§15.6 삼중적분Triple Integrals

Solution. The integrand $xyz^2$ is a product of one-variable factors $x$, $y$, $z^2$, and the domain $B$ is a product of intervals along the three axes, so integrating inward in any order permitted by Theorem 4 lets each factor slide out past the integrals that don't involve it. The triple integral therefore splits into a product of three single integrals.

Evaluating the three factors,

so

Hence $\displaystyle\iiint_B xyz^2\,dV=\dfrac{27}{4}$.

Solution. The first-octant condition gives $x,y,z\ge0$, and lying below $z=12xy$ makes that surface the upper limit for $z$. The remaining conditions $y=x$ and $x=1$ cut out a shadow $D$ with vertices $(0,0)$, $(1,0)$, $(1,1)$, trapped between the $x$-axis (the line $y=0$, where the first-octant condition $y\ge0$ binds) and the line $y=x$. Taking $y$ as the outer variable gives the Type II description $D=\{(x,y)\mid 0\le y\le1,\ y\le x\le1\}$, so $E$ is the type 1 solid

In Equation (6) the inner integral treats $x,y$ as fixed and runs over $z$ alone, so

Evaluating the inner $x$-integral,

and the remaining $y$-integral is

Hence $\displaystyle\iiint_E z\,dV=4$.

Solution. The paraboloid $y=x^2+z^2$ depends only on $x,z$ and is rotationally symmetric about the $y$-axis, and the integrand $\sqrt{x^2+z^2}$ is likewise symmetric in $x,z$ alone. This makes it natural to leave $x,z$ in the projection and let $y$ be the variable caught between the two surfaces — a type 3 description. The two boundaries $y=x^2+z^2$ and $y=4$ meet where $x^2+z^2=4$, so the projection onto the $xz$-plane is the disk $D_3=\{(x,z)\mid x^2+z^2\le4\}$, and

In (11) the inner integral holds $x,z$ fixed and runs over $y$ alone; since the integrand does not involve $y$,

Since $D_3$ is a disk and the integrand is a function of the distance to the origin alone, switch to polar coordinates in the $xz$-plane: $x=r\cos\theta$, $z=r\sin\theta$. Then $\sqrt{x^2+z^2}=r$, $dA=r\,dr\,d\theta$, and $D_3=\{0\le r\le2,\ 0\le\theta\le2\pi\}$, so

Evaluating the inner $r$-integral,

and integrating over $\theta$ multiplies this constant by $2\pi$, so

Hence $\displaystyle\iiint_E \sqrt{x^2+z^2}\,dV=\dfrac{128\pi}{15}$. (Setting $E$ up as a type 1 solid instead would leave the inner $z$-integral with irrational limits $\pm\sqrt{y-x^2}$, which makes the computation all but unworkable — the type 3 description paired with polar coordinates is what exploits the symmetry.)

Solution. Reading the limits of the given iterated integral from the inside out: with $x,y$ fixed, $z$ runs from $0$ to $y$; with $x$ fixed, $y$ runs from $0$ to $x^2$; and finally $x$ runs from $0$ to $1$. Hence

a type 1 solid in which $z$ is trapped between the planes $z=0$ and $z=y$ over the $xy$-projection $D_1=\{(x,y)\mid 0\le x\le1,\ 0\le y\le x^2\}$. Rewriting the order requires projecting $E$ onto the other two coordinate planes as well; together with the Type II description $\{0\le y\le1,\ \sqrt y\le x\le1\}$ of $D_1$, this uses $D_2$ (the $yz$-projection) and $D_3$ (the $xz$-projection) shown in the figure above.

$D_2$ is read off directly from the condition $0\le z\le y\le1$ that defines $E$: $D_2=\{(y,z)\mid 0\le y\le1,\ 0\le z\le y\}$ (figure above). $D_3$ comes from eliminating $y$: since $0\le z\le y\le x^2$ requires only $z\le x^2$, $D_3=\{(x,z)\mid 0\le x\le1,\ 0\le z\le x^2\}$ (figure above).

(a) The order $dx\,dz\,dy$ has $y$ outermost and $z$ in the middle, so write $D_2$ with $y$ outer: $D_2=\{0\le y\le1,\ 0\le z\le y\}$. The innermost limits for $x$ come from the condition $y\le x^2$ (that is, $x\ge\sqrt y$) together with $x\le1$, neither of which involves $z$. So $E$ is also the type 2 solid $\{(y,z)\in D_2,\ \sqrt y\le x\le1\}$, and

(b) The order $dy\,dx\,dz$ has $z$ outermost and $x$ in the middle, so write $D_3$ with $z$ outer: $D_3=\{0\le z\le1,\ \sqrt z\le x\le1\}$. The innermost limits for $y$ come directly from the original condition $z\le y\le x^2$ on $E$. So $E$ is also the type 3 solid $\{(x,z)\in D_3,\ z\le y\le x^2\}$, and

Solution. Solving $x+2y+z=2$ for $z$ gives $z=2-x-2y$, which is the top of the tetrahedron; the bottom is $z=0$. The shadow $D$ is the triangle enclosed in the $xy$-plane by the line $x+2y=2$, along which the top and the bottom meet, together with the two remaining faces $x=0$ and $x=2y$. Now $x=2y$ gives $y=x/2$ and $x+2y=2$ gives $y=1-x/2$; the two lines meet where $x/2=1-x/2$, i.e. at $x=1$ (where $y=1/2$). So

(with vertices $(0,0)$, $(0,1)$, $(1,\tfrac12)$), and $T$ is the type 1 solid $\{(x,y,z)\mid (x,y)\in D,\ 0\le z\le2-x-2y\}$. By Equation (12),

For the inner integral, write $b=1-\tfrac x2$, $a=\tfrac x2$, so $b-a=1-x$ and $a+b=1$, and

The remaining $x$-integral is

Hence $V(T)=\dfrac13$.

Solution. Since $z$ lies between $z=0$ and $z=x$, $E$ is the type 1 solid $\{(x,y,z)\mid (x,y)\in D,\ 0\le z\le x\}$ over its $xy$-projection $D$, where $D$ is what the trace $x=y^2$ of the parabolic cylinder and the plane $x=1$ cut out of the $xy$-plane: $D=\{(x,y)\mid -1\le y\le1,\ y^2\le x\le1\}$.

Computing $m$, $M_{yz}$ and $M_{xy}$ each begins with an inner integral in $z$ that leaves $\rho x$, $\rho x^2$, or $\tfrac12\rho x^2$ respectively, followed by integrating a power of $x$ over $D$. Exploiting this common pattern, record once and for all, for $n=1,2$,

(the last step follows because the bracket equals $1-\tfrac1{2n+3}$ at $y=1$ and $-1+\tfrac1{2n+3}$ at $y=-1$, whose difference is $2-\tfrac2{2n+3}=\tfrac{4n+4}{2n+3}$, times $\tfrac1{n+1}$). The mass now uses the inner $z$-integral $\int_0^x\rho\,dz=\rho x$:

The moment about the $yz$-plane uses $\int_0^x x\rho\,dz=x^2\rho$:

The moment about the $xz$-plane has inner integral $\int_0^x y\rho\,dz=xy\rho$, requiring $\iint_D xy\,dA$; but $D$ is symmetric under $y\mapsto-y$ (its boundary involves only $y^2$), while $xy$ changes sign under that symmetry, so the integral cancels:

The moment about the $xy$-plane uses $\int_0^x z\rho\,dz=\tfrac12x^2\rho$:

By Equation (15), the coordinates of the center of mass are

Hence the center of mass is $\left(\dfrac57,\,0,\,\dfrac5{14}\right)$.