10. 삼중적분Triple Integrals

Solution. $D$ is the region lying above the plane $z=3$ and inside the sphere $S:x^2+y^2+z^2=25$, i.e., the cap cut from the ball of radius $5$ by the plane $z=3$.

Projecting $D$ onto the $xy$-plane, at $z=3$ we have $x^2+y^2+9=25$, so $x^2+y^2=16$, the disk of radius $4$. For each point $(x,y)$ in this disk, $z$ runs from the plane $z=3$ up to the sphere $z=\sqrt{25-x^2-y^2}$ (the upper hemisphere).

Hence the triple integral as an iterated integral is

(As a check, substituting $F\equiv 1$ gives the volume of the cap, $\dfrac{52\pi}{3}$.)

Solution. Finding the planes containing the four faces of the tetrahedron with vertices $A(0,0,0)$, $B(1,1,0)$, $C(0,1,0)$, $D(0,1,1)$: face $ACD$ lies on $x=0$, face $ABC$ on $z=0$, face $BCD$ on $y=1$, and face $ABD$ on $x-y+z=0$ (since its normal is $\overrightarrow{AB}\times\overrightarrow{AD}=(1,1,0)\times(0,1,1)=(1,-1,1)$). Substituting the centroid $\left(\frac14,\frac34,\frac14\right)$ into each plane shows the interior of the tetrahedron is

i.e., $x\ge0$, $y\ge x$ (since $z\le y-x$ and $z\ge0$), $y\le1$, $0\le z\le y-x$.

(1) For fixed $x$, $y$ runs from $x$ to $1$ and $z$ runs from $0$ to $y-x$, so

(2) Since $y\ge x+z$ and $y\le1$, for fixed $x$, $z$ runs from $0$ to $1-x$ and $y$ runs from $x+z$ to $1$, so

(3) When $F=1$, both iterated integrals compute the volume of $D$. The scalar triple product $\overrightarrow{AB}\cdot(\overrightarrow{AC}\times\overrightarrow{AD}) = (1,1,0)\cdot(1,0,0)=1$ gives volume $\frac16|1|=\frac16$. Indeed, (1) gives $\int_0^1\int_x^1(y-x)\,dy\,dx=\int_0^1\frac{(1-x)^2}{2}dx=\frac16$, and (2) gives the same value $\frac16$ by an analogous computation, so the two results agree.

Solution.

(1) Innermost first: $\int_0^\pi \cos(u+v+w)\,du = \big[\sin(u+v+w)\big]_0^\pi = \sin(\pi+v+w)-\sin(v+w) = -2\sin(v+w)$. Next, $\int_0^\pi (-2\sin(v+w))\,dv = \big[2\cos(v+w)\big]_0^\pi = 2\cos(\pi+w)-2\cos w = -4\cos w$, and finally $\int_0^\pi(-4\cos w)\,dw=\big[-4\sin w\big]_0^\pi=0$. Hence the value is $0$.

(2) Inner $z$-integral: $\int_{x^2+3y^2}^{8-x^2-y^2}dz = 8-2x^2-4y^2$. Then for $x$ (with $0\le x\le 3y$),

Finally $\int_0^{\sqrt2}(24y-30y^3)\,dy=\big[12y^2-\tfrac{15}{2}y^4\big]_0^{\sqrt2}=12\cdot2-\tfrac{15}{2}\cdot4=24-30=-6$. (The printed bounds are used as given; the sign is simply the result of the formal computation.)

(3) Since $\cos(x^2)$ has no elementary antiderivative in $x$, we change the order of integration. The $z$-part separates from the $x,y$-part, so

$\int_0^4 z^{-1/2}dz=\big[2\sqrt z\big]_0^4=4$. The region $y:0\to1,\ x:2y\to2$ is the same as $x:0\to2,\ y:0\to x/2$, so

Hence the total value is $4\cdot\dfrac{\sin4}{2}=2\sin4$.

(4) Since the $z$- and $x$-parts do not separate cleanly from $y$, we swap the order of $x$ and $y$. The region $x:0\to1,\ y:x^2\to1$ becomes $y:0\to1,\ x:0\to\sqrt y$.

(using $\int_0^{\sqrt y}12xz\,dx = 6zy$). Substituting $u=zy^2$, $du=2zy\,dy$,

Finally, integrating over $z$: $\int_0^1 3(e^z-1)\,dz=3\big[e^z-z\big]_0^1=3\big((e-1)-1\big)=3e-6$.

(5) The $p$-integral treats $q$ as a constant: $\int_0^{\sqrt{4-q^2}}dp=\sqrt{4-q^2}$. Since $q/(r+1)$ separates in $r$ and $q$,

Here $\int_0^7\frac{dr}{r+1}=\big[\ln(r+1)\big]_0^7=\ln8=3\ln2$, and substituting $u=4-q^2$ gives $\int_0^2 q\sqrt{4-q^2}\,dq=\left[-\frac13(4-q^2)^{3/2}\right]_0^2=\frac{8}{3}$. Hence the value is $3\ln2\cdot\dfrac83=8\ln2$.

Solution. The inner $z$-integral gives $\int_a^{4-x^2-y}dz = (4-x^2-y)-a$. Setting $b=4-a$, the upper limit of $y$ is $4-a-x^2=b-x^2$, and the integrand is $(b-x^2)-y$, so

Hence the given expression equals

and setting this equal to $\dfrac{4}{15}$ gives

so $b=1$ or $b=-\dfrac13$. But for $D$ to be a genuine region, the upper limit $b-x^2$ of $y$ must be nonnegative for $x\in[0,1]$, forcing $b\ge1$. The value $b=1$ satisfies $1-x^2\ge0$ on $[0,1]$, while $b=-\frac13$ makes the upper limit negative for every $x$ and must be rejected. Hence $b=4-a=1$, i.e., $a=3$.

(Check: with $a=3$, $b=1$, $\frac12\int_0^1(1-x^2)^2dx=\frac12\left(1-\frac23+\frac15\right)=\frac12\cdot\frac{8}{15}=\frac{4}{15}$, as required.)

Solution.

(1) The two surfaces intersect where $x^2+3y^2=8-x^2-y^2$, i.e., $2x^2+4y^2=8$, or $x^2+2y^2=4$; this is the boundary of the elliptical region onto which $D$ projects in the $xy$-plane. The volume is

Substituting $x=2r\cos\theta$, $y=\sqrt2\,r\sin\theta$ ($0\le r\le1$), the Jacobian is $2\sqrt2\,r$, and $2x^2+4y^2=8r^2\cos^2\theta+8r^2\sin^2\theta=8r^2$, so the integrand simplifies to $8-8r^2=8(1-r^2)$. Hence

(2) $D$ is the common region of the cylinders $x^2+y^2\le1$ and $x^2+z^2\le1$, restricted to the first octant. Fixing $0\le x\le1$, $y$ runs from $0$ to $\sqrt{1-x^2}$, and (from the cylinder $x^2+z^2\le1$, independent of $y$) $z$ runs from $0$ to $\sqrt{1-x^2}$. Hence

Solution. Since $D=[0,2]\times[0,2]\times[0,2]$, its volume is $V=8$. As $F(x,y,z)=xyz$ separates in $x,y,z$,

Hence the average value is $\dfrac{1}{V}\displaystyle\iiint_D F\,dV=\dfrac{8}{8}=1$.

Solution. The surface $x^2+(y-1)^2=1$ is the cylinder of radius $1$ centered at $(0,1)$; in polar coordinates it is $x^2+y^2=2y$, i.e., $r^2=2r\sin\theta$, so $r=2\sin\theta$. This circle passes through the origin and lies in the upper half-plane $y\ge0$, so $\theta$ runs from $0$ to $\pi$.

$D$ is the region between $z=0$ and the paraboloid $z=x^2+y^2=r^2$, inside this cylinder. That is, over the disk $r\le2\sin\theta$, $z$ runs from $0$ to $r^2$ at each point. Hence

Solution. $D$ is the region over the disk $x^2+y^2\le4$ between $z=0$ and the paraboloid $z=x^2+y^2=r^2$, so in cylindrical coordinates $0\le\theta\le2\pi$, $0\le r\le2$, $0\le z\le r^2$.

(1), (2) $D$ is symmetric under $x\mapsto-x$ and under $y\mapsto-y$ respectively, while the integrands $x=r\cos\theta$ and $y=r\sin\theta$ are odd under these same reflections, so the integrals cancel to $0$.

(3)

Solution.

(1) The inner $z$-integral just multiplies by a factor depending on $x$: $\int_0^x(x^2+y^2)\,dz=x(x^2+y^2)$. The remaining $(x,y)$-region is the right half-disk $x^2+y^2\le1,\ x\ge0$, so with $x=r\cos\theta,\ y=r\sin\theta$ ($-\pi/2\le\theta\le\pi/2$, $0\le r\le1$), the integrand is $x(x^2+y^2)=r\cos\theta\cdot r^2=r^3\cos\theta$, and

(2) The inner $z$-integral has height $(x^2+y^2)-(-(x^2+y^2))=2(x^2+y^2)$, so the integrand becomes $42xy^2(x^2+y^2)$. The $(x,y)$-region is the same right half-disk as in (1), so with the same polar substitution, $42xy^2(x^2+y^2)=42\,r\cos\theta\cdot r^2\sin^2\theta\cdot r^2=42\,r^5\cos\theta\sin^2\theta$, and

Substituting $u=\sin\theta$, $\int_{-\pi/2}^{\pi/2}\cos\theta\sin^2\theta\,d\theta=\int_{-1}^1u^2du=\frac23$, and $\int_0^1r^6dr=\frac17$, so the value is $42\cdot\frac23\cdot\frac17=4$.

Note (anchor discrepancy). For (1), recomputing by two methods (direct Cartesian integration and the polar computation above) both give $\dfrac25$. For (2), the $x$-integration runs over $[0,1]$ (not symmetric), so there is no odd-function cancellation; both the Cartesian and polar computations agree on the value $4$.

Solution.

(1) Expanding $x^2+y^2+(z-1)^2=1$ gives $x^2+y^2+z^2-2z+1=1$, i.e., $x^2+y^2+z^2=2z$. Substituting $x^2+y^2+z^2=\rho^2$ and $z=\rho\cos\phi$ gives $\rho^2=2\rho\cos\phi$, so (for $\rho\ne0$) $\rho=2\cos\phi$.

(2) For $z=\sqrt{x^2+y^2}$, we have $z=\rho\cos\phi$ and $\sqrt{x^2+y^2}=\rho\sin\phi$ (since $\sin\phi\ge0$ for $0\le\phi\le\pi$), so $\rho\cos\phi=\rho\sin\phi$, i.e., (for $\rho\ne0$) $\tan\phi=1$. Since this is the upper nappe of the cone, $0\le\phi\le\pi/2$, so $\phi=\dfrac{\pi}{4}$.

Solution.

(1) $\rho=3$ is the set of points always at distance $3$ from the origin, so it is the sphere of radius $3$ centered at the origin.

(2) $\rho=\csc\phi$ is equivalent to $\rho\sin\phi=1$. Since $\rho\sin\phi=\sqrt{x^2+y^2}$ (the cylindrical radius $r$), this is $r=1$, i.e., the (infinite) cylinder $x^2+y^2=1$ of radius $1$ about the $z$-axis.

(3) $\rho=4\sec\phi$ is equivalent to $\rho\cos\phi=4$, i.e., the plane $z=4$. Hence $4\sec\phi\le\rho\le5$ describes the region above the plane $z=4$ (farther from the origin) and inside the sphere $\rho=5$ (of radius $5$) — that is, the cap cut from the ball of radius $5$ by the plane $z=4$.

Solution.

(1) $D$ is the region $0\le\rho\le1$, $0\le\phi\le\pi/3$ (an "ice cream cone" shape: the ball of radius $1$ restricted to the interior of the cone of half-angle $\pi/3$ about the $z$-axis), with $\theta$ ranging freely from $0$ to $2\pi$. Since the volume element in spherical coordinates is $\rho^2\sin\phi\,d\rho\,d\phi\,d\theta$,

(2) Since $x^2+y^2=\rho^2\sin^2\phi$,

Here $\displaystyle\int_0^1\rho^4d\rho=\frac15$, and $\displaystyle\int_0^{\pi/3}\sin^3\phi\,d\phi=\int_0^{\pi/3}(1-\cos^2\phi)\sin\phi\,d\phi=\left[-\cos\phi+\frac{\cos^3\phi}{3}\right]_0^{\pi/3}$. Since $\cos(\pi/3)=\frac12$, this equals $\left(-\frac12+\frac{1}{24}\right)-\left(-1+\frac13\right)=-\frac{11}{24}+\frac{16}{24}=\frac{5}{24}$. Hence

Solution. Note. The phrase "above $\rho=2\cos\phi$ and below $z=\sqrt{x^2+y^2}$" is ambiguous and admits two readings. Here (following the classical setup from Thomas' text) we take the sphere $\rho=2\cos\phi$ as the upper boundary of $D$ and the cone $z=\sqrt{x^2+y^2}$ ($\phi=\pi/4$) as its side/lower boundary. That is, $D$ is the "ice cream cone" region lying inside the cone ($0\le\phi\le\pi/4$) and inside the sphere ($0\le\rho\le2\cos\phi$) simultaneously.

The sphere $\rho=2\cos\phi$ is the sphere $x^2+y^2+(z-1)^2=1$ from Example 03(1), and the cone $z=\sqrt{x^2+y^2}$ is $\phi=\pi/4$ from Example 03(2). With $\theta$ from $0$ to $2\pi$, $\phi$ from $0$ to $\pi/4$, and, for each $(\theta,\phi)$, $\rho$ from $0$ to $2\cos\phi$,

Substituting $u=\cos\phi$, $du=-\sin\phi\,d\phi$; as $\phi:0\to\pi/4$, $u:1\to1/\sqrt2$, so

Hence

(For reference, reading the phrase the other way — "below the cone, inside the sphere," i.e. $\pi/4\le\phi\le\pi/2$, $0\le\rho\le2\cos\phi$ — gives $V=\pi/3$ by the same method; this equals the volume $4\pi/3$ of the whole ball $\rho\le2\cos\phi$ minus the ice-cream cone $\pi$ above.)

Solution. $D$ is the region $0\le\rho\le1$, $0\le\phi\le\pi/2$, $0\le\theta\le2\pi$, i.e., the upper half of the ball of radius $1$. First, the volume is

(This agrees with half the volume $\frac{4\pi}{3}$ of the unit ball.) Next, the integral of $f$ is

Since $\displaystyle\int_0^1\rho^3d\rho=\frac14$ and $\displaystyle\int_0^{\pi/2}\sin\phi\cos\phi\,d\phi=\left[\frac{\sin^2\phi}{2}\right]_0^{\pi/2}=\frac12$,

Hence the average value is

Solution.

(1) The $(x,y)$-region is the unit disk $x^2+y^2\le1$, and at each point $z$ runs from the cone $z=\sqrt{x^2+y^2}$ up to the plane $z=1$. So this integral is the volume between the plane $z=1$ and the cone (with vertex at the origin). Switching to polar coordinates $x=r\cos\theta,\ y=r\sin\theta$,

(2) The $(x,y)$-region is $x:0\to2$, $y:0\to\sqrt{2x-x^2}$; since $y^2\le2x-x^2$ is $(x-1)^2+y^2\le1$, this is the upper half ($y\ge0$) of the disk of radius $1$ centered at $(1,0)$. In polar coordinates this circle is $r=2\cos\theta$, and the upper half with $x\ge0,\ y\ge0$ corresponds to $0\le\theta\le\pi/2$. Here $z$ runs from the bottom to the top of the sphere $x^2+y^2+z^2=4$ (every point of this disk lies inside the sphere of radius $2$, since its farthest point $(2,0)$ lies exactly on the boundary $x^2+y^2=4$), so

Substituting $u=4-r^2$, $du=-2r\,dr$, in the inner integral,

(since $\sin\theta\ge0$ for $0\le\theta\le\pi/2$, $(4-4\cos^2\theta)^{3/2}=(4\sin^2\theta)^{3/2}=8\sin^3\theta$.) Hence

Since $\displaystyle\int_0^{\pi/2}\sin^3\theta\,d\theta=\frac23$,

Note (anchor discrepancy). The GROUND-TRUTH note estimated the value as "$16(\pi-2)/9$" (numerically about $2.03$), but the computation above — confirmed independently both by the polar integration and by a separate numerical double integration — gives $\dfrac{8(3\pi-4)}{9}=\dfrac{24\pi-32}{9}\approx4.822$, which is the value adopted here.

Solution. Solving the given transformation $u=\dfrac{2x-y}{2}=x-\dfrac{y}{2}$, $v=\dfrac{y}{2}$, $w=\dfrac{z}{3}$ for $x,y,z$ gives $y=2v$, $z=3w$, $x=u+v$, so the Jacobian is

so $dV=dx\,dy\,dz=6\,du\,dv\,dw$.

Transforming the bounds, $y/2\le x\le1+y/2$ becomes $0\le u\le1$, $0\le y\le4$ becomes $0\le v\le2$, and $0\le z\le3$ becomes $0\le w\le1$. The integrand becomes $\dfrac{2x-y}{2}+\dfrac{z}{3}=u+w$. Hence

The inner integral over $u,w$ is

so the total value is $6\cdot\displaystyle\int_0^2 1\,dv=6\cdot2=12$.

(Direct computation in Cartesian coordinates agrees. In the inner $x$-integral, setting $u=x-y/2$ gives $\displaystyle\int_{y/2}^{1+y/2}\left(x-\frac y2+\frac z3\right)dx=\int_0^1\left(u+\frac z3\right)du=\frac12+\frac z3$, independent of $y$. Then $\displaystyle\int_0^4\left(\frac12+\frac z3\right)dy=2+\frac{4z}{3}$, and finally $\displaystyle\int_0^3\left(2+\frac{4z}{3}\right)dz=6+\frac{4}{3}\cdot\frac{9}{2}=6+6=12$, in agreement.)

Note (anchor discrepancy). The GROUND-TRUTH note mistakenly computed the Jacobian as $3$ and estimated the value as $6$; as shown above, the correct Jacobian is $6$ (also confirmed by direct Cartesian computation), and the correct value is $12$.

Solution.

(1) With the transformation $x=au,\ y=bv,\ z=cw$, $D$ corresponds to the unit ball $u^2+v^2+w^2\le1$, and the Jacobian is

so the volume of $D$ equals the volume $\dfrac{4\pi}{3}$ of the unit ball times $abc$.

(2) With the same transformation,

Splitting the unit ball into eight octants, the integral of $|uvw|$ over each octant is equal by symmetry, so the total integral is $8$ times its value over the first octant ($u,v,w\ge0$). Using spherical coordinates $u=\rho\sin\phi\cos\theta,\ v=\rho\sin\phi\sin\theta,\ w=\rho\cos\phi$ ($0\le\rho\le1$, $0\le\phi\le\pi/2$, $0\le\theta\le\pi/2$), $uvw=\rho^3\sin^2\phi\cos\phi\sin\theta\cos\theta$, so

Hence over the whole unit ball this is $8\cdot\dfrac{1}{48}=\dfrac16$, and

Solution. Since $1\le x\le2$ on $D$, we have $x>0$, so $0\le xy\le2$ is equivalent to $0\le y\le\dfrac2x$. That is,

First integrating over $z$,

Next, integrating over $y$ (with $0\le y\le2/x$),

Finally, integrating over $x$,

Hence the value is $2+3\ln2$ (equivalently $2+\ln8$).

Note (anchor discrepancy). The GROUND-TRUTH note estimated the value as "$2+\dfrac{9}{2}\ln2$" (numerically about $5.12$), but the computation above (direct calculation, confirmed by sympy) gives the correct value $2+3\ln2=2+\ln8\approx4.0794$.