Applications of Integration

(a) Find the crossings, then split. The curves meet where $x^3-x=3x$, that is $x^3-4x=x(x-2)(x+2)=0$, so at $x=-2$, $0$ and $2$. Three crossings mean two pieces, and the curves swap places at $x=0$.

On $[0,2]$ test $x=1$: the line gives $3$ and the cubic gives $0$, so the line is on top. On $[-2,0]$ test $x=-1$: the line gives $-3$ and the cubic gives $0$, so the cubic is on top. (Two continuous curves that do not cross inside a piece cannot swap, so one test point per piece settles it.)

Both functions are odd, so the region is symmetric through the origin and the two pieces have equal area. Hence

$$A=2{\int }_0^2[3x-(x^3-x)]dx=2{\int }_0^2\left(4x-x^3\right)dx=2{\left[2x^2-\frac{x^4}{4}\right]}_0^2=2(8-4)=8.$$

If you prefer not to trust the symmetry, the left piece is ${\int }_{-2}^0\left(x^3-4x\right)dx={\left[\frac{x^4}{4}-2x^2\right]}_{-2}^0=0-(4-8)=4$, the same as the right piece.

(b) Integrate in $y$. Both boundaries are already given as $x$ in terms of $y$, and each is a single formula for the whole region, so horizontal strips are the natural choice. The curves meet where $y^2=3-2y$, that is $y^2+2y-3=(y+3)(y-1)=0$, so at $y=-3$ and $y=1$. At $y=0$ the line gives $x=3$ and the parabola $x=0$, so the line is on the right. Right minus left:

$$A={\int }_{-3}^1[(3-2y)-y^2]dy={\left[3y-y^2-\frac{y^3}{3}\right]}_{-3}^1=\left(3-1-\frac{1}{3}\right)-\left(-9-9+9\right)=\frac{5}{3}+9=\frac{32}{3}.$$

Why not $x$? The crossings are $(1,1)$ and $(9,-3)$. For $0\le x\le 1$ a vertical strip runs from $y=-\sqrt{x}$ to $y=\sqrt{x}$, but for $1\le x\le 9$ it runs from $y=-\sqrt{x}$ up to the line $y=\frac{3-x}{2}$: the top boundary changes formula at $x=1$, so the $x$-route is two integrals with square roots. It does give the same answer: ${\int }_0^{1}2\sqrt{x}dx+{\int }_1^9\left(\frac{3-x}{2}+\sqrt{x}\right)dx=\frac{4}{3}+\frac{45}{4}-\frac{23}{12}=\frac{32}{3}$.

(c) It is a signed area. The integrand $3x-(x^3-x)=4x-x^3$ is "top minus bottom" only on $[0,2]$; on $[-2,0]$ it is negative, so the left piece is counted with a minus sign. Since $4x-x^3$ is odd, the two pieces cancel exactly:

$${\int }_{-2}^2\left(4x-x^3\right)dx={\left[2x^2-\frac{x^4}{4}\right]}_{-2}^2=(8-4)-(8-4)=0.$$

A region of area $8$ reported as $0$ is the signed-area trap: locate the crossings first, then integrate $|f-g|$ piece by piece.

(a) Slice in $y$. The slices are perpendicular to the $y$-axis, so they are indexed by $y$, and $\mathcal{B}$ occupies $0\le y\le 4$. At height $y$ the base is the horizontal chord of the parabola, from $x=-\sqrt{4-y}$ to $x=\sqrt{4-y}$, so the triangle's side is $s=2\sqrt{4-y}$. An equilateral triangle of side $s$ has area $\frac{\sqrt{3}}{4}{s}^2$, so

$$A(y)=\frac{\sqrt{3}}{4}\cdot 4(4-y)=\sqrt{3}(4-y).$$

By the volume-by-slicing formula,

$$V={\int }_0^4\sqrt{3}(4-y)dy=\sqrt{3}{\left[4y-\frac{y^2}{2}\right]}_0^4=\sqrt{3}(16-8)=8\sqrt{3}.$$

(b) Slice in $x$. Now the slices are indexed by $x$, with $-2\le x\le 2$. At position $x$ the base runs from $y=0$ to $y=4-x^2$; that length is the diameter of the semicircle, so the radius is $\frac{1}{2}(4-x^2)$ and

$$A(x)=\frac{1}{2}\pi {\left(\frac{4-x^2}{2}\right)}^2=\frac{\pi }{8}{\left(4-x^2\right)}^2.$$

The integrand is even, so integrate over $[0,2]$ and double:

$$V=2\cdot \frac{\pi }{8}{\int }_0^2\left(16-8x^2+x^4\right)dx=\frac{\pi }{4}{\left[16x-\frac{8x^3}{3}+\frac{x^5}{5}\right]}_0^2=\frac{\pi }{4}\left(32-\frac{64}{3}+\frac{32}{5}\right)=\frac{\pi }{4}\cdot \frac{480-320+96}{15}=\frac{64\pi }{15}.$$

Halving the width to get the radius is the step most often skipped; leaving it out multiplies the answer by $4$.

(c) Bound by the largest slice. A solid of length $\ell$ whose slices all have area at most $A_{\max }$ has volume at most $A_{\max }\ell$, because the integrand is bounded by $A_{\max }$ throughout.

In (a) the largest triangle is at $y=0$, with side $4$ and area $4\sqrt{3}$; the solid runs over $0\le y\le 4$, so $V\le 16\sqrt{3}$. Our $8\sqrt{3}$ is exactly half, which is right: $A(y)$ decreases linearly to $0$, so its average is half its maximum.

In (b) the largest semicircle is at $x=0$, with diameter $4$ and area $\frac{\pi }{8}\cdot 16=2\pi$; the solid runs over $-2\le x\le 2$, so $V\le 8\pi$. Our $\frac{64\pi }{15}\approx 0.53\times 8\pi$ is comfortably inside.

(a) The region, the axis, and one slice. The curves meet where $x^2=2x$, so at $x=0$ and $x=2$: the points $(0,0)$ and $(2,4)$. Between them the line is on top (at $x=1$, $2>1$), so $\mathcal{R}$ is the thin lens between the straight segment from $(0,0)$ to $(2,4)$ above and the parabolic arc below, widest near $x=1$ where it runs from $y=1$ to $y=2$.

The line $y=4$ is horizontal, passes through the upper corner $(2,4)$, and otherwise lies above the region, so $\mathcal{R}$ sits entirely below the axis and touches it at one point. Draw it dashed.

A vertical slice at $x$ runs from the parabola at $(x,x^2)$ up to the line at $(x,2x)$. Rotated about $y=4$ it sweeps a washer whose radii are the distances from the axis to the two ends of the slice. The lower end is farther from the axis, so the parabola gives the outer radius and the line the inner one:

$$R(x)=4-x^2,r(x)=4-2x.$$

The rule is "farther from the axis is outer", not "higher is outer"; when the axis is above the region those two disagree. On the sketch, draw the washer as two concentric circles seen edge-on, centred on the dashed line, with $R$ reaching down to the parabola and $r$ reaching down to the line.

(b) Washers about $y=4$. By the washer method, slicing along the axis of rotation,

$$V=\pi {\int }_0^2\left[(4-x^2{)}^2-(4-2x{)}^2\right]dx=\pi {\int }_0^2\left[(16-8x^2+x^4)-(16-16x+4x^2)\right]dx=\pi {\int }_0^2\left(x^4-12x^2+16x\right)dx.$$

Hence

$$V=\pi {\left[\frac{x^5}{5}-4x^3+8x^2\right]}_0^2=\pi \left(\frac{32}{5}-32+32\right)=\frac{32\pi }{5}.$$

(c) Washers about the $x$-axis. Now the axis is below the region, so the higher curve is the farther one: the line gives $R=2x$ and the parabola $r=x^2$. Then

$$V=\pi {\int }_0^2\left(4x^2-x^4\right)dx=\pi {\left[\frac{4x^3}{3}-\frac{x^5}{5}\right]}_0^2=\pi \left(\frac{32}{3}-\frac{32}{5}\right)=\frac{64\pi }{15}.$$

This is smaller than $\frac{32\pi }{5}=\frac{96\pi }{15}$, and it should be: the bulk of the lens lies low, nearer the $x$-axis than the line $y=4$, so the same area is carried around a shorter circle. The Theorem of Pappus makes this exact — the region's centroid is at height $\frac{8}{5}$, which is $\frac{8}{5}$ from the $x$-axis and $\frac{12}{5}$ from $y=4$, and the ratio of the volumes is $\frac{12/5}{8/5}=\frac{3}{2}=\frac{96/15}{64/15}$.

(a) Shells about the $y$-axis. $4x-x^2=x(4-x)$ is non-negative exactly on $[0,4]$, so that is the region's extent. The axis is vertical, so use vertical strips. The strip at $x$ has distance $x$ from the axis, height $4x-x^2$, and thickness $dx$; rotated, it is a shell of volume $2\pi x(4x-x^2)dx$. By the shell method,

$$V={\int }_0^{4}2\pi x\left(4x-x^2\right)dx=2\pi {\int }_0^4\left(4x^2-x^3\right)dx=2\pi {\left[\frac{4x^3}{3}-\frac{x^4}{4}\right]}_0^4=2\pi \left(\frac{256}{3}-64\right)=\frac{128\pi }{3}.$$

(b) Shells about $x=5$. The axis is again vertical and lies to the right of the region, so the strip at $x$ is at distance $5-x$ from it (positive throughout $[0,4]$, as a radius must be). Height and thickness are unchanged:

$$V={\int }_0^{4}2\pi (5-x)\left(4x-x^2\right)dx=2\pi {\int }_0^4\left(20x-9x^2+x^3\right)dx=2\pi {\left[10x^2-3x^3+\frac{x^4}{4}\right]}_0^4=2\pi (160-192+64)=64\pi ,$$

using $(5-x)(4x-x^2)=20x-5x^2-4x^2+x^3$. Sanity check: the region is symmetric about $x=2$, so it balances at distance $2$ from the $y$-axis and $3$ from $x=5$; the ratio of the volumes is $\frac{64\pi }{128\pi /3}=\frac{3}{2}$, exactly the ratio of those distances, as the Theorem of Pappus predicts.

(c) Washers, the hard way. For washers about a vertical axis the slices are horizontal, indexed by $y$ from $0$ to $4$, and each radius is an $x$-value on the curve. That means solving $y=4x-x^2$ for $x$: the quadratic $x^2-4x+y=0$ gives

$$x=2\pm \sqrt{4-y}.$$

A horizontal line at height $y$ crosses the arch twice, and the slice of $\mathcal{R}$ runs between the two crossings, so the washer's radii come from the two different branches: outer $R=2+\sqrt{4-y}$, inner $r=2-\sqrt{4-y}$. That is what makes it awkward: an inversion, two branches, and square roots. Carrying on,

$$R^2-r^2=(R-r)(R+r)=2\sqrt{4-y}\cdot 4=8\sqrt{4-y},$$

so

$$V=\pi {\int }_0^{4}8\sqrt{4-y}dy=8\pi {\left[-\frac{2}{3}(4-y{)}^{3/2}\right]}_0^4=8\pi \cdot \frac{2}{3}\cdot 8=\frac{128\pi }{3},$$

in agreement with (a). Shells and washers always agree; the choice is only about which cut leaves the algebra simplest.

(a) Step 1 is wrong; Step 2 inherits the error. The disk method applies when the slice of the region runs from the axis of rotation out to the curve, so that rotating it fills a solid disk. Here the vertical slice at $x$ runs from $(x,0)$ up to $(x,\sqrt{x})$, and the axis $y=3$ lies above both ends. The slice never reaches the axis, so rotating it sweeps a ring with a hole, and the method needed is washers. "The region lies under a curve" describes the region's relation to the $x$-axis, not to the axis of rotation, and it is the axis of rotation that decides disk versus washer.

Given that, Step 2's quantity $3-\sqrt{x}$ is a correct distance but the wrong role: it is the distance from the axis to the nearer end of the slice, so it is the inner radius. The outer radius is the distance from the axis to the farther end, the point on the $x$-axis, which is $3-0=3$.

The student's integral $\pi {\int }_0^4(3-\sqrt{x}{)}^{2}dx$ is a correct disk-method computation for a different region: the one whose slice at $x$ runs from the curve $y=\sqrt{x}$ up to the axis $y=3$. That is the region between $y=\sqrt{x}$ and $y=3$ for $0\le x\le 4$, the complement of $\mathcal{R}$ inside the rectangle $[0,4]\times [0,3]$. So $12\pi$ is the volume of the solid generated by everything in the rectangle except $\mathcal{R}$.

(b) Washers. Slice vertically at $x$, $0\le x\le 4$; outer radius $R=3$, inner radius $r=3-\sqrt{x}$. By the washer method,

$$V=\pi {\int }_0^4\left[3^2-{\left(3-\sqrt{x}\right)}^2\right]dx=\pi {\int }_0^4\left(6\sqrt{x}-x\right)dx=\pi {\left[4x^{3/2}-\frac{x^2}{2}\right]}_0^4=\pi (32-8)=24\pi .$$

A check that also explains the student's number: the whole rectangle $[0,4]\times [0,3]$ rotated about $y=3$ is a cylinder of radius $3$ and length $4$, of volume $36\pi$, and $\mathcal{R}$ and its complement fill it between them: $24\pi +12\pi =36\pi$.

(c) Shells. For a horizontal axis, shells come from horizontal strips, indexed by $y$ from $0$ to $2$ (the curve reaches $y=2$ at $x=4$). The strip at height $y$ runs from the curve, where $x=y^2$, to the line $x=4$, so its length is $4-y^2$; its distance from the axis $y=3$ is $3-y$. By the shell method,

$$V={\int }_0^{2}2\pi (3-y)\left(4-y^2\right)dy=2\pi {\int }_0^2\left(12-4y-3y^2+y^3\right)dy=2\pi {\left[12y-2y^2-y^3+\frac{y^4}{4}\right]}_0^2=2\pi (24-8-8+4)=24\pi .$$

The two methods agree, and neither of them is the student's $12\pi$.

(a) Slice the motion; find $k$ first. Hooke's Law says $F(x)=kx$ with $x$ the extension beyond the natural length. One measurement fixes $k$: $12=k\cdot 0.05$, so $k=240$ N/m. Over a short step $dx$ at extension $x$ the force is nearly constant, contributing $240xdx$, so

$$W={\int }_{0.05}^{0.15}240xdx=120{\left[x^2\right]}_{0.05}^{0.15}=120(0.0225-0.0025)=2.4\text{J}.$$

Sanity check: the force runs from $12$ N to $36$ N over $0.1$ m, so the work lies between $1.2$ J and $3.6$ J; because the force is linear it is the midpoint, $2.4$ J.

(b) Two contributions, sliced two different ways. Let $h$ be the height of the bucket above its starting point, $0\le h\le 20$.

The water: slice the motion. The water drains linearly from $10$ kg to $0$ kg over the lift, so at height $h$ its mass is $10-\frac{h}{2}$ and its weight is $9.8\left(10-\frac{h}{2}\right)$ N. A step $dh$ contributes that weight times $dh$:

$$W_{\text{water}}={\int }_0^{20}9.8\left(10-\frac{h}{2}\right)dh=9.8{\left[10h-\frac{h^2}{4}\right]}_0^{20}=9.8(200-100)=980\text{J}.$$

The rope: slice the rope. The rope weighs $0.5\times 9.8=4.9$ N per metre. The piece of rope initially $x$ metres below the winch, of length $dx$, weighs $4.9dx$ and is lifted exactly $x$ metres, contributing $4.9xdx$:

$$W_{\text{rope}}={\int }_0^{20}4.9xdx=4.9\cdot \frac{400}{2}=980\text{J}.$$

(Slicing the motion instead gives the same thing: when the bucket is at height $h$ the hanging rope is $20-h$ long, and ${\int }_0^{20}4.9(20-h)dh=980$.)

Total: $W=980+980=1960$ J.

Sanity check: a non-leaking bucket alone would take $10\times 9.8\times 20=1960$ J; leaking to empty halves that, since the average mass carried is $5$ kg. The $10$ kg of rope rises through its centre of mass, $10$ m: $98\times 10=980$ J.

(c) Slice the liquid into horizontal layers. Put the origin at the bottom vertex with $y$ measured upward, so the water occupies $0\le y\le 2$ and the spout is at $y=3$. By similar triangles the width of the triangle at height $y$ is $w=y$ (it is $2$ at $y=2$), so a layer at height $y$ has horizontal area $A(y)=6y$ and volume $6ydy$, hence weight $9800\cdot 6ydy$. It must be lifted $d(y)=3-y$. Check the extremes: the top layer ($y=2$) rises $1$ m and the bottom layer rises $3$ m. Then

$$W={\int }_0^{2}9800\cdot 6y(3-y)dy=58,800{\int }_0^2\left(3y-y^2\right)dy=58,800{\left[\frac{3y^2}{2}-\frac{y^3}{3}\right]}_0^2=58,800\left(6-\frac{8}{3}\right)=58,800\cdot \frac{10}{3}=196,000\text{J}.$$

Sanity check: the water's volume is $\frac{1}{2}\cdot 2\cdot 2\cdot 6=12{\text{m}}^3$, weight $117,600$ N. A triangle's centroid is two thirds of the way from the vertex to the base, at $y=\frac{4}{3}$, so the whole mass rises on average $3-\frac{4}{3}=\frac{5}{3}$ m, and $117,600\times \frac{5}{3}=196,000$ J.

(a) A non-negative continuous function with zero integral is zero. Put $h=g-f$. Then $h$ is continuous, $h\ge 0$ on $[a,b]$, and

$${\int }_a^{b}hdx={\int }_a^{b}gdx-{\int }_a^{b}fdx=(b-a)\left(g_{\text{avg}}-f_{\text{avg}}\right)=0.$$

Suppose, for contradiction, that $h(c)>0$ for some $c\in [a,b]$. Continuity of $h$ at $c$ with $\epsilon =\frac{1}{2}h(c)$ gives $\delta >0$ such that $|h(x)-h(c)|<\frac{1}{2}h(c)$, hence $h(x)>\frac{1}{2}h(c)$, for every $x\in [a,b]$ with $|x-c|<\delta$. Let $[p,q]=\left[\max \left(a,c-\frac{\delta }{2}\right),\min \left(b,c+\frac{\delta }{2}\right)\right]$; this is a subinterval of $[a,b]$ of positive length on which $h>\frac{1}{2}h(c)$. By additivity and the comparison property (using $h\ge 0$ on the two outer pieces and $h>\frac{1}{2}h(c)$ on the middle one),

$${\int }_a^{b}hdx={\int }_a^{p}hdx+{\int }_p^{q}hdx+{\int }_q^{b}hdx\ge 0+\frac{h(c)}{2}(q-p)+0>0,$$

contradicting ${\int }_a^{b}h=0$. Hence $h(x)=0$, that is $f(x)=g(x)$, for every $x\in [a,b]$. $■$

(b) One point is not enough without continuity. On $[0,1]$ let $f\equiv 0$ and let $g(x)=1$ if $x=\frac{1}{2}$, $g(x)=0$ otherwise. Then $f\le g$ everywhere, and ${\int }_0^{1}gdx=0$: every Riemann sum for $g$ has at most one nonzero term, of size at most $\Delta x$, which tends to $0$. So $f_{\text{avg}}=g_{\text{avg}}=0$, yet $g\left(\frac{1}{2}\right)\ne f\left(\frac{1}{2}\right)$.

The proof breaks at the sentence "continuity of $h$ at $c$ gives $\delta >0$ such that $h(x)>\frac{1}{2}h(c)$ on a whole interval around $c$". That is the only place continuity was used, and it is essential: it converts a positive value at a point, which has no length and contributes nothing to an integral, into a positive value on an interval, which does. For the step function $h=g$ there is no such interval.

(c) Use the Mean Value Theorem for derivatives. Define $F(x)={\int }_a^{x}f(t)dt$ for $x\in [a,b]$. Since $f$ is continuous, the Fundamental Theorem of Calculus says $F$ is differentiable on $[a,b]$ with $F^{\prime }(x)=f(x)$; in particular $F$ is continuous on $[a,b]$ and differentiable on $(a,b)$, which are exactly the hypotheses of the Mean Value Theorem. That theorem supplies $c\in (a,b)$ — the open interval — with

$$F(b)-F(a)=F^{\prime }(c)(b-a),\text{that is}{\int }_a^{b}f(x)dx=f(c)(b-a).$$

Dividing by $b-a$ gives $f(c)=f_{\text{avg}}$ with $c$ strictly between $a$ and $b$. $■$

The closed-interval proof, via the Extreme and Intermediate Value Theorems, cannot promise this: the IVT is silent about where its point lies. Routing through $F$ trades that for the MVT's guarantee of an interior point.

(a) Complete the square. $y^{\prime }=\frac{x^3}{2}-\frac{1}{2x^3}$, so

$$(y^{\prime }{)}^2=\frac{x^6}{4}-\frac{1}{2}+\frac{1}{4x^6},1+(y^{\prime }{)}^2=\frac{x^6}{4}+\frac{1}{2}+\frac{1}{4x^6}={\left(\frac{x^3}{2}+\frac{1}{2x^3}\right)}^2.$$

The added $1$ flips the sign of the cross term, which is the whole trick. For $x>0$ both $\frac{x^3}{2}$ and $\frac{1}{2x^3}$ are positive, so the square root of the square is the expression itself, not its negative. By the arc length formula,

$$L={\int }_1^2\left(\frac{x^3}{2}+\frac{1}{2x^3}\right)dx={\left[\frac{x^4}{8}-\frac{1}{4x^2}\right]}_1^2=\left(2-\frac{1}{16}\right)-\left(\frac{1}{8}-\frac{1}{4}\right)=\frac{31}{16}+\frac{1}{8}=\frac{33}{16}.$$

Sanity check: the endpoints are $\left(1,\frac{3}{8}\right)$ and $\left(2,\frac{33}{16}\right)$, at straight-line distance $\sqrt{1+{\left(\frac{27}{16}\right)}^2}=\frac{\sqrt{985}}{16}\approx 1.96$; the curve, at $\frac{33}{16}\approx 2.06$, is a little longer, as it must be.

(b) The same trick in $y$. The curve is given as $x$ in terms of $y$, so use $L=\int \sqrt{1+(dx/dy{)}^2}dy$. Here $\frac{dx}{dy}=y^2-\frac{1}{4y^2}$, and

$$1+{\left(\frac{dx}{dy}\right)}^2=1+y^4-\frac{1}{2}+\frac{1}{16y^4}=y^4+\frac{1}{2}+\frac{1}{16y^4}={\left(y^2+\frac{1}{4y^2}\right)}^2,$$

with the bracket positive for $y>0$. Hence

$$L={\int }_1^3\left(y^2+\frac{1}{4y^2}\right)dy={\left[\frac{y^3}{3}-\frac{1}{4y}\right]}_1^3=\left(9-\frac{1}{12}\right)-\left(\frac{1}{3}-\frac{1}{4}\right)=\frac{107}{12}-\frac{1}{12}=\frac{53}{6}.$$

Sanity check: the endpoints are $\left(\frac{7}{12},1\right)$ and $\left(\frac{109}{12},3\right)$, whose chord has length $\sqrt{{\left(\frac{17}{2}\right)}^2+4}=\frac{\sqrt{305}}{2}\approx 8.73$, slightly under $\frac{53}{6}\approx 8.83$.

(c) The arc length function. With the root already removed in (a),

$$s(x)={\int }_1^x\left(\frac{t^3}{2}+\frac{1}{2t^3}\right)dt={\left[\frac{t^4}{8}-\frac{1}{4t^2}\right]}_1^x=\frac{x^4}{8}-\frac{1}{4x^2}+\frac{1}{8},$$

and as a check $s(2)=2-\frac{1}{16}+\frac{1}{8}=\frac{33}{16}$, the answer to (a). By the Fundamental Theorem of Calculus, $s^{\prime }(x)=\frac{x^3}{2}+\frac{1}{2x^3}$, so $s^{\prime }(1)=\frac{1}{2}+\frac{1}{2}=1$.

The meaning: $s^{\prime }(x)=\sqrt{1+(y^{\prime }{)}^2}\ge 1$ always, with equality exactly where $y^{\prime }=0$. And indeed $y^{\prime }(1)=\frac{1}{2}-\frac{1}{2}=0$: the tangent at $x=1$ is horizontal, so a small horizontal step $dx$ there produces a step $ds$ of the same length along the curve; at that instant the curve is no longer than its shadow on the $x$-axis. Since $y^{\prime }<0$ for $0<x<1$ and $y^{\prime }>0$ for $x>1$, this is the lowest point of the curve, $\left(1,\frac{3}{8}\right)$.

(a) A substitution kills the root. Rotation about the $x$-axis, so the radius is $y=\frac{x^3}{3}$, and with $y^{\prime }=x^2$ the arc length element in $x$ is $ds=\sqrt{1+x^4}dx$. By the surface area formula,

$$S={\int }_0^{1}2\pi \cdot \frac{x^3}{3}\sqrt{1+x^4}dx.$$

The factor $x^3$ is, up to a constant, the derivative of $1+x^4$. Substitute $u=1+x^4$, $du=4x^{3}dx$; the limits $x=0,1$ become $u=1,2$:

$$S=\frac{2\pi }{3}\cdot \frac{1}{4}{\int }_1^2\sqrt{u}du=\frac{\pi }{6}\cdot \frac{2}{3}[u^{3/2}{]}_1^2=\frac{\pi }{9}\left(2\sqrt{2}-1\right).$$

Sanity check: on $[0,1]$ the root lies between $1$ and $\sqrt{2}$, so $S$ lies between $2\pi {\int }_0^1\frac{x^3}{3}dx=\frac{\pi }{6}\approx 0.52$ and $\sqrt{2}$ times that, $\approx 0.74$; our $\frac{\pi }{9}(2\sqrt{2}-1)\approx 0.64$ is between them.

(b) The perfect-square family. Radius $y=\frac{x^3}{3}+\frac{1}{4x}$, and $y^{\prime }=x^2-\frac{1}{4x^2}$ has the shape $u-\frac{1}{4u}$ with $u=x^2$, so

$$1+(y^{\prime }{)}^2=x^4+\frac{1}{2}+\frac{1}{16x^4}={\left(x^2+\frac{1}{4x^2}\right)}^2,ds=\left(x^2+\frac{1}{4x^2}\right)dx$$

(the bracket is positive for $x>0$). Multiply out the integrand:

$$\left(\frac{x^3}{3}+\frac{1}{4x}\right)\left(x^2+\frac{1}{4x^2}\right)=\frac{x^5}{3}+\frac{x}{12}+\frac{x}{4}+\frac{1}{16x^3}=\frac{x^5}{3}+\frac{x}{3}+\frac{1}{16x^3}.$$

Hence

$$S=2\pi {\int }_1^2\left(\frac{x^5}{3}+\frac{x}{3}+\frac{1}{16x^3}\right)dx=2\pi {\left[\frac{x^6}{18}+\frac{x^2}{6}-\frac{1}{32x^2}\right]}_1^2.$$

At $x=2$ the bracket is $\frac{32}{9}+\frac{2}{3}-\frac{1}{128}$; at $x=1$ it is $\frac{1}{18}+\frac{1}{6}-\frac{1}{32}$. Subtracting term by term: $\frac{32}{9}-\frac{1}{18}=\frac{63}{18}=\frac{7}{2}$, $\frac{2}{3}-\frac{1}{6}=\frac{1}{2}$, $\frac{1}{32}-\frac{1}{128}=\frac{3}{128}$. The difference is $4+\frac{3}{128}=\frac{515}{128}$, so

$$S=2\pi \cdot \frac{515}{128}=\frac{515\pi }{64}.$$

Sanity check: the arc has length ${\left[\frac{x^3}{3}-\frac{1}{4x}\right]}_1^2=\frac{59}{24}\approx 2.46$ and its radii run from $\frac{7}{12}$ to $\frac{67}{24}$. A single frustum with those radii and that slant height has area $\pi \left(\frac{7}{12}+\frac{67}{24}\right)\frac{59}{24}\approx 8.30\pi$, a shade above our $\frac{515}{64}\pi \approx 8.05\pi$, as expected for a curve that sags below its chord.

(c) The zone: the root cancels. Generate the sphere by rotating $y=\sqrt{a^2-x^2}$ about the $x$-axis, and let the planes be $x=c$ and $x=c+h$ with $-a\le c<c+h\le a$. The radius is $y$, and

$$y^{\prime }=\frac{-x}{\sqrt{a^2-x^2}},1+(y^{\prime }{)}^2=\frac{a^2}{a^2-x^2},ds=\frac{a}{\sqrt{a^2-x^2}}dx.$$

The root in the radius cancels the root in $ds$ exactly:

$$S={\int }_c^{c+h}2\pi \sqrt{a^2-x^2}\cdot \frac{a}{\sqrt{a^2-x^2}}dx=2\pi a{\int }_c^{c+h}dx=2\pi ah.$$

The answer contains no $c$, which is the claim: the integrand is the constant $2\pi a$, so every slab of thickness $dx$ carries the same surface area wherever it is cut. If one plane is tangent at a pole ($c=-a$ or $c+h=a$) the derivative $y^{\prime }$ blows up there, but the integrand simplified to a constant before the endpoint was reached, so the improper integral is harmless. As a check, $h=2a$ gives the full sphere, $4\pi a^2$.

(a) Horizontal strips down the trapezoid. Let $y$ be the depth below the surface, $0\le y\le 6$; since the water reaches the top of the face, $d(y)=y$. The width shrinks linearly from $20$ to $10$ over $6$ m:

$$w(y)=20-\frac{10}{6}y=20-\frac{5y}{3},w(0)=20,w(6)=10.$$

The strip at depth $y$ has area $w(y)dy$ and feels pressure $\rho gy$, so by the hydrostatic force formula

$$F={\int }_0^{6}9800y\left(20-\frac{5y}{3}\right)dy=9800{\left[10y^2-\frac{5y^3}{9}\right]}_0^6=9800(360-120)=9800\times 240=2,352,000\text{N},$$

about $2.35\times 10^6$ N.

(b) Strips across the window. Let $y$ be the signed vertical distance below the window's centre, $-1\le y\le 1$. Then $d(y)=5+y$: the top strip ($y=-1$) is at depth $4$ and the bottom strip at depth $6$. The width is the chord of the unit circle, $w(y)=2\sqrt{1-y^2}$. So

$$F=9800{\int }_{-1}^1(5+y)\cdot 2\sqrt{1-y^2}dy=9800\left[5{\int }_{-1}^{1}2\sqrt{1-y^2}dy+{\int }_{-1}^{1}2y\sqrt{1-y^2}dy\right].$$

The first integral adds up the widths of the horizontal chords of the unit disc, so it is the disc's area, $\pi$. The second has an odd integrand on a symmetric interval, so it is $0$: the extra pressure on the lower half is exactly matched by the shortfall on the upper half. Hence

$$F=9800\times 5\pi =49,000\pi \text{N}\approx 1.54\times 10^5\text{N}.$$

(c) The centroid shortcut. Take $y$ to be depth below the surface, with the plate occupying $a\le y\le b$, so $d(y)=y$. The strip at depth $y$ has area $w(y)dy$, and its own centre lies at depth $y$, so its moment about the surface line is $yw(y)dy$. Summing strips, the plate's area and its moment about the surface are

$$A={\int }_a^{b}w(y)dy,M={\int }_a^{b}yw(y)dy,$$

and by definition the centroid's depth is the area-weighted average depth, $\bar{d}=M/A$. (This is the centroid formula with the strips cut horizontally rather than vertically; the derivation is the same, since a thin strip balances at its centre.) Therefore

$$F=\rho g{\int }_a^{b}yw(y)dy=\rho gM=\rho g\bar{d}A.■$$

The shortcut is exact, not an approximation, because the pressure is linear in depth.

Check of (b). A disc's centroid is its centre, at depth $5$, and its area is $\pi \cdot 1^2=\pi$, so $F=9800\times 5\times \pi =49,000\pi$ N, as before.

Centroid of the trapezoid. Its area is $A=\frac{20+10}{2}\cdot 6=90{\text{m}}^2$, so from (a)

$$\bar{d}=\frac{F}{\rho gA}=\frac{240}{90}=\frac{8}{3}\text{m}.$$

Plausibility: a $6$ m rectangle would balance at depth $3$; a triangle with its $20$ m side at the top and its vertex at the bottom would balance at depth $2$; the trapezoid, wider at the top than a rectangle would be relative to its bottom, sits between them at $\frac{8}{3}$.

(a) Centroid of the triangle. The hypotenuse joins $(b,0)$ to $(0,h)$, so the triangle is the region under $f(x)=h\left(1-\frac{x}{b}\right)$, above the $x$-axis, for $0\le x\le b$, and its area is $A=\frac{1}{2}bh$ (or, from the integral, ${\int }_0^{b}h\left(1-\frac{x}{b}\right)dx=h\left(b-\frac{b}{2}\right)=\frac{1}{2}bh$).

By the centroid formulas,

$$\bar{x}=\frac{1}{A}{\int }_0^{b}xh\left(1-\frac{x}{b}\right)dx=\frac{2}{bh}\cdot h{\left[\frac{x^2}{2}-\frac{x^3}{3b}\right]}_0^b=\frac{2}{b}\left(\frac{b^2}{2}-\frac{b^2}{3}\right)=\frac{2}{b}\cdot \frac{b^2}{6}=\frac{b}{3}.$$

For $\bar{y}$, substitute $u=1-\frac{x}{b}$, $dx=-bdu$, so that ${\int }_0^b{\left(1-\frac{x}{b}\right)}^{2}dx=b{\int }_0^1{u}^{2}du=\frac{b}{3}$. Then

$$\bar{y}=\frac{1}{A}{\int }_0^b\frac{1}{2}{\left[h\left(1-\frac{x}{b}\right)\right]}^{2}dx=\frac{2}{bh}\cdot \frac{h^2}{2}\cdot \frac{b}{3}=\frac{h}{3}.$$

So the centroid is $\left(\frac{b}{3},\frac{h}{3}\right)$. $■$ As a check, the median from $(0,h)$ to the midpoint $\left(\frac{b}{2},0\right)$ of the opposite side passes through this point (two thirds of the way along it), and a triangle's centroid is where its medians meet.

(b) Two cones by Pappus. The triangle lies in $x\ge 0$, touching the $y$-axis along one leg, so the hypothesis of the Theorem of Pappus holds for that axis; the distance from the centroid to the axis is $d=\bar{x}=\frac{b}{3}$. Hence

$$V=2\pi dA=2\pi \cdot \frac{b}{3}\cdot \frac{bh}{2}=\frac{1}{3}\pi b^{2}h,$$

the volume of a cone of base radius $b$ and height $h$: one third of base area times height.

Likewise the triangle lies in $y\ge 0$, touching the $x$-axis along its other leg, and $d=\bar{y}=\frac{h}{3}$, so

$$V=2\pi \cdot \frac{h}{3}\cdot \frac{bh}{2}=\frac{1}{3}\pi h^{2}b,$$

the cone of base radius $h$ and height $b$. Swapping which leg is the axis swaps the roles of $b$ and $h$, as it should.

(c) Pappus run backwards. Swapping $x$ and $y$ leaves the description $x^2+y^2\le r^2$, $x\ge 0$, $y\ge 0$ unchanged, so $\mathcal{Q}$ is symmetric about the line $y=x$, and by the symmetry principle its centroid lies on that line: $\bar{x}=\bar{y}$.

Now rotate $\mathcal{Q}$ about the $y$-axis. The region lies in $x\ge 0$ and touches the axis along a radius, so Pappus applies; the solid swept out is a hemisphere of radius $r$, of volume $\frac{1}{2}\cdot \frac{4}{3}\pi r^3=\frac{2}{3}\pi r^3$, and the area of $\mathcal{Q}$ is $\frac{1}{4}\pi r^2$. The Theorem of Pappus gives

$$\frac{2}{3}\pi r^3=2\pi \bar{x}\cdot \frac{\pi r^2}{4}=\frac{{\pi }^2{r}^2}{2}\bar{x},\text{so}\bar{x}=\frac{2\pi r^3/3}{{\pi }^2{r}^2/2}=\frac{4r}{3\pi },$$