Multiple Integrals

(a) Both orders. With $y$ inside, hold $x$ fixed:

$${\int }_0^2(2x+3y^2)dy=[2xy+y^3{]}_{y=0}^{y=2}=4x+8,{\int }_0^1(4x+8)dx=[2x^2+8x{]}_0^1=10.$$

With $x$ inside:

$${\int }_0^1(2x+3y^2)dx=[x^2+3x{y}^2{]}_{x=0}^{x=1}=1+3y^2,{\int }_0^2(1+3y^2)dy=[y+y^3{]}_0^2=10.$$

The agreement is Fubini's Theorem: the integrand is continuous on the rectangle, so either iterated integral equals ${\iint }_{R}fdA=10$.

(b) Integrate in $y$ first. With $x$ fixed, $\frac{\partial }{\partial y}{e}^{xy}=x{e}^{xy}$, so the factor $x$ is exactly what the $y$-antiderivative needs:

$${\int }_0^{1}x{e}^{xy}dy=[e^{xy}{]}_{y=0}^{y=1}=e^x-1,{\int }_0^1(e^x-1)dx=[e^x-x{]}_0^1=(e-1)-1=e-2\approx \mathrm{0.718.}$$

The other order starts with ${\int }_0^{1}x{e}^{xy}dx$, which needs integration by parts and produces $\frac{e^y}{y}-\frac{e^y-1}{y^2}$; that can only be integrated by recognising it as the derivative of $\frac{e^y-1}{y}$. Choosing the order well replaces all of that with two lines.

(c) Product form, then divide by the area. The integrand is $g(x)h(y)$ on a rectangle, so the integral is a product of single integrals:

$${\iint }_R{x}^2\sin ydA={\int }_0^2{x}^{2}dx{\int }_0^{\pi }\sin ydy=\frac{8}{3}\cdot 2=\frac{16}{3}.$$

The rectangle has area $A(R)=2\pi$, so

$$f_{\text{avg}}=\frac{1}{2\pi }\cdot \frac{16}{3}=\frac{8}{3\pi }\approx \mathrm{0.85.}$$

Sanity check: $0\le x^2\sin y\le 4$ on $R$, and the integrand is small over most of the rectangle (near $x=0$ and near $y=0$ and $y=\pi$), so an average well below the midpoint $2$ is right.

(a) Intersections and the type I description. The curves meet where $x^2=x+2$, that is $x^2-x-2=(x-2)(x+1)=0$, so at $x=-1$ and $x=2$: the points $(-1,1)$ and $(2,4)$. Between them the line is on top (at $x=0$ it gives $2>0$), so a vertical line at $x$ enters $D$ on the parabola and leaves on the line:

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

The area is the integral of $1$:

$$A(D)={\int }_{-1}^2{\int }_{x^2}^{x+2}dydx={\int }_{-1}^2(x+2-x^2)dx={\left[\frac{x^2}{2}+2x-\frac{x^3}{3}\right]}_{-1}^2=\left(2+4-\frac{8}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)=\frac{10}{3}+\frac{7}{6}=\frac{9}{2}.$$

(b) The moment, and what the quotient means. The inner integral simply multiplies the width by $x$:

$${\iint }_{D}xdA={\int }_{-1}^{2}x(x+2-x^2)dx={\int }_{-1}^2\left(x^2+2x-x^3\right)dx={\left[\frac{x^3}{3}+x^2-\frac{x^4}{4}\right]}_{-1}^2=\frac{8}{3}-\frac{5}{12}=\frac{27}{12}=\frac{9}{4}.$$

(At $x=2$ the bracket is $\frac{8}{3}+4-4=\frac{8}{3}$; at $x=-1$ it is $-\frac{1}{3}+1-\frac{1}{4}=\frac{5}{12}$.)

The quotient $\frac{9/4}{9/2}=\frac{1}{2}$ is $\bar{x}=M_y/m$ for a lamina of constant density $1$ on $D$: the $x$-coordinate of the centroid. It had to be $\frac{1}{2}$ because the vertical width of $D$,

$$w(x)=x+2-x^2=\frac{9}{4}-{\left(x-\frac{1}{2}\right)}^2,$$

is symmetric about $x=\frac{1}{2}$, the midpoint of $-1$ and $2$. Every strip at $\frac{1}{2}+t$ has a twin of equal area at $\frac{1}{2}-t$, so the region balances on the line $x=\frac{1}{2}$.

(c) Two type II pieces. For a type II description a horizontal line at height $y$ must enter and leave $D$ by a single formula each. For $0\le y\le 1$ the line meets only the parabola, entering at $x=-\sqrt{y}$ and leaving at $x=\sqrt{y}$. For $1\le y\le 4$ it enters on the line, at $x=y-2$, and leaves on the parabola, at $x=\sqrt{y}$. The left boundary changes formula at $y=1$, at the corner $(-1,1)$, so one description will not do:

$$D=\{0\le y\le 1,-\sqrt{y}\le x\le \sqrt{y}\}\cup \{1\le y\le 4,y-2\le x\le \sqrt{y}\}.$$

Check against (a): the areas of the pieces are ${\int }_0^{1}2\sqrt{y}dy=\frac{4}{3}$ and

$${\int }_1^4\left(\sqrt{y}-y+2\right)dy={\left[\frac{2}{3}{y}^{3/2}-\frac{y^2}{2}+2y\right]}_1^4=\frac{16}{3}-\frac{13}{6}=\frac{19}{6},$$

and $\frac{4}{3}+\frac{19}{6}=\frac{27}{6}=\frac{9}{2}$, as before.

(a) The region. The limits say $0\le y\le 2$ and $\frac{y}{2}\le x\le 1$. The left boundary $x=y/2$ is the line $y=2x$ through the origin with slope $2$; the right boundary is the vertical line $x=1$; the bottom is $y=0$; and at $y=2$ the left and right boundaries meet, since $y/2=1$ there. So the region is the triangle with corners $(0,0)$, $(1,0)$ and $(1,2)$: it lies below the line $y=2x$, to the left of $x=1$ and above the $x$-axis.

The sketch must show the three lines labelled $y=2x$, $x=1$ and $y=0$, the shaded triangle between them, the three corners with their coordinates, and a horizontal arrow at some height $y$ running from the slanted line (entry at $x=y/2$) to the vertical wall (exit at $x=1$). That arrow is the inner integral as written: at each fixed height, $x$ sweeps from the line to the wall.

(b) Why the given order is stuck. The inner integral is $\int \cos (x^2)dx$, and $\cos (x^2)$ has no elementary antiderivative — it is in the same family as $\sin (y^2)$ and $e^{x^2}$. So the inner integration cannot be written down in closed form, whatever the limits are.

(c) Reverse and evaluate. Sweep the triangle with vertical lines instead: $x$ runs from $0$ to $1$, and the vertical line at $x$ enters on the $x$-axis at $y=0$ and leaves on the slanted line at $y=2x$. Hence

$$I={\int }_0^1{\int }_0^{2x}\cos (x^2)dydx={\int }_0^{1}2x\cos (x^2)dx.$$

The inner integral was trivial because the integrand does not involve $y$, and it produced the factor $2x$ that the substitution $u=x^2$, $du=2xdx$, needs:

$$I=[\sin (x^2){]}_0^1=\sin 1\approx \mathrm{0.841.}$$

Sanity check: the triangle has area $1$, and on it $\cos (x^2)$ lies between $\cos 1\approx 0.54$ and $1$, so $I$ must lie between $0.54$ and $1$. It does. The tempting wrong reversal ${\int }_{y/2}^1{\int }_0^2\cos (x^2)dydx$ leaves a $y$ in an outer limit, which is meaningless — the limits change shape, not just position.

(a) A quarter disk. The limits say $0\le x\le 1$ and $0\le y\le \sqrt{1-x^2}$: the part of the unit disk in the first quadrant, which in polar coordinates is $0\le r\le 1$, $0\le \theta \le \frac{\pi }{2}$. With $x^2+y^2=r^2$ and $dA=rdrd\theta$,

$${\int }_0^{\pi /2}{\int }_0^1{r}^2\cdot rdrd\theta =\frac{\pi }{2}{\left[\frac{r^4}{4}\right]}_0^1=\frac{\pi }{8}.$$

Sanity check: the quarter disk has area $\frac{\pi }{4}$ and the integrand lies between $0$ and $1$; $\frac{\pi }{8}$ is half the area, which is right for an integrand that is near $0$ at the origin and equals $1$ on the arc.

(b) An annulus. $D$ is $1\le r\le \sqrt{3}$, $0\le \theta \le 2\pi$, and the integrand is $\frac{1}{1+r^2}$:

$${\iint }_D\frac{dA}{1+x^2+y^2}={\int }_0^{2\pi }{\int }_1^{\sqrt{3}}\frac{1}{1+r^2}rdrd\theta .$$

The factor $r$ supplied by $dA$ is what the substitution $u=1+r^2$, $du=2rdr$, needs:

$${\int }_1^{\sqrt{3}}\frac{rdr}{1+r^2}=\frac{1}{2}[\ln (1+r^2){]}_1^{\sqrt{3}}=\frac{1}{2}(\ln 4-\ln 2)=\frac{1}{2}\ln 2,$$

so the integral is $2\pi \cdot \frac{1}{2}\ln 2=\pi \ln 2\approx 2.18$.

Sanity check: the annulus has area $\pi (3-1)=2\pi$, and on it the integrand lies between $\frac{1}{4}$ and $\frac{1}{2}$, so the answer must lie between $\frac{\pi }{2}\approx 1.57$ and $\pi \approx 3.14$.

(c) The disk $r=2\cos \theta$. Multiplying the equation by $r$ gives $r^2=2r\cos \theta$, that is $x^2+y^2=2x$, or $(x-1{)}^2+y^2=1$: the circle of radius $1$ centred at $(1,0)$, traced once for $-\frac{\pi }{2}\le \theta \le \frac{\pi }{2}$ (outside that range $2\cos \theta <0$, and $r$ is kept non-negative). The ray at angle $\theta$ enters at $r=0$ and leaves at $r=2\cos \theta$, and $x=r\cos \theta$:

$${\iint }_{D}xdA={\int }_{-\pi /2}^{\pi /2}{\int }_0^{2\cos \theta }r\cos \theta \cdot rdrd\theta ={\int }_{-\pi /2}^{\pi /2}\cos \theta \cdot \frac{8{\cos }^3\theta }{3}d\theta =\frac{8}{3}{\int }_{-\pi /2}^{\pi /2}{\cos }^4\theta d\theta .$$

Reduce the power twice: from ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$,

$${\cos }^4\theta =\frac{1+2\cos 2\theta +{\cos }^{2}2\theta }{4}=\frac{3}{8}+\frac{\cos 2\theta }{2}+\frac{\cos 4\theta }{8}.$$

Over $[-\frac{\pi }{2},\frac{\pi }{2}]$ the terms $\cos 2\theta$ and $\cos 4\theta$ integrate to $\frac{\sin 2\theta }{2}$ and $\frac{\sin 4\theta }{4}$, both zero at both ends, leaving $\frac{3}{8}\cdot \pi$. Hence

$${\iint }_{D}xdA=\frac{8}{3}\cdot \frac{3\pi }{8}=\pi .$$

Check: $D$ is a disk of area $\pi$ whose centroid is its centre $(1,0)$, so ${\iint }_{D}xdA=\bar{x}A(D)=1\cdot \pi =\pi$. Exactly.

(a) A continuous function attains its average. Let $\overline{f}=\frac{1}{A(R)}{\iint }_{R}fdA$ be the average value; the claim is that $f(x_0,y_0)=\overline{f}$ for some point of $R$.

Step 1: trap the average between the extreme values. $R$ is closed and bounded and $f$ is continuous on it, so by the Extreme Value Theorem $f$ attains a minimum value $m$ at some point $P\in R$ and a maximum value $M$ at some point $Q\in R$. Since $m\le f\le M$ on $R$, the bounds property gives $mA(R)\le {\iint }_{R}fdA\le MA(R)$, and dividing by $A(R)>0$,

$$m\le \overline{f}\le M.$$

Step 2: connect $P$ to $Q$ inside $R$. Define $g(t)=f(P+t(Q-P))$ for $0\le t\le 1$. The point $P+t(Q-P)$ lies in $R$ for every such $t$: its $x$-coordinate is $(1-t)x_P+t{x}_Q$, which lies between two numbers of $[a,b]$, and likewise its $y$-coordinate lies in $[c,d]$. This is where the rectangle is used: the whole segment from $P$ to $Q$ stays inside $R$ because a rectangle is convex. For a crescent-shaped region the segment could leave the region, and one would need a path inside it instead.

Step 3: apply the IVT. $g$ is the composition of the continuous map $t\mapsto P+t(Q-P)$ with the continuous $f$, so it is continuous on $[0,1]$, with $g(0)=f(P)=m$ and $g(1)=f(Q)=M$. Because $m\le \overline{f}\le M$, the Intermediate Value Theorem gives $t_0\in [0,1]$ with $g(t_0)=\overline{f}$. (If $m=M$ then $f$ is constant and $t_0=0$ works.) Setting $(x_0,y_0)=P+t_0(Q-P)\in R$,

$$f(x_0,y_0)=\overline{f},\text{that is}{\iint }_{R}fdA=f(x_0,y_0)A(R).■$$

(b) A positive value forces a positive integral. Write $\epsilon =\frac{1}{2}f(P)>0$. By continuity of $f$ at $P$ there is $\delta >0$ such that $|f(X)-f(P)|<\epsilon$ for every point $X$ of $R$ within distance $\delta$ of $P$; for those $X$, $f(X)>f(P)-\epsilon =\frac{1}{2}f(P)$. Since $P$ is an interior point, we may choose a small closed rectangle $R^{\prime }$ centred at $P$ that lies inside $R$ and inside the disk of radius $\delta$ about $P$. On $R^{\prime }$, then, $f\ge \frac{1}{2}f(P)$.

Now compare the integrals over $R^{\prime }$ and over $R$. Let $F$ be $f$ on $R^{\prime }$ and $0$ on the rest of $R$. By the definition of the integral over a general region, ${\iint }_{R^{\prime }}fdA={\iint }_{R}FdA$, and $F$ is integrable because its only discontinuities lie on the four edges of $R^{\prime }$. Because $f\ge 0$ on $R$, we have $F\le f$ at every point of $R$: on $R^{\prime }$ the two are equal, and off $R^{\prime }$ it is $0\le f$. Monotonicity therefore gives

$${\iint }_{R}fdA\ge {\iint }_{R}FdA={\iint }_{R^{\prime }}fdA\ge \frac{1}{2}f(P)A(R^{\prime })>0,$$

the second inequality being the lower bound property on $R^{\prime }$ with $m=\frac{1}{2}f(P)$. $■$

Deduction. Suppose $f\ge 0$ is continuous with ${\iint }_{R}fdA=0$. If $f$ were positive at any interior point, the integral would be positive — a contradiction — so $f=0$ at every interior point. A boundary point of $R$ is a limit of interior points and $f$ is continuous, so $f=0$ there too. Hence $f\equiv 0$ on $R$. Note that (a) alone would not give this: it only produces one point where $f$ equals its average $0$.

Throughout, $D$ is $0\le r\le 2$, $0\le \theta \le \frac{\pi }{2}$ in polar coordinates, the density is $\rho =r$, and $dA=rdrd\theta$ — so every integrand carries one factor of $r$ from the density and one from the area element.

(a) Mass. By definition $m={\iint }_D\rho dA$:

$$m={\int }_0^{\pi /2}{\int }_0^{2}r\cdot rdrd\theta =\frac{\pi }{2}{\left[\frac{r^3}{3}\right]}_0^2=\frac{\pi }{2}\cdot \frac{8}{3}=\frac{4\pi }{3}.$$

(b) Centre of mass. By definition $M_y={\iint }_{D}x\rho dA$ and $\bar{x}=M_y/m$. With $x=r\cos \theta$ the integrand becomes $r^3\cos \theta$, a product over a rectangle in the $r\theta$-plane:

$$M_y={\int }_0^{\pi /2}{\int }_0^{2}r\cos \theta \cdot r\cdot rdrd\theta ={\int }_0^2{r}^{3}dr{\int }_0^{\pi /2}\cos \theta d\theta =4\cdot 1=4,\bar{x}=\frac{4}{4\pi /3}=\frac{3}{\pi }.$$

The symmetry is the reflection $(x,y)\mapsto (y,x)$ in the line $y=x$: it maps $D$ onto itself and leaves $\rho =\sqrt{x^2+y^2}$ unchanged, while exchanging the integrands $x\rho$ and $y\rho$ of the two moments. Hence $M_x=M_y=4$ and $\bar{y}=\frac{3}{\pi }$ as well:

$$(\bar{x},\bar{y})=\left(\frac{3}{\pi },\frac{3}{\pi }\right)\approx (0.955,0.955).$$

Comparison: $\frac{3}{\pi }\approx 0.955>\frac{8}{3\pi }\approx 0.849$. The density grows with distance from the origin, so mass is shifted toward the arc, and the balance point moves outward along the axis of symmetry $y=x$ — the right direction. The point is inside $D$, since $2{\left(\frac{3}{\pi }\right)}^2=\frac{18}{{\pi }^2}\approx 1.82<4$.

(c) Moments of inertia. By definition $I_0={\iint }_D(x^2+y^2)\rho dA$, and $x^2+y^2=r^2$:

$$I_0={\int }_0^{\pi /2}{\int }_0^2{r}^2\cdot r\cdot rdrd\theta =\frac{\pi }{2}{\left[\frac{r^5}{5}\right]}_0^2=\frac{\pi }{2}\cdot \frac{32}{5}=\frac{16\pi }{5}.$$

The same reflection exchanges $I_x={\iint }_D{y}^2\rho dA$ and $I_y={\iint }_D{x}^2\rho dA$, so $I_x=I_y$; since $I_x+I_y=I_0$, each equals $\frac{8\pi }{5}$.

In terms of the mass, $I_0=\frac{16\pi /5}{4\pi /3}m=\frac{12}{5}m=2.4m$. A uniform disk of radius $a=2$ has $I_0=\frac{1}{2}m{a}^2=2m$, and so does a uniform quarter disk — the ratio $I_0/m$ does not see the angular extent. This lamina's ratio is larger for the same reason its centre of mass sits further out: more of its mass is at large $r$, and $I_0$ weights mass by $r^2$.

(a) A square base. $z_x=x^{1/2}$ and $z_y=y^{1/2}$, so the stretch factor is

$$\sqrt{1+z_x^2+z_y^2}=\sqrt{1+x+y}.$$

Over the unit square,

$$A={\int }_0^1{\int }_0^1\sqrt{1+x+y}dydx={\int }_0^1{\left[\frac{2}{3}(1+x+y{)}^{3/2}\right]}_{y=0}^{y=1}dx=\frac{2}{3}{\int }_0^1\left[(2+x{)}^{3/2}-(1+x{)}^{3/2}\right]dx.$$

Each term integrates by the power rule, the exponent rising to $\frac{5}{2}$:

$$A=\frac{2}{3}\cdot \frac{2}{5}{\left[(2+x{)}^{5/2}-(1+x{)}^{5/2}\right]}_0^1=\frac{4}{15}\left[\left(3^{5/2}-2^{5/2}\right)-\left(2^{5/2}-1\right)\right]=\frac{4}{15}\left(9\sqrt{3}-8\sqrt{2}+1\right)\approx 1.41,$$

using $3^{5/2}=9\sqrt{3}$ and $2^{5/2}=4\sqrt{2}$.

Sanity check: the stretch factor runs from $1$ at the origin to $\sqrt{3}$ at $(1,1)$ over a base of area $1$, so $1<A<\sqrt{3}\approx 1.73$.

(b) The saddle over a disk. $z_x=y$ and $z_y=x$, so the factor is $\sqrt{1+x^2+y^2}=\sqrt{1+r^2}$. Region and integrand both ask for polar coordinates:

$$A={\int }_0^{2\pi }{\int }_0^1\sqrt{1+r^2}rdrd\theta =2\pi {\left[\frac{1}{3}{\left(1+r^2\right)}^{3/2}\right]}_0^1=\frac{2\pi }{3}\left(2\sqrt{2}-1\right)\approx 3.83,$$

the $r$ from $dA$ being exactly what the substitution $u=1+r^2$, $du=2rdr$, requires.

Sanity check: the factor lies between $1$ and $\sqrt{2}$ over a disk of area $\pi$, so $\pi <A<\pi \sqrt{2}\approx 4.44$.

(c) Shifting and steepening. The surface-area integrand contains only the partial derivatives, never $z$ itself. Adding $5$ changes neither $z_x$ nor $z_y$, so $z=xy+5$ over the same disk has the same integrand over the same region and hence the same area — as it must, since a vertical translation is a rigid motion.

For $z=2xy$ the partials double: $z_x=2y$, $z_y=2x$, and the factor becomes $\sqrt{1+4r^2}$. Since $\sqrt{1+4r^2}\ge \sqrt{1+r^2}$ everywhere, with strict inequality except at the origin, monotonicity of the double integral makes the new area larger. Computing it, with $u=1+4r^2$, $du=8rdr$,

$$A^{\prime }={\int }_0^{2\pi }{\int }_0^1\sqrt{1+4r^2}rdrd\theta =2\pi \cdot \frac{1}{8}\cdot \frac{2}{3}{\left[{\left(1+4r^2\right)}^{3/2}\right]}_0^1=\frac{\pi }{6}\left(5\sqrt{5}-1\right)\approx \mathrm{5.33.}$$

Indeed $5.33>3.83$.

(a) Floor, ceiling, shadow. The plane $y+z=4$ is $z=4-y$, and the solid lies between it and $z=0$. Since $4-y\ge 0$ exactly when $y\le 4$, the floor is $z=0$, the ceiling is $z=4-y$, and they meet along the line $y=4$ of the $xy$-plane. The parabolic cylinder $y=x^2$ is vertical and supplies the other side of the shadow, $y\ge x^2$. So

$$D=\{(x,y):-2\le x\le 2,x^2\le y\le 4\},$$

the region between the parabola and the line $y=4$, whose ends are at $x=\pm 2$ (where $x^2=4$). The shadow is not a choice: the ceiling drops to height zero exactly along $y=4$, and a larger shadow would count negative heights.

$$V={\int }_{-2}^2{\int }_{x^2}^4{\int }_0^{4-y}dzdydx={\int }_{-2}^2{\int }_{x^2}^4(4-y)dydx.$$

Inner integral: ${\left[4y-\frac{y^2}{2}\right]}_{y=x^2}^{y=4}=(16-8)-\left(4x^2-\frac{x^4}{2}\right)=8-4x^2+\frac{x^4}{2}$. This is even in $x$, so

$$V=2{\int }_0^2\left(8-4x^2+\frac{x^4}{2}\right)dx=2{\left[8x-\frac{4x^3}{3}+\frac{x^5}{10}\right]}_0^2=2\left(16-\frac{32}{3}+\frac{16}{5}\right)=2\cdot \frac{240-160+48}{15}=\frac{256}{15}\approx \mathrm{17.1.}$$

Sanity check: the shadow has area ${\int }_{-2}^2(4-x^2)dx=\frac{32}{3}$ and the height $4-y$ lies between $0$ and $4$, so $0<V<\frac{128}{3}\approx 42.7$. The height is $4$ only along $y=0$ and falls to $0$ along $y=4$, so a value under half of that bound is right.

(b) The order $dxdzdy$. With $x$ innermost, project $E$ onto the $yz$-plane. The conditions on $(y,z)$ that do not involve $x$ are $0\le z\le 4-y$ and, because $y\ge x^2\ge 0$, $y\ge 0$; together they give the triangle $0\le y\le 4$, $0\le z\le 4-y$, with corners $(y,z)=(0,0)$, $(4,0)$ and $(0,4)$. For a fixed $(y,z)$ in that triangle the only remaining constraint is $x^2\le y$, so the pencil parallel to the $x$-axis enters at $x=-\sqrt{y}$ and leaves at $x=\sqrt{y}$. The parabola, which bounded the old shadow, now sits in the innermost limits, while the plane, the old ceiling, now bounds the projection:

$$V={\int }_0^4{\int }_0^{4-y}{\int }_{-\sqrt{y}}^{\sqrt{y}}dxdzdy={\int }_0^4{\int }_0^{4-y}2\sqrt{y}dzdy={\int }_0^{4}2\sqrt{y}(4-y)dy.$$

Then

$$V={\int }_0^4\left(8y^{1/2}-2y^{3/2}\right)dy={\left[\frac{16}{3}{y}^{3/2}-\frac{4}{5}{y}^{5/2}\right]}_0^4=\frac{16}{3}\cdot 8-\frac{4}{5}\cdot 32=\frac{128}{3}-\frac{128}{5}=\frac{256}{15},$$

in agreement with (a). Both descriptions pass the shape test: outer limits constant, middle limits depending only on the outer variable, inner limits on the outer two.

(a) Where the surfaces meet. In cylindrical coordinates the cone is $z=r$ and the paraboloid is $z=6-r^2$. They meet where $r=6-r^2$, that is $r^2+r-6=(r+3)(r-2)=0$; since $r\ge 0$ the only solution is $r=2$, at height $z=2$. The curve of intersection is the circle $x^2+y^2=4$ in the plane $z=2$, and the shadow of $E$ is the disk $r\le 2$.

Inside that disk the paraboloid is on top: at $r=0$ it is at height $6$ while the cone is at $0$, and two continuous heights that agree only at $r=2$ cannot change order before then. So the floor is the cone and the ceiling the paraboloid:

$$E=\{(r,\theta ,z):0\le \theta \le 2\pi ,0\le r\le 2,r\le z\le 6-r^2\}.$$

(b) Volume. With $dV=rdzdrd\theta$,

$$V={\int }_0^{2\pi }{\int }_0^2{\int }_r^{6-r^2}rdzdrd\theta =2\pi {\int }_0^2\left(6-r^2-r\right)rdr=2\pi {\int }_0^2\left(6r-r^3-r^2\right)dr,$$

so

$$V=2\pi {\left[3r^2-\frac{r^4}{4}-\frac{r^3}{3}\right]}_0^2=2\pi \left(12-4-\frac{8}{3}\right)=2\pi \cdot \frac{16}{3}=\frac{32\pi }{3}\approx \mathrm{33.5.}$$

Sanity check by cutting along the plane $z=2$. Below it, $E$ is the solid cone $r\le z\le 2$ with base radius $2$ and height $2$, volume $\frac{1}{3}\pi \cdot 4\cdot 2=\frac{8\pi }{3}$. Above it, $E$ is the paraboloid cap $2\le z\le 6-r^2$, which fills half of its circumscribed cylinder of radius $2$ and height $4$, volume $8\pi$. Total $\frac{8\pi }{3}+8\pi =\frac{32\pi }{3}$.

(c) Height of the centroid. With constant density, $\bar{z}=\frac{1}{V}{\iiint }_{E}zdV$. The inner integral now has integrand $z$:

$${\iiint }_{E}zdV={\int }_0^{2\pi }{\int }_0^2{\int }_r^{6-r^2}zrdzdrd\theta =2\pi {\int }_0^2\frac{(6-r^2{)}^2-r^2}{2}rdr=\pi {\int }_0^2\left(36-13r^2+r^4\right)rdr,$$

since $(6-r^2{)}^2-r^2=36-12r^2+r^4-r^2$. Hence

$${\iiint }_{E}zdV=\pi {\left[18r^2-\frac{13r^4}{4}+\frac{r^6}{6}\right]}_0^2=\pi \left(72-52+\frac{32}{3}\right)=\frac{92\pi }{3},\bar{z}=\frac{92\pi /3}{32\pi /3}=\frac{23}{8}=\mathrm{2.875.}$$

Sanity check: $E$ spans $0\le z\le 6$, and the cap above $z=2$ carries three quarters of the volume, so $\bar{z}$ should be comfortably above $2$ and well below $6$. The cut from (b) pins it down: the cone (volume $\frac{8\pi }{3}$) has its centroid three quarters of the way up from its apex, at $z=\frac{3}{2}$, and the paraboloid cap (volume $8\pi$) has its centroid two thirds of the way from its vertex at $z=6$ to its base, at $z=6-\frac{8}{3}=\frac{10}{3}$ — both are short computations of the same kind as this one. Combining,

$$\bar{z}=\frac{\frac{8\pi }{3}\cdot \frac{3}{2}+8\pi \cdot \frac{10}{3}}{\frac{32\pi }{3}}=\frac{4\pi +\frac{80\pi }{3}}{\frac{32\pi }{3}}=\frac{92}{32}=\frac{23}{8}.$$

(a) Reading the limits. The outer two limits, $-1\le x\le 1$ and $-\sqrt{1-x^2}\le y\le \sqrt{1-x^2}$, describe the unit disk $x^2+y^2\le 1$: the shadow of $E$. The inner limits run from $z=\sqrt{x^2+y^2}$, the upper half of the cone $z^2=x^2+y^2$, up to $z=\sqrt{2-x^2-y^2}$, the upper hemisphere of the sphere $x^2+y^2+z^2=2$. Cone and sphere meet where $z^2=x^2+y^2$ and $x^2+y^2+z^2=2$ together, so $2(x^2+y^2)=2$: the circle $x^2+y^2=1$ at height $z=1$, whose projection is exactly the unit disk. The shadow is consistent with the surfaces, and $E$ is the solid inside the sphere of radius $\sqrt{2}$ and above the cone.

In spherical coordinates the sphere is $\rho =\sqrt{2}$, and on the cone $z=r$ gives $\tan \varphi =\frac{r}{z}=1$, so the cone is $\varphi =\frac{\pi }{4}$; above the cone means smaller $\varphi$. Hence

$$E=\left\{0\le \rho \le \sqrt{2},0\le \theta \le 2\pi ,0\le \varphi \le \frac{\pi }{4}\right\},$$

a spherical wedge with constant limits.

(b) The integral. With $z=\rho \cos \varphi$ and $dV={\rho }^2\sin \varphi d\rho d\theta d\varphi$, the integrand is a product and all limits are constant, so

$$J={\int }_0^{\pi /4}{\int }_0^{2\pi }{\int }_0^{\sqrt{2}}\rho \cos \varphi \cdot {\rho }^2\sin \varphi d\rho d\theta d\varphi ={\int }_0^{\sqrt{2}}{\rho }^{3}d\rho {\int }_0^{2\pi }d\theta {\int }_0^{\pi /4}\sin \varphi \cos \varphi d\varphi .$$

The three factors are ${\left[\frac{{\rho }^4}{4}\right]}_0^{\sqrt{2}}=\frac{4}{4}=1$, then $2\pi$, then ${\left[\frac{{\sin }^2\varphi }{2}\right]}_0^{\pi /4}=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$. So

$$J=1\cdot 2\pi \cdot \frac{1}{4}=\frac{\pi }{2}.$$

Cross-check in cylindrical coordinates, where $E$ is $0\le r\le 1$, $r\le z\le \sqrt{2-r^2}$:

$$J=2\pi {\int }_0^1\frac{(2-r^2)-r^2}{2}rdr=\pi {\int }_0^1\left(2r-2r^3\right)dr=\pi \left(1-\frac{1}{2}\right)=\frac{\pi }{2}.$$

(c) Volume and centroid height. Integrate $1$ over the same wedge:

$$V={\int }_0^{\sqrt{2}}{\rho }^{2}d\rho {\int }_0^{2\pi }d\theta {\int }_0^{\pi /4}\sin \varphi d\varphi =\frac{2\sqrt{2}}{3}\cdot 2\pi \cdot \left(1-\frac{\sqrt{2}}{2}\right)=\frac{4\sqrt{2}\pi }{3}-\frac{4\pi }{3}=\frac{4\pi }{3}\left(\sqrt{2}-1\right)\approx 1.73,$$

using $(\sqrt{2}{)}^3=2\sqrt{2}$ and ${\left[-\cos \varphi \right]}_0^{\pi /4}=1-\frac{\sqrt{2}}{2}$. Then

$$\bar{z}=\frac{J}{V}=\frac{\pi /2}{\frac{4\pi }{3}(\sqrt{2}-1)}=\frac{3}{8(\sqrt{2}-1)}=\frac{3(\sqrt{2}+1)}{8}\approx 0.905,$$

rationalising with $(\sqrt{2}-1)(\sqrt{2}+1)=1$.

Plausibility: $E$ is the solid cone of height $1$ (apex at the origin, base the unit circle at $z=1$) with a spherical cap on top reaching $z=\sqrt{2}\approx 1.41$. The cone alone has volume $\frac{\pi }{3}\approx 1.05$ and centroid height $\frac{3}{4}$; the cap adds volume $\approx 0.69$ lying entirely above $z=1$. So $\bar{z}$ must lie between $\frac{3}{4}$ and $\sqrt{2}$, and somewhat under $1$ since the cone is the larger piece — which is exactly where $0.905$ sits.