} f\,dx\,dy$ , is meaningless. The dependence always runs outward: inner limits may involve the outer variable, never the reverse.

Exercises for §12.2 →

12.3Double integrals in polar coordinates

A disk described in Cartesian coordinates is $-\sqrt{a^{2} - x^{2}} \le y \le \sqrt{a^{2} - x^{2}}$ , and integrals over it fill up with square roots. In polar coordinates the same disk is $0 \le r \le a$ , $0 \le \theta \le 2\pi$ , a rectangle in the $(r,\theta)$ -plane. The price of that simplicity is a factor $r$ in the area element, and it is important to see exactly where the factor comes from.

Recall that polar coordinates $(r,\theta)$ are related to Cartesian coordinates by

x = r\cos\theta, \qquad y = r\sin\theta, \qquad r^2 = x^2 + y^2, \qquad \tan\theta = \frac{y}{x}.

A polar rectangle is a region of the form $R = \{(r,\theta) : a \le r \le b,\ \alpha \le \theta \le \beta\}$ , the part of an annulus cut out by two rays. Divide $[a,b]$ into $m$ subintervals of width $\Delta r$ and $[\alpha,\beta]$ into $n$ subintervals of width $\Delta\theta$ . The circles $r = r_i$ and the rays $\theta = \theta_j$ cut $R$ into polar subrectangles $R_{ij}$ . These are not congruent: the ones far from the origin are wider. The area of a sector of angle $\Delta\theta$ and radius $r$ is $\frac12 r^{2}\Delta\theta$ , so the area of $R_{ij}$ , a difference of two sectors, is

\Delta A_{ij} = \frac12 r_i^2\,\Delta\theta - \frac12 r_{i-1}^2\,\Delta\theta = \frac12 (r_i + r_{i-1})(r_i - r_{i-1})\,\Delta\theta = r_i^*\,\Delta r\,\Delta\theta,

where $r_i^* = \frac12(r_i + r_{i-1})$ is the midpoint radius. Taking the sample point $(r_i^*\cos\theta_j^*, r_i^*\sin\theta_j^*)$ in each subrectangle, the Riemann sum for $\iint_R f\,dA$ becomes

\sum_{i=1}^{m}\sum_{j=1}^{n} f(r_i^*\cos\theta_j^*, r_i^*\sin\theta_j^*)\,r_i^*\,\Delta r\,\Delta\theta,

which is an ordinary Riemann sum for the function $g(r,\theta) = r\,f(r\cos\theta, r\sin\theta)$ over the ordinary rectangle $[a,b]\times[\alpha,\beta]$ in the $r\theta$ -plane. Letting $m,n \to\infty$ proves the following.

Theorem 12.23 (Change to polar coordinates in a double integral). If $f$ is continuous on the polar rectangle $R = \{a \le r \le b,\ \alpha \le \theta \le \beta\}$ with $0 \le a$ and $0 \le \beta - \alpha \le 2\pi$ , then

\iint_R f(x,y)\,dA = \int_\alpha^\beta\int_a^b f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta.

Proof. The derivation above: the area of a polar subrectangle is $r_i^*\Delta r\Delta\theta$ , so the polar Riemann sums for $f$ are the Cartesian Riemann sums for $r f(r\cos\theta, r\sin\theta)$ over $[a,b]\times[\alpha,\beta]$ , and the two limits coincide.∎

In words: replace $x$ and $y$ by $r\cos\theta$ and $r\sin\theta$ , use polar limits, and replace $dA$ by $r\,dr\,d\theta$ . The factor $r$ is not optional and not decorative; it is the area of a polar cell, and without it the integral of $1$ over a disk of radius $a$ would come out as $2\pi a$ instead of $\pi a^{2}$ .

The theorem extends, exactly as the type I formula extended Fubini, to regions bounded by two polar curves. If

D = \{(r,\theta) : \alpha \le \theta \le \beta,\ h_1(\theta) \le r \le h_2(\theta)\},

then

\iint_D f(x,y)\,dA = \int_\alpha^\beta\int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta.

With $f = 1$ this recovers the area formula for polar curves, $A = \int_\alpha^\beta \frac12\left[h_2(\theta)^2 - h_1(\theta)^2\right]d\theta$ , since $\int_{h_1}^{h_2} r\,dr = \frac12(h_2^2 - h_1^2)$ .

Intuition. Cut a pizza into the usual wedges, then cut each wedge across into an inner piece and an outer piece. The outer pieces are bigger, even though both have the same angular width and the same radial thickness. A polar cell at radius $r$ has arc length $r\,\Delta\theta$ and thickness $\Delta r$ , so its area is about $r\,\Delta r\,\Delta\theta$ : the factor $r$ records that the outer pieces are worth more.

Example 12.24 (A polynomial over half an annulus). Evaluate $\iint_R (3x + 4y^{2})\,dA$ , where $R$ is the region in the upper half-plane between the circles $x^{2} + y^{2} = 1$ and $x^{2} + y^{2} = 4$ .

Solution.

In polar coordinates $R = \{1 \le r \le 2,\ 0 \le \theta \le \pi\}$ .
Substitute and add the factor $r$ :

\iint_R (3x + 4y^2)\,dA = \int_0^\pi\int_1^2 (3r\cos\theta + 4r^2\sin^2\theta)\,r\,dr\,d\theta = \int_0^\pi\int_1^2 (3r^2\cos\theta + 4r^3\sin^2\theta)\,dr\,d\theta.

Inner integral: $\left[r^{3}\cos\theta + r^{4}\sin^{2}\theta\right]_1^2 = 7\cos\theta + 15\sin^{2}\theta$ .
Outer integral, using $\sin^{2}\theta = \frac12(1 - \cos 2\theta)$ :

\int_0^\pi (7\cos\theta + 15\sin^2\theta)\,d\theta = 0 + \frac{15}{2}\left[\theta - \frac{\sin 2\theta}{2}\right]_0^\pi = \frac{15\pi}{2}.

The $3x$ term contributes nothing because the region is symmetric under $x \mapsto -x$ and $3x$ is odd in $x$ ; the polar computation shows this as $\int_0^\pi\cos\theta\,d\theta = 0$ .□

Example 12.25 (A sector of an annulus). Evaluate $\iint_D xy\,dA$ over the quarter-annulus $D = \{1 \le x^{2} + y^{2} \le 9,\ x \ge 0,\ y \ge 0\}$ .

Solution.

The region is $1 \le r \le 3$ , $0 \le \theta \le \pi/2$ , and $xy = r^{2}\cos\theta\sin\theta$ .
With the factor $r$ the integrand becomes $r^{3}\cos\theta\sin\theta$ , which is a product of a function of $r$ and a function of $\theta$ over a rectangle in the $r\theta$ -plane, so the product corollary applies:

\iint_D xy\,dA = \int_1^3 r^3\,dr\int_0^{\pi/2}\cos\theta\sin\theta\,d\theta = \frac{81 - 1}{4}\cdot\left[\frac{\sin^2\theta}{2}\right]_0^{\pi/2} = 20\cdot\frac12 = 10.

Sanity check: $xy$ is at most $\frac92$ on $D$ (at $r = 3$ , $\theta = \pi/4$ ) and the region has area $\frac{\pi}{4}(9 - 1) = 2\pi \approx 6.28$ , so the integral cannot exceed $28$ ; also $xy > 0$ throughout, so the answer must be positive. Both hold.

□

Example 12.26 (Volume under a paraboloid). Find the volume of the solid bounded by the paraboloid $z = 1 - x^{2} - y^{2}$ and the plane $z = 0$ .

Solution.

The paraboloid meets the plane $z = 0$ on the circle $x^{2} + y^{2} = 1$ , so the solid sits over the unit disk $D$ and $V = \iint_D (1 - x^{2} - y^{2})\,dA$ .
In polar coordinates $1 - x^{2} - y^{2} = 1 - r^{2}$ and $D = \{0 \le r \le 1,\ 0 \le \theta \le 2\pi\}$ :

V = \int_0^{2\pi}\int_0^1 (1 - r^2)\,r\,dr\,d\theta = 2\pi\left[\frac{r^2}{2} - \frac{r^4}{4}\right]_0^1 = 2\pi\cdot\frac14 = \frac{\pi}{2}.

Sanity check: the solid is inscribed in the cylinder of radius $1$ and height $1$ , volume $\pi$ , and contains the cone of the same base and height, volume $\pi/3$ . Indeed $\pi/3 < \pi/2 < \pi$ ; the paraboloid holds exactly half the cylinder.

□

Example 12.27 (A region between two circles). Find the area of the region that lies inside the circle $r = 2\cos\theta$ and outside the circle $r = 1$ .

Solution.

The curve $r = 2\cos\theta$ is the circle of radius $1$ centred at $(1,0)$ , traced for $-\pi/2 \le \theta \le \pi/2$ . It meets $r = 1$ where $2\cos\theta = 1$ , so at $\theta = \pm\pi/3$ .
For $-\pi/3 \le \theta \le \pi/3$ the ray at angle $\theta$ enters the region at $r = 1$ and leaves at $r = 2\cos\theta$ . So

A = \int_{-\pi/3}^{\pi/3}\int_1^{2\cos\theta} r\,dr\,d\theta = \int_{-\pi/3}^{\pi/3}\frac12\left(4\cos^2\theta - 1\right)d\theta.

Using $4\cos^{2}\theta = 2 + 2\cos 2\theta$ , the integrand is $\frac12 + \cos 2\theta$ , so

A = \left[\frac{\theta}{2} + \frac{\sin 2\theta}{2}\right]_{-\pi/3}^{\pi/3} = \frac{\pi}{3} + \sin\frac{2\pi}{3} = \frac{\pi}{3} + \frac{\sqrt 3}{2} \approx 1.91.

Sanity check: the circle $r = 2\cos\theta$ has area $\pi \approx 3.14$ , and the unit disk removes less than half of it, so the answer should be a little more than $1.57$ . It is.

□

Example 12.28 (A Gaussian over a disk). Evaluate $\iint_D e^{-(x^{2} + y^{2})}\,dA$ , where $D$ is the disk $x^{2} + y^{2} \le 4$ .

Solution.

The integrand depends only on $x^{2} + y^{2} = r^{2}$ and the region is a disk: polar coordinates are compulsory. The disk is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ .
Write the integral as $\int_0^{2\pi}\int_0^2 e^{-r^{2}}\,r\,dr\,d\theta$ . The $\theta$ integral contributes the factor $2\pi$ .
For the $r$ integral substitute $u = -r^{2}$ , $du = -2r\,dr$ ; the factor $r$ that polar coordinates supplied is precisely what the substitution needs:

\int_0^2 r e^{-r^2}\,dr = -\frac12\int_0^{-4} e^u\,du = \frac12\left(1 - e^{-4}\right).

The integral is $2\pi\cdot\frac12(1 - e^{-4}) = \pi(1 - e^{-4}) \approx 3.08$ .

Note that $\int e^{-x^{2}}\,dx$ has no elementary antiderivative, so the Cartesian version of this integral cannot be started. Polar coordinates do not merely simplify it; they make it possible.□

The same trick evaluates one of the most important integrals in mathematics.

Example 12.29 (The Gaussian integral). Show that $\displaystyle\int_{-\infty}^{\infty} e^{-x^{2}}\,dx = \sqrt{\pi}$ .

Solution.

Let $I = \int_{-\infty}^\infty e^{-x^{2}}\,dx$ ; the integral converges because $e^{-x^{2}} \le e^{-|x|}$ for $|x| \ge 1$ . Since the name of the variable is immaterial, also $I = \int_{-\infty}^\infty e^{-y^{2}}\,dy$ .
Multiply the two, and use the product corollary in reverse to turn the product into a double integral over the whole plane:

I^2 = \int_{-\infty}^\infty e^{-x^2}\,dx\int_{-\infty}^\infty e^{-y^2}\,dy = \iint_{\mathbb R^2} e^{-(x^2+y^2)}\,dA.

The double integral over the plane is the limit as $a \to\infty$ of the integral over the disk of radius $a$ , which the previous example computed:

\iint_{x^2+y^2\le a^2} e^{-(x^2+y^2)}\,dA = \pi\left(1 - e^{-a^2}\right) \to \pi.

Hence $I^{2} = \pi$ , and since $I > 0$ , $I = \sqrt\pi$ .

The square of the integral could be done although the integral itself could not, because squaring turned a line into a plane, where the symmetry of $e^{-r^{2}}$ becomes usable. Rescaling gives $\int_{-\infty}^\infty e^{-x^{2}/2}\,dx = \sqrt{2\pi}$ , the normalising constant of the normal distribution.□

Pitfall. The single most common error in this chapter is writing $\iint_D f\,dA = \int\int f(r\cos\theta, r\sin\theta)\,dr\,d\theta$ without the factor $r$ . Check every polar integral for it before integrating. A second trap: the substitution $r^{2} = x^{2} + y^{2}$ is safe, but $\sqrt{x^{2} + y^{2}} = r$ only because $r \ge 0$ by convention; keep $r \ge 0$ and describe regions that cross the origin by their $\theta$ -range rather than by allowing negative $r$ .

Exercises for §12.3 →

12.4Applications: mass, centre of mass, moments and probability

The volume interpretation is only the first use of a double integral. Whenever a quantity is spread over a plane region with a known density, the total is a double integral, and the same integral with an extra factor of $x$ or $y$ or $x^{2} + y^{2}$ measures how that quantity is distributed.

Definition 12.30 (Mass and moments of a lamina). A lamina occupies a region $D$ of the plane with density $\rho(x,y)$ (mass per unit area), continuous on $D$ . Its mass, its moment about the $x$ -axis, and its moment about the $y$ -axis are

m = \iint_D \rho(x,y)\,dA, \qquad M_x = \iint_D y\,\rho(x,y)\,dA, \qquad M_y = \iint_D x\,\rho(x,y)\,dA.

Its centre of mass is the point $(\bar x, \bar y)$ with

\bar x = \frac{M_y}{m}, \qquad \bar y = \frac{M_x}{m}.

When $\rho$ is constant the centre of mass is called the centroid of $D$ .

The mass formula is a Riemann sum argument: a small cell of area $\Delta A$ at $(x^*, y^*)$ has mass about $\rho(x^*, y^*)\Delta A$ , and the sum of these tends to the integral. The moments encode leverage. A point mass $m_0$ at $(x, y)$ exerts a turning tendency $m_0 y$ about the $x$ -axis — its mass times its lever arm — and a lamina's moment is the sum of those over all its cells. The centre of mass is the point where a single particle of mass $m$ would produce the same two moments, which is why $m\bar x = M_y$ and $m\bar y = M_x$ . Physically it is the balance point: the lamina rests level on a pin placed there. Notice the crossed subscripts, which trip up everyone once: $M_x$ , the moment about the $x$ -axis, involves the distance $y$ from that axis, and it is $M_x$ that gives $\bar y$ .

Intuition. Two children sit on a see-saw. The heavier one sits closer to the pivot, and it balances when mass times distance is the same on both sides. A lamina is a see-saw in two dimensions with infinitely many children, one per cell, each weighing $\rho\,\Delta A$ . The moment adds up every child's mass times distance from the axis, and the centre of mass is where the pivot must go so that the total moment on each side cancels.

Example 12.31 (Centre of mass of a triangular lamina). A lamina occupies the triangle with vertices $(0,0)$ , $(1,0)$ and $(0,2)$ and has density $\rho(x,y) = 1 + 3x + y$ . Find its mass and centre of mass.

Solution.

The hypotenuse joins $(1,0)$ to $(0,2)$ , so it is the line $y = 2 - 2x$ , and the triangle is $D = \{0 \le x \le 1,\ 0 \le y \le 2 - 2x\}$ .
Mass:

m = \int_0^1\int_0^{2-2x} (1 + 3x + y)\,dy\,dx = \int_0^1\left[(1+3x)y + \frac{y^2}{2}\right]_{y=0}^{y=2-2x} dx = \int_0^1 \left(4 - 4x^2\right) dx = \frac{8}{3},

where the bracket simplifies because $(1+3x)(2-2x) + 2(1-x)^2 = (2 + 4x - 6x^{2}) + (2 - 4x + 2x^{2}) = 4 - 4x^{2}$ . 3. Moment about the $y$ -axis (multiply the density by $x$ ): the inner integral is $x$ times the bracket above, so

M_y = \int_0^1 x(4 - 4x^2)\,dx = 2 - 1 = 1, \qquad \bar x = \frac{M_y}{m} = \frac{3}{8}.

Moment about the $x$ -axis:

M_x = \int_0^1\int_0^{2-2x} y(1 + 3x + y)\,dy\,dx = \int_0^1\left[(1+3x)\frac{y^2}{2} + \frac{y^3}{3}\right]_0^{2-2x} dx.

With $u = 1 - x$ (so $1 + 3x = 4 - 3u$ and $2 - 2x = 2u$ ), the integrand becomes $2(4 - 3u)u^{2} + \frac{8u^{3}}{3}$ , and

M_x = \int_0^1\left(8u^2 - 6u^3 + \frac{8u^3}{3}\right) du = \frac83 - \frac32 + \frac23 = \frac{11}{6}, \qquad \bar y = \frac{M_x}{m} = \frac{11}{6}\cdot\frac{3}{8} = \frac{11}{16}.

The centre of mass is $\left(\frac38, \frac{11}{16}\right)$ . Sanity check: the centroid of the triangle is the average of its vertices, $\left(\frac13, \frac23\right)$ . The density increases with $x$ and with $y$ , so the centre of mass should sit slightly to the right of and above the centroid — and $\frac38 > \frac13$ , $\frac{11}{16} > \frac23$ .

□

Example 12.32 (Centroid of a half-disk). Find the centroid of the half-disk $D = \{x^{2} + y^{2} \le a^{2},\ y \ge 0\}$ .

Solution.

By symmetry $\bar x = 0$ : the region and the constant density are unchanged by $x \mapsto -x$ , which changes the sign of $M_y$ , so $M_y = 0$ .
Take $\rho = 1$ , so $m = A(D) = \frac12\pi a^{2}$ .
For $M_x$ use polar coordinates, where $y = r\sin\theta$ and the half-disk is $0 \le r \le a$ , $0 \le \theta \le \pi$ :

M_x = \iint_D y\,dA = \int_0^\pi\int_0^a r\sin\theta\cdot r\,dr\,d\theta = \int_0^a r^2\,dr\int_0^\pi\sin\theta\,d\theta = \frac{a^3}{3}\cdot 2 = \frac{2a^3}{3}.

Therefore

\bar y = \frac{M_x}{m} = \frac{2a^3/3}{\pi a^2/2} = \frac{4a}{3\pi} \approx 0.42a.

Sanity check: the centroid must lie inside the half-disk, so $0 < \bar y < a$ , and since more of the area is near the diameter than near the top, $\bar y$ should be below $a/2$ . Both hold.

□

The moment of inertia measures the resistance of a rotating body to changes in its spin, in the way mass measures resistance to changes in velocity. It weights each cell not by distance from the axis but by distance squared.

Definition 12.33 (Moments of inertia). For a lamina with density $\rho$ on $D$ , the moments of inertia about the $x$ -axis, about the $y$ -axis, and about the origin (the polar moment of inertia) are

I_x = \iint_D y^2\rho(x,y)\,dA, \qquad I_y = \iint_D x^2\rho(x,y)\,dA, \qquad I_0 = \iint_D (x^2 + y^2)\rho(x,y)\,dA = I_x + I_y.

Example 12.34 (Moments of inertia of a uniform disk). Find $I_0$ , $I_x$ and $I_y$ for a disk of radius $a$ centred at the origin with constant density $\rho$ , and express them in terms of the mass $m = \rho\pi a^{2}$ .

Solution.

In polar coordinates $x^{2} + y^{2} = r^{2}$ , so

I_0 = \int_0^{2\pi}\int_0^a \rho\,r^2\cdot r\,dr\,d\theta = 2\pi\rho\cdot\frac{a^4}{4} = \frac{\pi\rho a^4}{2} = \frac12 m a^2.

The disk is symmetric under the swap of $x$ and $y$ , so $I_x = I_y$ , and since $I_x + I_y = I_0$ each equals $\frac14 m a^{2}$ .

The factor $\frac12$ in $I_0 = \frac12 ma^{2}$ is the familiar one from the physics of a spinning disk; a thin hoop of the same mass and radius would have $I_0 = ma^{2}$ , because all its mass sits at the full distance $a$ .□

The last application replaces mass by probability. If $X$ and $Y$ are continuous random variables, their joint behaviour is described by a joint density function.

Definition 12.35 (Joint density and expected values). A joint density function of two random variables $X$ and $Y$ is a function $f$ with $f(x,y) \ge 0$ and $\iint_{\mathbb R^{2}} f(x,y)\,dA = 1$ such that, for every region $D$ ,

P\big((X,Y)\in D\big) = \iint_D f(x,y)\,dA.

The expected values of $X$ and $Y$ are

\mu_1 = \iint_{\mathbb R^2} x f(x,y)\,dA, \qquad \mu_2 = \iint_{\mathbb R^2} y f(x,y)\,dA.

The expected values are exactly the coordinates of the centre of mass of a lamina of total mass $1$ with density $f$ . If $X$ and $Y$ are independent the joint density factors as $f(x,y) = f_1(x)f_2(y)$ , and integrals over rectangles then split by the product corollary; the interesting cases are the regions that are not rectangles.

Example 12.36 (Working with a joint density). Let $f(x,y) = x + y$ for $0 \le x \le 1$ , $0 \le y \le 1$ , and $f = 0$ elsewhere. Verify that $f$ is a joint density, find $P(X + Y \le 1)$ , and find $\mu_1$ .

Solution.

$f \ge 0$ on its support, and $\iint f\,dA = \int_0^1\int_0^1 (x + y)\,dy\,dx = \int_0^1 \left(x + \frac12\right) dx = 1$ , so $f$ is a density.
The event $X + Y \le 1$ is the triangle $D = \{0 \le x \le 1,\ 0 \le y \le 1 - x\}$ :

P(X + Y \le 1) = \int_0^1\int_0^{1-x} (x+y)\,dy\,dx = \int_0^1\left[x(1-x) + \frac{(1-x)^2}{2}\right] dx = \int_0^1 \left(\frac12 - \frac{x^2}{2}\right) dx = \frac12 - \frac16 = \frac13.

The expected value of $X$ :

\mu_1 = \int_0^1\int_0^1 x(x+y)\,dy\,dx = \int_0^1 \left(x^2 + \frac{x}{2}\right) dx = \frac13 + \frac14 = \frac{7}{12}.

Sanity check: the triangle is half the square, but the density is larger in the far corner where $x + y$ is big, so $P(X + Y \le 1)$ should be less than $\frac12$ . Likewise the density grows with $x$ , so $\mu_1$ should exceed $\frac12$ . Both $\frac13$ and $\frac{7}{12}$ agree.

□

Pitfall. $M_x$ is the moment about the $x$ -axis and it contains $y$ , not $x$ : distance from the $x$ -axis is measured by $y$ . And $\bar x = M_y/m$ , not $M_x/m$ . Write the definitions down before computing rather than trusting the subscript, and check the answer against symmetry: a region symmetric about the $y$ -axis with a density symmetric under $x \mapsto -x$ has $\bar x = 0$ , whatever the integrals seem to say.

Exercises for §12.4 →

12.5Surface area

The arc length of a curve $y = f(x)$ is $\int_a^b\sqrt{1 + f'(x)^2}\,dx$ : the length of the graph exceeds the length of the interval beneath it by a factor that grows with the slope. The area of a surface $z = f(x,y)$ has the same structure, and the same derivation.

Let $S$ be the graph of $z = f(x,y)$ over a region $D$ , where $f$ has continuous partial derivatives. Partition $D$ into small rectangles $R_{ij}$ of area $\Delta A = \Delta x\,\Delta y$ , and let $(x_i, y_j)$ be the corner of $R_{ij}$ nearest the origin with $P_{ij} = (x_i, y_j, f(x_i,y_j))$ the point of $S$ above it. The part of $S$ above $R_{ij}$ is approximated by the part of the tangent plane at $P_{ij}$ above $R_{ij}$ , which is a parallelogram. Its sides are the vectors along the tangent plane in the $x$ - and $y$ -directions,

\mathbf a = \Delta x\,\mathbf i + f_x(x_i,y_j)\,\Delta x\,\mathbf k, \qquad \mathbf b = \Delta y\,\mathbf j + f_y(x_i,y_j)\,\Delta y\,\mathbf k,

because moving $\Delta x$ in the $x$ -direction along the tangent plane raises $z$ by $f_x\Delta x$ . The area of the parallelogram is $|\mathbf a\times\mathbf b|$ , and

\mathbf a\times\mathbf b = \begin{vmatrix}\mathbf i & \mathbf j & \mathbf k\\ \Delta x & 0 & f_x\Delta x\\ 0 & \Delta y & f_y\Delta y\end{vmatrix} = \left(-f_x\,\mathbf i - f_y\,\mathbf j + \mathbf k\right)\Delta x\,\Delta y,

so the parallelogram has area

\Delta T_{ij} = \sqrt{f_x(x_i,y_j)^2 + f_y(x_i,y_j)^2 + 1}\,\Delta A.

Definition 12.37 (Surface area). The area of the surface $S$ with equation $z = f(x,y)$ , $(x,y) \in D$ , is

A(S) = \lim_{m,n\to\infty}\sum_{i=1}^m\sum_{j=1}^n \Delta T_{ij},

the limit of the total area of the tangent-plane parallelograms as the partition of $D$ is refined.

Theorem 12.38 (Surface area formula). If $f$ has continuous partial derivatives on $D$ , the area of the surface $z = f(x,y)$ over $D$ is

A(S) = \iint_D \sqrt{1 + \left(\frac{\partial z}{\partial x}\right)^2 + \left(\frac{\partial z}{\partial y}\right)^2}\;dA.

Proof. The sum $\sum\sum \Delta T_{ij} = \sum\sum\sqrt{1 + f_x^{2} + f_y^{2}}\,\Delta A$ is a Riemann sum for the continuous function $\sqrt{1 + f_x^{2} + f_y^{2}}$ over $D$ , so its limit is the double integral.∎

The integrand is a stretch factor, always at least $1$ : it equals $1$ exactly where the surface is horizontal and grows with the steepness. Geometrically, if $\gamma$ is the angle between the upward normal $(-f_x, -f_y, 1)$ and the vertical, then $\cos\gamma = 1/\sqrt{1 + f_x^{2} + f_y^{2}}$ , and the formula says $dS = dA/\cos\gamma$ : a tilted patch is larger than its shadow by the secant of the tilt.

Intuition. A flat roof over a $10\times 10$ metre house uses $100$ square metres of tiles. A pitched roof over the same house covers the same floor plan but uses more tiles, and the steeper the pitch the more it uses: at a $45^\circ$ pitch the factor is $\sqrt 2$ . The surface-area integrand is that factor, computed patch by patch, for a roof whose pitch varies from place to place.

Example 12.39 (Surface area over a triangle). Find the area of the part of the surface $z = x^{2} + 2y$ that lies above the triangle $T$ with vertices $(0,0)$ , $(1,0)$ and $(1,1)$ .

Solution.

$z_x = 2x$ and $z_y = 2$ , so the integrand is $\sqrt{1 + 4x^{2} + 4} = \sqrt{5 + 4x^{2}}$ .
The triangle is $0 \le x \le 1$ , $0 \le y \le x$ . Integrating in $y$ first costs nothing because the integrand does not involve $y$ :

A = \int_0^1\int_0^x \sqrt{5 + 4x^2}\,dy\,dx = \int_0^1 x\sqrt{5 + 4x^2}\,dx.

Substitute $u = 5 + 4x^{2}$ , $du = 8x\,dx$ :

A = \frac18\int_5^9 u^{1/2}\,du = \frac18\cdot\frac23\left[u^{3/2}\right]_5^9 = \frac{1}{12}\left(27 - 5\sqrt 5\right) \approx 1.32.

Sanity check: the triangle has area $\frac12$ , and the stretch factor ranges from $\sqrt5 \approx 2.24$ to $3$ over it, so $A$ must lie between $1.12$ and $1.5$ .

□

Example 12.40 (Surface area of a piece of a paraboloid). Find the area of the part of the paraboloid $z = x^{2} + y^{2}$ that lies under the plane $z = 9$ .

Solution.

The surface lies over the disk $D$ : $x^{2} + y^{2} \le 9$ . With $z_x = 2x$ and $z_y = 2y$ ,

A = \iint_D \sqrt{1 + 4x^2 + 4y^2}\,dA.

The region and the integrand both call for polar coordinates: $1 + 4x^{2} + 4y^{2} = 1 + 4r^{2}$ , so

A = \int_0^{2\pi}\int_0^3 \sqrt{1 + 4r^2}\,r\,dr\,d\theta.

The $\theta$ integral gives $2\pi$ . For the $r$ integral substitute $u = 1 + 4r^{2}$ , $du = 8r\,dr$ :

\int_0^3 r\sqrt{1 + 4r^2}\,dr = \frac18\cdot\frac23\left[(1 + 4r^2)^{3/2}\right]_0^3 = \frac{1}{12}\left(37\sqrt{37} - 1\right).

So $A = \frac{\pi}{6}\left(37\sqrt{37} - 1\right) \approx 117.3$ .
Sanity check: the surface covers a disk of area $9\pi \approx 28.3$ and is steep near the rim (stretch factor $\sqrt{37} \approx 6.1$ at $r = 3$ ), so an answer several times $28$ is right.

□

Example 12.41 (Surface area of a hemisphere). Use the surface area formula to find the area of the hemisphere $z = \sqrt{a^{2} - x^{2} - y^{2}}$ .

Solution.

Here $z_x = -x/\sqrt{a^{2} - x^{2} - y^{2}}$ and $z_y = -y/\sqrt{a^{2} - x^{2} - y^{2}}$ , so

1 + z_x^2 + z_y^2 = 1 + \frac{x^2 + y^2}{a^2 - x^2 - y^2} = \frac{a^2}{a^2 - x^2 - y^2}.

In polar coordinates over the disk $r \le a$ ,

A = \int_0^{2\pi}\int_0^a \frac{a}{\sqrt{a^2 - r^2}}\,r\,dr\,d\theta.

The integrand blows up at $r = a$ , where the hemisphere is vertical, so the inner integral is improper; it converges:

\int_0^a \frac{a r}{\sqrt{a^2 - r^2}}\,dr = \lim_{t\to a^-}\left[-a\sqrt{a^2 - r^2}\right]_0^t = a\cdot a = a^2.

Hence $A = 2\pi a^{2}$ , half the area $4\pi a^{2}$ of the sphere, as it should be.

□

Pitfall. Surface area is not $\iint_D f\,dA$ . That integral is the volume under the surface. Nor is it $\iint_D 1\,dA$ , which is the area of the shadow $D$ . The surface area integrand is the stretch factor $\sqrt{1 + f_x^{2} + f_y^{2}}$ , and the integrand does not contain $f$ itself at all, only its derivatives — a surface shifted vertically by a constant has the same area.

Exercises for §12.5 →

12.6Triple integrals

A triple integral is defined over a solid region exactly as a double integral is over a plane region. Begin with a rectangular box

B = [a,b]\times[c,d]\times[r,s] = \{(x,y,z) : a \le x \le b,\ c \le y \le d,\ r \le z \le s\},

divide it into $lmn$ sub-boxes $B_{ijk}$ of volume $\Delta V = \Delta x\,\Delta y\,\Delta z$ , choose a sample point $(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)$ in each, and form the triple Riemann sum.

Definition 12.42 (Triple integral over a box).

\iiint_B f(x,y,z)\,dV = \lim_{l,m,n\to\infty}\sum_{i=1}^l\sum_{j=1}^m\sum_{k=1}^n f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)\,\Delta V,

whenever the limit exists independently of the choice of sample points. Every function continuous on $B$ is integrable.

There is no longer a picture of the integral as a volume under a graph, because that graph would live in four dimensions. The pictures that remain are the useful ones: if $f = 1$ the integral is the volume of the solid; if $f$ is a density (mass, charge, probability per unit volume) the integral is the total amount; and if $f$ is a temperature the integral divided by the volume is the average temperature.

Theorem 12.43 (Fubini's Theorem for triple integrals). If $f$ is continuous on the box $B = [a,b]\times[c,d]\times[r,s]$ , then

\iiint_B f(x,y,z)\,dV = \int_r^s\int_c^d\int_a^b f(x,y,z)\,dx\,dy\,dz,

and the same holds for each of the other five orders of integration.

The proof is the same slicing argument as before, applied twice, and the reading rule is the same: innermost differential first, outer variables held constant.

Example 12.44 (A triple integral over a box). Evaluate $\iiint_B xyz^{2}\,dV$ , where $B = [0,1]\times[-1,2]\times[0,3]$ .

Solution.

The integrand is a product $x\cdot y\cdot z^{2}$ over a box, so the triple integral is a product of three single integrals, for the same reason as in the double case.
$\int_0^1 x\,dx = \frac12$ , $\int_{-1}^2 y\,dy = \frac{4 - 1}{2} = \frac32$ , and $\int_0^3 z^{2}\,dz = 9$ .
The integral is $\frac12\cdot\frac32\cdot 9 = \frac{27}{4}$ .

□

Solids that are not boxes are handled by the extension-by-zero trick and then described by inequalities. The cleanest description is: a plane region $D$ for two of the variables, and for the third a range between two surfaces over $D$ .

Definition 12.45 (Type 1, 2 and 3 solid regions). A solid region $E$ is of type 1 if it lies between two continuous surfaces over a plane region $D$ in the $xy$ -plane:

E = \{(x,y,z) : (x,y)\in D,\ u_1(x,y) \le z \le u_2(x,y)\}.

It is of type 2 if the same holds with $D$ in the $yz$ -plane and $x$ between $u_1(y,z)$ and $u_2(y,z)$ , and of type 3 if $D$ lies in the $xz$ -plane and $y$ runs between $u_1(x,z)$ and $u_2(x,z)$ .

Theorem 12.46 (Triple integrals over general solids). If $f$ is continuous on the type 1 region $E$ above, then

\iiint_E f(x,y,z)\,dV = \iint_D\left[\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\right] dA,

and the remaining double integral over $D$ is evaluated as in the section on general regions. The analogous formulas hold for type 2 and type 3 regions with $x$ or $y$ as the inner variable.

Proof. Enclose $E$ in a box, extend $f$ by zero, apply Fubini's Theorem with $z$ innermost, and observe that for fixed $(x,y)$ in $D$ the extended function vanishes outside $u_1(x,y) \le z \le u_2(x,y)$ — the same argument as for type I plane regions, one dimension up.∎

If $D$ is itself of type I, the type 1 formula unfolds into three nested integrals:

\iiint_E f\,dV = \int_a^b\int_{g_1(x)}^{g_2(x)}\int_{u_1(x,y)}^{u_2(x,y)} f(x,y,z)\,dz\,dy\,dx,

with the now-familiar shape: the innermost limits may depend on both outer variables, the middle limits on the outermost variable only, and the outer limits are constants. The region $D$ is the projection (shadow) of $E$ on the $xy$ -plane; when $E$ is bounded by two surfaces that intersect, $D$ is bounded by the curve of intersection.

Intuition. A triple integral over a type 1 solid is a stack of vertical pencils. Over each point $(x,y)$ of the shadow $D$ , stand a thin pencil that enters the solid through the floor surface $z = u_1(x,y)$ and leaves through the ceiling $z = u_2(x,y)$ ; the inner integral adds up $f$ along that pencil. The outer double integral then adds up the pencils over the whole shadow. Choosing which axis the pencils are parallel to is choosing between types 1, 2 and 3, and the right choice is the one whose floor and ceiling are each a single formula.

Example 12.47 (A triple integral over a tetrahedron). Evaluate $\iiint_E z\,dV$ , where $E$ is the solid tetrahedron bounded by the four planes $x = 0$ , $y = 0$ , $z = 0$ and $x + y + z = 1$ .

Solution.

The floor is $z = 0$ and the ceiling is $z = 1 - x - y$ . The shadow $D$ on the $xy$ -plane is where the ceiling is above the floor, $x + y \le 1$ with $x, y \ge 0$ : the triangle $0 \le x \le 1$ , $0 \le y \le 1 - x$ .
Innermost integral:

\int_0^{1-x-y} z\,dz = \frac{(1 - x - y)^2}{2}.

Middle integral, with $w = 1 - x - y$ so that $dy = -dw$ :

\int_0^{1-x}\frac{(1 - x - y)^2}{2}\,dy = \int_0^{1-x}\frac{w^2}{2}\,dw = \frac{(1-x)^3}{6}.

Outer integral: $\int_0^1 \frac{(1-x)^3}{6}\,dx = \frac{1}{24}$ .
Sanity check: the tetrahedron has volume $\frac16$ and $z$ ranges over $[0,1]$ with most of the solid near $z = 0$ , so the answer should be well below $\frac16\cdot\frac12 = \frac{1}{12}$ ; indeed the average value of $z$ over $E$ is $\frac{1/24}{1/6} = \frac14$ , the height of the centroid.

□

Example 12.48 (Volume of a tetrahedron as a triple integral). Use a triple integral to find the volume of the tetrahedron with vertices $(0,0,0)$ , $(2,0,0)$ , $(0,3,0)$ and $(0,0,6)$ .

Solution.

The slanted face passes through $(2,0,0)$ , $(0,3,0)$ , $(0,0,6)$ , so its equation is $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$ , that is $z = 6 - 3x - 2y$ . The shadow on the $xy$ -plane is the triangle $0 \le x \le 2$ , $0 \le y \le 3 - \frac32 x$ .
Set up with $z$ innermost:

V = \int_0^2\int_0^{3 - 3x/2}\int_0^{6 - 3x - 2y} dz\,dy\,dx = \int_0^2\int_0^{3 - 3x/2} (6 - 3x - 2y)\,dy\,dx.

Inner integral: with $h = 3 - \frac32 x$ the upper limit, $\int_0^h (2h - 2y)\,dy = 2h^{2} - h^{2} = h^{2}$ , since $6 - 3x = 2h$ . So the inner integral is $\left(3 - \frac32 x\right)^2 = 9\left(1 - \frac{x}{2}\right)^2$ .
Outer integral: $V = 9\int_0^2 \left(1 - \frac{x}{2}\right)^2 dx = 9\left[-\frac23\left(1 - \frac x2\right)^3\right]_0^2 = 9\cdot\frac23 = 6$ .
Sanity check: $V = \frac13(\text{base})(\text{height}) = \frac13\cdot\frac12\cdot 2\cdot 3\cdot 6 = 6$ .

□

Example 12.49 (Choosing the projection). Evaluate $\iiint_E \sqrt{x^{2} + z^{2}}\,dV$ , where $E$ is the solid bounded by the paraboloid $y = x^{2} + z^{2}$ and the plane $y = 4$ .

Solution.

The paraboloid opens along the $y$ -axis. Projecting on the $xy$ -plane would give a floor and ceiling in $z$ of $\pm\sqrt{y - x^{2}}$ and a messy integrand. Projecting on the $xz$ -plane instead (type 3) gives $y$ from the paraboloid up to the plane: $x^{2} + z^{2} \le y \le 4$ , over the shadow $D_3 = \{x^{2} + z^{2} \le 4\}$ where the surfaces meet.
Inner integral in $y$ :

\iiint_E \sqrt{x^2 + z^2}\,dV = \iint_{D_3}\left(4 - x^2 - z^2\right)\sqrt{x^2 + z^2}\,dA.

The remaining double integral over a disk in the $xz$ -plane is done in polar coordinates $x = r\cos\theta$ , $z = r\sin\theta$ :

\int_0^{2\pi}\int_0^2 (4 - r^2)\,r\cdot r\,dr\,d\theta = 2\pi\int_0^2 (4r^2 - r^4)\,dr = 2\pi\left(\frac{32}{3} - \frac{32}{5}\right) = \frac{128\pi}{15}.

□

The last skill is changing the order of a triple integral, which is needed when an inner integral is impossible in the given order, or when a region is more naturally sliced another way.

Example 12.50 (Changing the order of a triple integral). Rewrite $\displaystyle\int_0^1\int_0^{x^{2}}\int_0^y f(x,y,z)\,dz\,dy\,dx$ as an iterated integral in the order $dx\,dy\,dz$ , and verify the two descriptions agree by computing the volume of the region.

Solution.

The limits describe $E = \{0 \le x \le 1,\ 0 \le y \le x^{2},\ 0 \le z \le y\}$ . Its projection on the $xy$ -plane is the region under the parabola $y = x^{2}$ for $0 \le x \le 1$ , and over each point $z$ runs from $0$ up to $y$ .
In the new order $x$ is innermost, so we need the projection of $E$ on the $yz$ -plane and then, for each $(y,z)$ , the range of $x$ . The $yz$ -projection is $\{0 \le z \le y \le 1\}$ , the triangle bounded by $z = 0$ , $y = 1$ and the line $z = y$ . For a point in it, $x$ must satisfy $y \le x^{2}$ , that is $\sqrt y \le x \le 1$ . So

\int_0^1\int_0^{x^2}\int_0^y f\,dz\,dy\,dx = \int_0^1\int_z^1\int_{\sqrt y}^1 f\,dx\,dy\,dz.

Verification with $f = 1$ . Original order: $\int_0^1\int_0^{x^{2}} y\,dy\,dx = \int_0^1 \frac{x^{4}}{2}\,dx = \frac{1}{10}$ . New order:

\int_0^1\int_z^1 \left(1 - \sqrt y\right) dy\,dz = \int_0^1\left[(1 - z) - \frac23\left(1 - z^{3/2}\right)\right] dz = \frac12 - \frac23\cdot\frac35 = \frac{1}{10}.

Both give $\frac{1}{10}$ .□

Every application of double integrals has a triple-integral version, with $dV$ in place of $dA$ and one more coordinate.

Proposition 12.51 (Volume, mass and centre of mass of a solid). For a solid $E$ with density $\rho(x,y,z)$ ,

V(E) = \iiint_E 1\,dV, \qquad m = \iiint_E \rho\,dV, \qquad \bar x = \frac{1}{m}\iiint_E x\rho\,dV, \quad \bar y = \frac{1}{m}\iiint_E y\rho\,dV, \quad \bar z = \frac{1}{m}\iiint_E z\rho\,dV.

The moment of inertia about the $z$ -axis is $I_z = \iiint_E (x^{2} + y^{2})\rho\,dV$ , with $I_x$ and $I_y$ defined by the analogous squared distances.

Thus the earlier computation $\iiint_E z\,dV = \frac{1}{24}$ over the unit tetrahedron says that its centroid has $\bar z = \frac{1/24}{1/6} = \frac14$ , and by symmetry the centroid is $\left(\frac14, \frac14, \frac14\right)$ .

Pitfall. When the two surfaces bounding a solid intersect, the shadow $D$ is not something you choose; it is determined by the curve where the floor and the ceiling meet. Setting $u_1 = u_2$ and simplifying gives the boundary of $D$ . Writing a triple integral over a shadow that is too large (a square instead of the disk where a paraboloid meets a plane, say) makes the inner integral negative on the excess, and the volume comes out wrong with no sign of trouble in the arithmetic.

Exercises for §12.6 →

12.7Triple integrals in cylindrical coordinates

Cylindrical coordinates are polar coordinates in the $xy$ -plane with $z$ left alone. A point $P$ is written $(r, \theta, z)$ , where $(r,\theta)$ are the polar coordinates of its projection on the $xy$ -plane and $z$ is its height:

x = r\cos\theta, \qquad y = r\sin\theta, \qquad z = z, \qquad r^2 = x^2 + y^2.

The coordinate surfaces are the cylinder $r = c$ about the $z$ -axis, the half-plane $\theta = c$ through the $z$ -axis, and the horizontal plane $z = c$ , which is why these coordinates suit any solid with an axis of symmetry: cylinders, cones, paraboloids, and anything bounded by them.

The volume element follows at once from the polar area element. A small cylindrical cell has base a polar rectangle of area $r\,\Delta r\,\Delta\theta$ and height $\Delta z$ , so

\Delta V \approx r\,\Delta r\,\Delta\theta\,\Delta z, \qquad dV = r\,dz\,dr\,d\theta.

Theorem 12.52 (Triple integral in cylindrical coordinates). Let $E$ be a type 1 solid $\{(x,y)\in D,\ u_1(x,y) \le z \le u_2(x,y)\}$ whose shadow $D$ is a polar region $\{\alpha \le \theta \le \beta,\ h_1(\theta) \le r \le h_2(\theta)\}$ . If $f$ is continuous on $E$ , then

\iiint_E f(x,y,z)\,dV = \int_\alpha^\beta\int_{h_1(\theta)}^{h_2(\theta)}\int_{u_1(r\cos\theta, r\sin\theta)}^{u_2(r\cos\theta, r\sin\theta)} f(r\cos\theta, r\sin\theta, z)\,r\,dz\,dr\,d\theta.

Proof. By the type 1 formula, $\iiint_E f\,dV = \iint_D\left[\int_{u_1}^{u_2} f\,dz\right]dA$ . The bracket is a continuous function of $(x,y)$ on $D$ , and converting the outer double integral to polar coordinates by the polar change-of-coordinates theorem supplies the factor $r$ and the polar limits.∎

The recipe is: convert the integrand and the $z$ -limits with $x = r\cos\theta$ , $y = r\sin\theta$ ; describe the shadow in polar coordinates; and multiply by $r$ . Expressions such as $x^{2} + y^{2}$ and $\sqrt{x^{2} + y^{2}}$ collapse to $r^{2}$ and $r$ , which is usually the whole point.

Intuition. Picture a stack of coins whose diameters vary with height — a solid of revolution. Cylindrical coordinates describe it by height $z$ , distance $r$ from the axle, and angle $\theta$ around it. Nothing about the shape depends on $\theta$ , so the $\theta$ integral is a bare $2\pi$ , and the real work is the two-variable integral in $r$ and $z$ that describes one cross-section through the axis.

Example 12.53 (Volume between a paraboloid and a plane). Find the volume of the solid bounded by the paraboloid $z = x^{2} + y^{2}$ and the plane $z = 4$ .

Solution.

The floor is the paraboloid $z = r^{2}$ and the ceiling is the plane $z = 4$ ; they meet where $r^{2} = 4$ , so the shadow is the disk $r \le 2$ .
In cylindrical coordinates,

V = \int_0^{2\pi}\int_0^2\int_{r^2}^4 r\,dz\,dr\,d\theta = \int_0^{2\pi}\int_0^2 (4 - r^2)\,r\,dr\,d\theta = 2\pi\left[2r^2 - \frac{r^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi.

Sanity check: the solid sits inside the cylinder of radius $2$ and height $4$ , volume $16\pi$ , and a paraboloid fills exactly half its circumscribed cylinder, as the earlier polar example also found. $8\pi$ is half of $16\pi$ .

□

Example 12.54 (Mass of a solid with density proportional to distance from the axis). A solid lies inside the cylinder $x^{2} + y^{2} = 1$ , below the plane $z = 4$ and above the paraboloid $z = 1 - x^{2} - y^{2}$ . Its density at a point is proportional to the point's distance from the $z$ -axis. Find its mass.

Solution.

The density is $\rho = K\sqrt{x^{2} + y^{2}} = Kr$ for a constant $K$ . The solid is $0 \le r \le 1$ , $0 \le \theta \le 2\pi$ , $1 - r^{2} \le z \le 4$ .
Mass:

m = \int_0^{2\pi}\int_0^1\int_{1 - r^2}^{4} Kr\cdot r\,dz\,dr\,d\theta = 2\pi K\int_0^1 r^2\left(4 - (1 - r^2)\right) dr = 2\pi K\int_0^1 (3r^2 + r^4)\,dr = 2\pi K\left(1 + \frac15\right) = \frac{12\pi K}{5}.

Note the two factors of $r$ in the integrand: one from the density, one from $dV$ .□

Example 12.55 (Converting a Cartesian triple integral). Evaluate $\displaystyle\int_{-2}^{2}\int_{-\sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}}\int_{\sqrt{x^{2} + y^{2}}}^{2} (x^{2} + y^{2})\,dz\,dy\,dx$ .

Solution.

Read off the region. The outer two limits say $(x,y)$ lies in the disk $x^{2} + y^{2} \le 4$ ; the inner limits say $z$ runs from the cone $z = \sqrt{x^{2} + y^{2}}$ up to the plane $z = 2$ . The solid is the region inside the cone and below the plane, and the integrand is $r^{2}$ .
In cylindrical coordinates the cone is $z = r$ and the disk is $r \le 2$ :

\int_0^{2\pi}\int_0^2\int_r^2 r^2\cdot r\,dz\,dr\,d\theta = 2\pi\int_0^2 r^3(2 - r)\,dr = 2\pi\left[\frac{r^4}{2} - \frac{r^5}{5}\right]_0^2 = 2\pi\left(8 - \frac{32}{5}\right) = \frac{16\pi}{5}.

Sanity check: the solid is a cone of base radius $2$ and height $2$ , volume $\frac{8\pi}{3}$ , and $r^{2}$ averages somewhat under $2$ over it (the wide end at the top is where $r$ is large, and it carries most of the volume), so an answer near $\frac{16\pi}{5} \approx 1.2\cdot\frac{8\pi}{3}$ is reasonable.

□

Pitfall. Cylindrical $r$ is the distance from the $z$ -axis, $\sqrt{x^{2} + y^{2}}$ , not the distance from the origin. A sphere $x^{2} + y^{2} + z^{2} = a^{2}$ becomes $r^{2} + z^{2} = a^{2}$ in cylindrical coordinates, not $r = a$ . If a problem involves distance from the origin, spherical coordinates are probably the right tool.

Exercises for §12.7 →

12.8Triple integrals in spherical coordinates

Spherical coordinates locate a point by its distance from the origin and two angles: the longitude $\theta$ of cylindrical coordinates, and the angle $\phi$ down from the positive $z$ -axis, which plays the role of co-latitude.

Definition 12.56 (Spherical coordinates). The spherical coordinates of a point $P$ are $(\rho, \theta, \phi)$ , where $\rho = |OP| \ge 0$ is the distance from the origin, $\theta$ is the same angle as in cylindrical coordinates, and $\phi$ is the angle between the positive $z$ -axis and the segment $OP$ , with $0 \le \phi \le \pi$ . The conversion formulas are

x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi, \qquad \rho^2 = x^2 + y^2 + z^2.

The formulas come from two right triangles. Dropping a perpendicular from $P$ to the $z$ -axis gives $z = \rho\cos\phi$ and a horizontal leg of length $r = \rho\sin\phi$ ; that $r$ is the cylindrical radius, and $x = r\cos\theta$ , $y = r\sin\theta$ finishes the job. The coordinate surfaces are the sphere $\rho = c$ , the half-plane $\theta = c$ , and the half-cone $\phi = c$ (a cone opening upward for $\phi < \pi/2$ , the $xy$ -plane for $\phi = \pi/2$ , a cone opening downward for $\phi > \pi/2$ ). Anything bounded by spheres and cones — balls, shells, ice-cream cones — belongs in spherical coordinates.

The volume element takes more care than the cylindrical one. Consider the spherical wedge

E = \{(\rho,\theta,\phi) : a \le \rho \le b,\ \alpha \le \theta \le \beta,\ c \le \phi \le d\},

and cut it into small wedges by spheres $\rho = \rho_i$ , half-planes $\theta = \theta_j$ and cones $\phi = \phi_k$ . A small wedge is nearly a rectangular box. Its three edges are: a radial segment of length $\Delta\rho$ ; an arc of a meridian (a great circle of radius $\rho_i$ ) subtending angle $\Delta\phi$ , of length $\rho_i\,\Delta\phi$ ; and an arc of a circle of latitude subtending angle $\Delta\theta$ . That circle of latitude has radius $r = \rho_i\sin\phi_k$ , the cylindrical radius of the point, so the arc has length $\rho_i\sin\phi_k\,\Delta\theta$ . Multiplying the three edges,

\Delta V_{ijk} \approx \rho_i^2\sin\phi_k\,\Delta\rho\,\Delta\theta\,\Delta\phi.

A more careful computation gives the exact volume as $\frac13(\rho_i^{3} - \rho_{i-1}^3)(\cos\phi_{k-1} - \cos\phi_k)\,\Delta\theta$ , and the Mean Value Theorem turns this into $\tilde\rho_i^{2}\sin\tilde\phi_k\,\Delta\rho\,\Delta\theta\,\Delta\phi$ for some $\tilde\rho_i$ in $[\rho_{i-1}, \rho_i]$ and $\tilde\phi_k$ in $[\phi_{k-1}, \phi_k]$ , which is what makes the Riemann-sum argument rigorous.

Theorem 12.57 (Triple integral in spherical coordinates). If $f$ is continuous on the spherical wedge $E$ above, then

\iiint_E f(x,y,z)\,dV = \int_c^d\int_\alpha^\beta\int_a^b f(\rho\sin\phi\cos\theta,\ \rho\sin\phi\sin\theta,\ \rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\theta\,d\phi.

The formula extends to regions of the form $\{\alpha \le \theta \le \beta,\ c \le \phi \le d,\ g_1(\theta,\phi) \le \rho \le g_2(\theta,\phi)\}$ with the $\rho$ -limits replaced accordingly.

Proof. The Riemann sum $\sum f(x_{ijk}^*, y_{ijk}^*, z_{ijk}^*)\,\Delta V_{ijk}$ , with the sample point chosen at $(\tilde\rho_i, \theta_j, \tilde\phi_k)$ and $\Delta V_{ijk}$ written as above, is a Riemann sum for the function $\rho^{2}\sin\phi\, f(\rho\sin\phi\cos\theta, \rho\sin\phi\sin\theta, \rho\cos\phi)$ over the box $[a,b]\times[\alpha,\beta]\times[c,d]$ in $\rho\theta\phi$ -space. The two limits agree.∎

In practice: replace $x, y, z$ by their spherical expressions (remembering $x^{2} + y^{2} + z^{2} = \rho^{2}$ and $x^{2} + y^{2} = \rho^{2}\sin^{2}\phi$ ), use spherical limits, and replace $dV$ by $\rho^{2}\sin\phi\,d\rho\,d\theta\,d\phi$ . The order of the three differentials is immaterial when all six limits are constants.

Intuition. On a globe, the grid squares between neighbouring lines of latitude and longitude are large at the equator and shrink to slivers near the poles: a degree of longitude spans far less distance in Svalbard than in Singapore. The factor $\sin\phi$ is that shrinking, since the circle of latitude at co-latitude $\phi$ has radius $\rho\sin\phi$ . The factor $\rho^{2}$ says that a larger globe has proportionally larger grid squares in both directions.

Example 12.58 (Volume of a ball). Find the volume of the ball $x^{2} + y^{2} + z^{2} \le a^{2}$ .

Solution.

The ball is the spherical wedge $0 \le \rho \le a$ , $0 \le \theta \le 2\pi$ , $0 \le \phi \le \pi$ , and its volume is the integral of $1$ :

V = \int_0^\pi\int_0^{2\pi}\int_0^a \rho^2\sin\phi\,d\rho\,d\theta\,d\phi.

All limits are constant and the integrand is a product, so

V = \int_0^a \rho^2\,d\rho\int_0^{2\pi} d\theta\int_0^\pi\sin\phi\,d\phi = \frac{a^3}{3}\cdot 2\pi\cdot 2 = \frac43\pi a^3.

The factor $2 = \int_0^\pi\sin\phi\,d\phi$ is worth remembering; a hemisphere ( $0 \le \phi \le \pi/2$ ) gives $1$ instead.□

Example 12.59 (An integrand that only spherical coordinates can handle). Evaluate $\iiint_B e^{(x^{2} + y^{2} + z^{2})^{3/2}}\,dV$ , where $B$ is the unit ball.

Solution.

The exponent is $(\rho^{2})^{3/2} = \rho^{3}$ , and the ball is $0 \le \rho \le 1$ with the full ranges of $\theta$ and $\phi$ :

\iiint_B e^{\rho^3}\rho^2\sin\phi\,dV = \int_0^\pi\sin\phi\,d\phi\int_0^{2\pi}d\theta\int_0^1 \rho^2 e^{\rho^3}\,d\rho.

The $\rho$ integral is set up for $u = \rho^{3}$ , $du = 3\rho^{2}\,d\rho$ — again the coordinate factor supplies exactly the $\rho^{2}$ the substitution needs — giving $\frac13(e - 1)$ .
The total is $2\cdot 2\pi\cdot\frac13(e - 1) = \frac{4\pi}{3}(e - 1) \approx 7.2$ .
Sanity check: the integrand lies between $1$ and $e$ on the ball, whose volume is $\frac43\pi \approx 4.19$ , so the answer lies between $4.19$ and $11.4$ . In Cartesian coordinates this integral cannot even be started.

□

Example 12.60 (Volume inside a sphere and above a cone). Find the volume of the solid that lies inside the sphere $x^{2} + y^{2} + z^{2} = 4$ and above the cone $z = \sqrt{3(x^{2} + y^{2})}$ .

Solution.

The sphere is $\rho = 2$ . On the cone, $z = \sqrt 3\,r$ , so $\tan\phi = r/z = 1/\sqrt 3$ and the cone is $\phi = \pi/6$ . Above the cone means $0 \le \phi \le \pi/6$ .
The solid is the wedge $0 \le \rho \le 2$ , $0 \le \theta \le 2\pi$ , $0 \le \phi \le \pi/6$ :

V = \int_0^{\pi/6}\int_0^{2\pi}\int_0^2 \rho^2\sin\phi\,d\rho\,d\theta\,d\phi = \frac{8}{3}\cdot 2\pi\cdot\Big[-\cos\phi\Big]_0^{\pi/6} = \frac{16\pi}{3}\left(1 - \frac{\sqrt 3}{2}\right) = \frac{8\pi}{3}\left(2 - \sqrt 3\right) \approx 2.24.

Sanity check: the whole ball has volume $\frac{32\pi}{3} \approx 33.5$ , and a narrow cone of half-angle $30^\circ$ should cut off a small fraction of it: the fraction is $\frac12(1 - \cos\frac\pi6) \approx 0.067$ , so about $2.24$ .

In general the solid inside $\rho = a$ and above the cone $\phi = \phi_0$ has volume $\frac{2\pi a^{3}}{3}(1 - \cos\phi_0)$ .□

Example 12.61 (The ice-cream cone). Find the volume of the solid that lies above the cone $z = \sqrt{x^{2} + y^{2}}$ and inside the sphere $x^{2} + y^{2} + z^{2} = z$ .

Solution.

Completing the square, the sphere is $x^{2} + y^{2} + (z - \frac12)^2 = \frac14$ : centre $(0,0,\frac12)$ , radius $\frac12$ , resting on the origin. In spherical coordinates $x^{2} + y^{2} + z^{2} = z$ reads $\rho^{2} = \rho\cos\phi$ , so the sphere is $\rho = \cos\phi$ . The cone $z = r$ has $\tan\phi = 1$ , so it is $\phi = \pi/4$ .
The solid — the scoop of ice cream sitting in the cone — is described by $0 \le \theta \le 2\pi$ , $0 \le \phi \le \pi/4$ , $0 \le \rho \le \cos\phi$ . The upper $\rho$ -limit depends on $\phi$ , so $\rho$ must be integrated first:

V = \int_0^{2\pi}\int_0^{\pi/4}\int_0^{\cos\phi} \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\int_0^{\pi/4}\frac{\cos^3\phi}{3}\sin\phi\,d\phi.

Substitute $u = \cos\phi$ , $du = -\sin\phi\,d\phi$ :

V = \frac{2\pi}{3}\left[-\frac{\cos^4\phi}{4}\right]_0^{\pi/4} = \frac{2\pi}{3}\cdot\frac14\left(1 - \frac14\right) = \frac{\pi}{8}.

Sanity check: the whole sphere has volume $\frac43\pi\left(\frac12\right)^3 = \frac{\pi}{6}$ , and the solid is a substantial part of it (the cone cuts away only the lower rim), so $\frac{\pi}{8} = 0.75\cdot\frac{\pi}{6}$ is plausible.

□

Pitfall. Three separate errors haunt spherical coordinates. The factor is $\rho^{2}\sin\phi$ , with the polar angle $\phi$ , never $\sin\theta$ . The range of $\phi$ is $[0,\pi]$ , never $[0, 2\pi]$ — a $\phi$ -range of $2\pi$ covers the ball twice. And a surface such as $z = \sqrt{x^{2} + y^{2}}$ is a cone $\phi =$ const, not a sphere; a surface such as $x^{2} + y^{2} + z^{2} = 2z$ is a sphere $\rho = 2\cos\phi$ that is not centred at the origin, so its $\rho$ -limit depends on $\phi$ .

Exercises for §12.8 →

12.9Change of variables and the Jacobian

The substitution rule in one variable, with $x = g(u)$ ,

\int_a^b f(x)\,dx = \int_c^d f(g(u))\,g'(u)\,du,

replaces $dx$ by $g'(u)\,du$ : the factor $g'(u)$ is the amount by which $g$ stretches a small interval of $u$ -values. Polar coordinates replaced $dA$ by $r\,dr\,d\theta$ , spherical coordinates replaced $dV$ by $\rho^{2}\sin\phi\,d\rho\,d\theta\,d\phi$ , and in each case the extra factor was the stretching of a small cell. The general theorem makes this precise for any change of coordinates.

Definition 12.62 (Transformation and Jacobian). A transformation $T$ from a region $S$ of the $uv$ -plane to the $xy$ -plane is a map $T(u,v) = (x(u,v), y(u,v))$ whose component functions have continuous first partial derivatives. The image of $S$ is $R = T(S)$ . The Jacobian of $T$ is the determinant

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v}\\[1.2ex] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v}\end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}.

To see why the determinant appears, take a small rectangle $S_0$ in the $uv$ -plane with corner $(u_0, v_0)$ and sides $\Delta u$ , $\Delta v$ , and write the transformation as a position vector $\mathbf r(u,v) = x(u,v)\,\mathbf i + y(u,v)\,\mathbf j$ . The image of the bottom edge of $S_0$ is the curve $\mathbf r(u, v_0)$ for $u_0 \le u \le u_0 + \Delta u$ , and by the tangent-line approximation its displacement is

\mathbf r(u_0 + \Delta u, v_0) - \mathbf r(u_0, v_0) \approx \mathbf r_u\,\Delta u, \qquad \mathbf r_u = \frac{\partial x}{\partial u}\mathbf i + \frac{\partial y}{\partial u}\mathbf j,

and similarly the left edge maps to approximately $\mathbf r_v\,\Delta v$ . So the image of $S_0$ is approximately the parallelogram spanned by $\mathbf r_u\Delta u$ and $\mathbf r_v\Delta v$ , whose area is $|\mathbf r_u\times\mathbf r_v|\,\Delta u\,\Delta v$ . Computing the cross product with the vectors regarded as lying in the $xy$ -plane of space,

\mathbf r_u\times\mathbf r_v = \left(\frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}\right)\mathbf k = \frac{\partial(x,y)}{\partial(u,v)}\,\mathbf k,

so the image has area approximately $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|\Delta u\,\Delta v$ . The absolute value of the Jacobian is the local area magnification factor of $T$ .

Theorem 12.63 (Change of variables in a double integral). Let $T$ be a transformation from $S$ onto $R$ that is one-to-one except possibly on the boundary of $S$ , with continuous partial derivatives and nonzero Jacobian on the interior of $S$ . If $f$ is continuous on $R$ and $S$ and $R$ are type I or type II regions, then

\iint_R f(x,y)\,dA = \iint_S f\big(x(u,v), y(u,v)\big)\left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv.

Proof. The idea, with the full justification left to real analysis. Divide $S$ into small rectangles $S_{ij}$ with corners $(u_i, v_j)$ ; their images $R_{ij}$ tile $R$ . Approximate $\iint_R f\,dA$ by $\sum_{i,j} f(x_i, y_j)\,\Delta A_{ij}$ , where $(x_i, y_j) = T(u_i, v_j)$ and $\Delta A_{ij}$ is the area of $R_{ij}$ . By the estimate above, $\Delta A_{ij} \approx |J(u_i, v_j)|\,\Delta u\,\Delta v$ with $J$ the Jacobian, so the sum is approximately

\sum_{i,j} f\big(x(u_i, v_j), y(u_i,v_j)\big)\,|J(u_i, v_j)|\,\Delta u\,\Delta v,

which is a Riemann sum for the right-hand side of the theorem. As the partition is refined, both sides converge, and the approximation errors, which are of smaller order than $\Delta u\,\Delta v$ , vanish in the limit.∎

Compare the theorem with the one-variable rule. There the factor is $g'(u)$ without absolute value, because a decreasing $g$ reverses the limits of integration and the two sign changes cancel; here regions have no orientation, so the absolute value is needed. The condition that the Jacobian be nonzero says $T$ does not crush any area to zero, so that it can be inverted; one-to-one says no part of $R$ is counted twice.

Intuition. Print a map on a rubber sheet and stretch it unevenly. A small square of the original around the point $(u,v)$ becomes a small parallelogram on the stretched sheet, and the Jacobian at $(u,v)$ is the ratio of the parallelogram's area to the square's — the local magnification. To find how much ink is on the stretched map, add up ink density times area over the stretched parallelograms, which is the same as adding up ink density times magnification times original area over the squares. That equality of two bookkeeping methods is the change of variables theorem.

Example 12.64 (Polar coordinates as a change of variables). Show that the polar change-of-coordinates theorem is the case $x = r\cos\theta$ , $y = r\sin\theta$ of the change of variables theorem.

Solution.

With $u = r$ and $v = \theta$ ,

\frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r.

Since $r \ge 0$ , $|J| = r$ , and the theorem gives $\iint_R f\,dA = \iint_S f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$ — the polar formula, with the factor $r$ now explained as a Jacobian. The map is one-to-one on the interior of $S$ and the Jacobian vanishes only on the edge $r = 0$ , which the theorem allows.

□

Example 12.65 (A rotated square). Evaluate $\iint_R (x - y)^2\,dA$ , where $R$ is the square with vertices $(0,0)$ , $(1,1)$ , $(2,0)$ and $(1,-1)$ .

Solution.

The sides of $R$ lie on the lines $x + y = 0$ , $x + y = 2$ , $x - y = 0$ and $x - y = 2$ , which suggests the coordinates $u = x + y$ , $v = x - y$ . Solving, $x = \frac12(u + v)$ and $y = \frac12(u - v)$ .
The vertices map to $(u,v) = (0,0), (2,0), (2,2), (0,2)$ , so $S$ is the square $[0,2]\times[0,2]$ in the $uv$ -plane.
Jacobian:

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}\tfrac12 & \tfrac12\\ \tfrac12 & -\tfrac12\end{vmatrix} = -\frac14 - \frac14 = -\frac12, \qquad |J| = \frac12.

The integrand is $(x - y)^2 = v^{2}$ , so

\iint_R (x-y)^2\,dA = \int_0^2\int_0^2 v^2\cdot\frac12\,du\,dv = \frac12\cdot 2\cdot\frac{8}{3} = \frac83.

Sanity check: with integrand $1$ the same computation gives $\frac12\cdot 4 = 2$ , and $R$ is a square of side $\sqrt 2$ , area $2$ . The Jacobian $\frac12$ is right: the square $S$ of area $4$ is shrunk to area $2$ .

□

Example 12.66 (A trapezoid and an awkward integrand). Evaluate $\iint_R e^{(x + y)/(x - y)}\,dA$ , where $R$ is the trapezoid with vertices $(1,0)$ , $(2,0)$ , $(0,-2)$ and $(0,-1)$ .

Solution.

The integrand suggests $u = x + y$ , $v = x - y$ , so again $x = \frac12(u+v)$ , $y = \frac12(u-v)$ and $|J| = \frac12$ .
The vertices map to $(1,1), (2,2), (-2,2), (-1,1)$ . The sides of $R$ are the segments on $y = 0$ (which maps to $u = v$ ), $x = 0$ (which maps to $u = -v$ ), $x - y = 1$ ( $v = 1$ ) and $x - y = 2$ ( $v = 2$ ). So $S$ is the trapezoid $1 \le v \le 2$ , $-v \le u \le v$ , and it must be integrated with $u$ inside.
Then

\iint_R e^{(x+y)/(x-y)}\,dA = \int_1^2\int_{-v}^{v} e^{u/v}\cdot\frac12\,du\,dv = \frac12\int_1^2\Big[v e^{u/v}\Big]_{u=-v}^{u=v} dv = \frac12\left(e - e^{-1}\right)\int_1^2 v\,dv = \frac34\left(e - e^{-1}\right) \approx 1.76.

Sanity check: $R$ has area $\frac32$ (a trapezoid with parallel sides $\sqrt 2$ and $2\sqrt 2$ and height $1/\sqrt 2$ ), and on $R$ the exponent $(x+y)/(x-y)$ ranges over $[-1,1]$ , so the integrand lies between $e^{-1}$ and $e$ ; the answer must lie between $0.55$ and $4.08$ .

□

Example 12.67 (Area of an ellipse, and an integral over it). Find the area of the ellipse $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} \le 1$ , and evaluate $\iint_R x^{2}\,dA$ over the ellipse $\dfrac{x^{2}}{4} + y^{2} \le 1$ .

Solution.

The transformation $x = au$ , $y = bv$ maps the unit disk $S$ : $u^{2} + v^{2} \le 1$ onto the ellipse, since $x^{2}/a^{2} + y^{2}/b^{2} = u^{2} + v^{2}$ . Its Jacobian is

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix}a & 0\\ 0 & b\end{vmatrix} = ab,

a constant: the map magnifies every area by $ab$ . 2. Hence $A = \iint_S ab\,du\,dv = ab\cdot A(S) = \pi ab$ . With $a = b$ this is the area of a circle. 3. For the second part $a = 2$ , $b = 1$ , $x^{2} = 4u^{2}$ and $|J| = 2$ , so

\iint_R x^2\,dA = \iint_S 4u^2\cdot 2\,du\,dv = 8\iint_S u^2\,du\,dv.

By the symmetry of the disk, $\iint_S u^{2} = \iint_S v^{2} = \frac12\iint_S (u^{2} + v^{2}) = \frac12\int_0^{2\pi}\int_0^1 r^{3}\,dr\,d\theta = \frac{\pi}{4}$ . So the integral is $8\cdot\frac{\pi}{4} = 2\pi$ .□

Often the natural new variables are given as functions of $x$ and $y$ — $u = xy$ , say — and solving for $x$ and $y$ is unpleasant. The Jacobian of the inverse map rescues the situation.

Proposition 12.68 (Jacobian of the inverse transformation). If $T$ is one-to-one with nonzero Jacobian, then

\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{\dfrac{\partial(u,v)}{\partial(x,y)}}.

Proof. Composing $T$ with its inverse gives the identity map, whose Jacobian matrix is the identity. By the chain rule the Jacobian matrix of a composition is the product of the Jacobian matrices, and the determinant of a product is the product of the determinants; so the two Jacobians multiply to $1$ .∎

Example 12.69 (A region bounded by hyperbolas and lines). Find the area of the region $R$ in the first quadrant bounded by the hyperbolas $xy = 1$ , $xy = 4$ and the lines $y = x$ , $y = 4x$ .

Solution.

The four boundaries are level curves of $u = xy$ and $v = y/x$ , so in those variables $R$ becomes the rectangle $S = \{1 \le u \le 4,\ 1 \le v \le 4\}$ .
Rather than solve for $x$ and $y$ , compute the inverse Jacobian directly:

\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix}y & x\\ -y/x^2 & 1/x\end{vmatrix} = \frac{y}{x} + \frac{y}{x} = \frac{2y}{x} = 2v.

By the proposition, $\dfrac{\partial(x,y)}{\partial(u,v)} = \dfrac{1}{2v}$ , positive on $S$ . Therefore

A(R) = \iint_S \frac{1}{2v}\,du\,dv = \int_1^4 du\int_1^4\frac{dv}{2v} = 3\cdot\frac12\ln 4 = 3\ln 2 \approx 2.08.

Sanity check: solving gives $x = \sqrt{u/v}$ , $y = \sqrt{uv}$ , and differentiating these directly also yields $\frac{1}{2v}$ ; and $R$ fits inside the rectangle $[\frac12, 2]\times[1, 4]$ of area $4.5$ , so an area of about $2$ is plausible.

□

The triple-integral version is identical in form.

Theorem 12.70 (Change of variables in a triple integral). If $T(u,v,w) = (x,y,z)$ is a transformation from $S$ onto $E$ that is one-to-one with continuous partial derivatives and nonzero Jacobian on the interior of $S$ , then for $f$ continuous on $E$ ,

\iiint_E f(x,y,z)\,dV = \iiint_S f\big(x(u,v,w), y(u,v,w), z(u,v,w)\big)\left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right| du\,dv\,dw,

where the Jacobian is the $3\times 3$ determinant of the matrix of partial derivatives of $x, y, z$ with respect to $u, v, w$ .

The justification is the same as in two dimensions: the image of a small box is approximately the parallelepiped spanned by $\mathbf r_u\Delta u$ , $\mathbf r_v\Delta v$ and $\mathbf r_w\Delta w$ , and the volume of a parallelepiped is the absolute value of the determinant of its edge vectors.

Example 12.71 (The spherical volume element as a Jacobian). Compute the Jacobian of the spherical coordinate transformation $x = \rho\sin\phi\cos\theta$ , $y = \rho\sin\phi\sin\theta$ , $z = \rho\cos\phi$ .

Solution.

The matrix of partial derivatives with respect to $(\rho, \theta, \phi)$ is

\frac{\partial(x,y,z)}{\partial(\rho,\theta,\phi)} = \begin{vmatrix}\sin\phi\cos\theta & -\rho\sin\phi\sin\theta & \rho\cos\phi\cos\theta\\ \sin\phi\sin\theta & \rho\sin\phi\cos\theta & \rho\cos\phi\sin\theta\\ \cos\phi & 0 & -\rho\sin\phi\end{vmatrix}.

Expand along the bottom row. The $(3,1)$ cofactor is $\cos\phi$ times

\begin{vmatrix}-\rho\sin\phi\sin\theta & \rho\cos\phi\cos\theta\\ \rho\sin\phi\cos\theta & \rho\cos\phi\sin\theta\end{vmatrix} = -\rho^2\sin\phi\cos\phi\left(\sin^2\theta + \cos^2\theta\right) = -\rho^2\sin\phi\cos\phi,

and the $(3,3)$ entry $-\rho\sin\phi$ multiplies

\begin{vmatrix}\sin\phi\cos\theta & -\rho\sin\phi\sin\theta\\ \sin\phi\sin\theta & \rho\sin\phi\cos\theta\end{vmatrix} = \rho\sin^2\phi\left(\cos^2\theta + \sin^2\theta\right) = \rho\sin^2\phi.

The determinant is $-\rho^{2}\sin\phi\cos^{2}\phi - \rho^{2}\sin^{3}\phi = -\rho^{2}\sin\phi$ , so its absolute value is $\rho^{2}\sin\phi$ — the volume element found geometrically in the previous section, now derived by algebra. The sign is negative only because $(\rho,\theta,\phi)$ in that order is a left-handed system; listing the variables as $(\rho,\phi,\theta)$ gives $+\rho^{2}\sin\phi$ .

□

Example 12.72 (Volume of an ellipsoid). Find the volume of the solid ellipsoid $\dfrac{x^{2}}{a^{2}} + \dfrac{y^{2}}{b^{2}} + \dfrac{z^{2}}{c^{2}} \le 1$ .

Solution.

The map $x = au$ , $y = bv$ , $z = cw$ sends the unit ball $S$ onto the ellipsoid $E$ , and its Jacobian is the determinant of the diagonal matrix with entries $a, b, c$ , namely $abc$ .
So $V(E) = \iiint_S abc\,du\,dv\,dw = abc\cdot V(S) = \frac43\pi abc$ .

A sphere is the case $a = b = c$ . The same idea shows that a linear map with matrix $A$ multiplies every volume by $|\det A|$ , which is the geometric meaning of the determinant.□

Pitfall. Two errors account for most wrong Jacobian answers. First, forgetting the absolute value: a negative Jacobian is normal (it means the map reverses orientation) and the integrand must use $|J|$ . Second, using $\partial(u,v)/\partial(x,y)$ where $\partial(x,y)/\partial(u,v)$ is required. When the new variables are given as functions of the old, the Jacobian you can compute directly is the inverse of the one you need — take its reciprocal.

Summary. The double integral is a limit of Riemann sums, and for $f$ continuous on $R = [a,b]\times[c,d]$ Fubini's Theorem evaluates it as either iterated integral; over a type I region $\{a\le x\le b,\ g_1(x)\le y\le g_2(x)\}$ the inner limits are functions of $x$ and the outer limits are constants, and symmetrically for type II. Area is $A(D) = \iint_D 1\,dA$ , and for a lamina of density $\rho$ , $m = \iint_D \rho\,dA$ with $\bar x = M_y/m$ , $\bar y = M_x/m$ . The coordinate changes are the formulas to carry away:

dA = r\,dr\,d\theta, \qquad dV = r\,dz\,dr\,d\theta, \qquad dV = \rho^2\sin\phi\,d\rho\,d\theta\,d\phi,

for polar, cylindrical and spherical coordinates respectively, with $x = \rho\sin\phi\cos\theta$ , $y = \rho\sin\phi\sin\theta$ , $z = \rho\cos\phi$ and $0 \le \phi \le \pi$ . The surface $z = f(x,y)$ over $D$ , with $f$ having continuous partials, has area $\iint_D\sqrt{1 + z_x^{2} + z_y^{2}}\,dA$ . All of these are instances of the change of variables theorem: if $T$ is one-to-one except possibly on the boundary, has continuous partial derivatives and nonzero Jacobian on the interior of $S$ , and $f$ is continuous, then

\iint_R f(x,y)\,dA = \iint_S f\big(x(u,v), y(u,v)\big)\left|\frac{\partial(x,y)}{\partial(u,v)}\right| du\,dv,

with the absolute value of the Jacobian, and the analogous three-dimensional statement for triple integrals.

Exercises for §12.9 →

Common Mistakes

Attaching limits to the wrong variable in an iterated integral. The innermost differential goes with the innermost integral sign and its limits; read $\int_a^b\int_c^d f\,dy\,dx$ as $y$ from $c$ to $d$ inside, $x$ from $a$ to $b$ outside.
Letting an outer limit depend on a variable. Inner limits may involve outer variables; the outermost limits are always constants. An answer that still contains $x$ or $y$ signals a region described backwards.
Reversing the order of integration by swapping the limits without redrawing the region. The limits change shape, not just position: $\int_0^1\int_x^{1} f\,dy\,dx$ becomes $\int_0^1\int_0^y f\,dx\,dy$ , not $\int_x^{1}\int_0^1 f\,dx\,dy$ .
Dropping the factor $r$ in polar and cylindrical coordinates, or $\rho^{2}\sin\phi$ in spherical. The factor is the area or volume of a coordinate cell and is never optional.
Writing $\sin\theta$ instead of $\sin\phi$ in the spherical volume element, or letting $\phi$ range over $[0, 2\pi]$ . The polar angle $\phi$ runs from $0$ to $\pi$ only.
Confusing the cylindrical radius $r = \sqrt{x^{2} + y^{2}}$ with the spherical radius $\rho = \sqrt{x^{2} + y^{2} + z^{2}}$ . A sphere is $r^{2} + z^{2} = a^{2}$ in cylindrical coordinates and $\rho = a$ in spherical.
Choosing the shadow region by guesswork instead of from the intersection of the bounding surfaces. Setting floor equal to ceiling gives the boundary of the projection.
Swapping the moments: $M_x = \iint y\rho\,dA$ contains $y$ , and $\bar x = M_y/m$ . Check every centre of mass against the symmetry of the region.
Computing surface area as $\iint f\,dA$ . Surface area uses the stretch factor $\sqrt{1 + f_x^{2} + f_y^{2}}$ ; the function value itself never appears.
Forgetting the absolute value of the Jacobian, or using $\partial(u,v)/\partial(x,y)$ in place of its reciprocal $\partial(x,y)/\partial(u,v)$ .
Treating $\sqrt{x^{2} + y^{2}}$ as $\pm r$ . By convention $r \ge 0$ , and regions are described by their $\theta$ -range rather than by negative radii.
Evaluating a product-form integral as a product when the region is not a rectangle. $\iint_D g(x)h(y)\,dA$ splits into $\int g\int h$ only when the limits are all constants.

About the practice questions. They check that you can carry out this chapter's computations correctly, and each one is graded on a single answer. They are not proof exercises: working through them confirms the mechanics, not that you could prove the results yourself. For that, re-read the statements above and try to reconstruct their proofs with the page closed.

← Partial Derivatives Vector Calculus →

Helpful?