Exercises — Triple integrals in Cartesian, cylindrical, spherical coordinates

WHAT the region is: a round tube (cylinder) of radius $3$ sitting between two flat lids. Round in the $xy$-plane, straight up the $z$-axis — this is the textbook case for cylindrical coordinates.

WHY cylindrical: $x^2+y^2=9$ becomes the single clean statement $r=3$. In Cartesian we'd fight $y=\pm \sqrt{9-x^2}$ (a square root), which is uglier for no reason.

Limits:

• $r$ enters at the axis $r=0$, exits the wall $r=3$.
• $\theta$ sweeps a full circle: $0\to 2\pi$.
• $z$ runs between the lids: $0\to 4$.

${\int }_0^{2\pi }{\int }_0^3{\int }_0^{4}f\underset{d{V}_{\text{cyl}}}{\underbrace{rdzdrd\theta }}.$

WHAT: a ball of radius $5$, round in all directions — the signature of spherical coordinates.

WHY: $x^2+y^2+z^2\le 25$ becomes simply $\rho \le 5$. Every boundary is one constant.

Limits: $\rho :0\to 5$, $\varphi :0\to \pi$ (top pole to bottom pole), $\theta :0\to 2\pi$ (all the way around). ${\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^{5}f{\rho }^2\sin \varphi d\rho d\varphi d\theta .$

WHAT the region looks like — see the figure below. For a fixed distance $r$ from the axis, the solid lives between the slanted cone surface $z=r$ (bottom) and the flat cap $z=2$ (top). The cone meets the cap where $r=z=2$, so $r$ ranges $0\to 2$.

Set up with $f=1$ (volume) and $d{V}_{\text{cyl}}=rdzdrd\theta$: $V={\int }_0^{2\pi }{\int }_0^2{\int }_r^{2}rdzdrd\theta .$

Inner ($z$): the integrand $r$ is constant in $z$, so ${\int }_r^{2}rdz=r(2-r).$ WHY: integrating a constant over a length just multiplies by the length $(2-r)$.

Middle ($r$): ${\int }_0^2(2r-r^2)dr=[r^2-\frac{r^3}{3}{]}_0^2=4-\frac{8}{3}=\frac{4}{3}.$

Outer ($\theta$): multiply by ${\int }_0^{2\pi }d\theta =2\pi$. $V=2\pi \cdot \frac{4}{3}=\boxed{\frac{8\pi }{3}}.$ Sanity check: a cone of base radius $R$ and height $H$ has volume $\frac{1}{3}\pi R^{2}H$. Here $R=2,H=2$ gives $\frac{1}{3}\pi \cdot 4\cdot 2=\frac{8\pi }{3}$. ✓

WHY cylindrical: the integrand $x^2+y^2$ is exactly $r^2$ — the rotational symmetry lives in both region and integrand.

Rewrite: $(x^2+y^2)dV=r^2\cdot rdzdrd\theta =r^{3}dzdrd\theta$. Don't forget the Jacobian $r$: it multiplies the $r^2$ to give $r^3$.

${\int }_0^{2\pi }{\int }_0^1{\int }_0^3{r}^{3}dzdrd\theta .$ All limits constant and integrand factors, so multiply three 1D integrals: $\underset{2\pi }{\underbrace{{\int }_0^{2\pi }d\theta }}\underset{1/4}{\underbrace{{\int }_0^1{r}^{3}dr}}\underset{3}{\underbrace{{\int }_0^{3}dz}}=2\pi \cdot \frac{1}{4}\cdot 3=\boxed{\frac{3\pi }{2}}.$

WHAT the region is — see the figure. A bowl (paraboloid) with a flat lid at $z=4$. For fixed $(r,\theta )$ a vertical arrow enters the bowl at $z=r^2$ and exits the lid at $z=4$.

WHERE the shadow ends: the bowl meets the lid when $r^2=4$, i.e. $r=2$. So the $xy$-shadow is the disk $r\le 2$.

WHY cylindrical: $z=x^2+y^2$ is $z=r^2$ (clean), and the shadow is a disk.

$V={\int }_0^{2\pi }{\int }_0^2{\int }_{r^2}^{4}rdzdrd\theta .$ Inner ($z$): ${\int }_{r^2}^{4}rdz=r(4-r^2)$. Middle ($r$): ${\int }_0^2(4r-r^3)dr=[2r^2-\frac{r^4}{4}{]}_0^2=8-4=4.$ Outer ($\theta$): $\times 2\pi$. $V=2\pi \cdot 4=\boxed{8\pi }.$

WHAT: an "ice-cream-cone shell" — the wedge of a ball trapped between two cones (measured by polar angle $\varphi$) and out to radius $2$.

WHY spherical: boundaries are $\rho =2$, $\varphi =\pi /6$, $\varphi =\pi /3$ — all constant coordinate surfaces. Perfect.

Rewrite the integrand: $z=\rho \cos \varphi$, and $dV={\rho }^2\sin \varphi d\rho d\varphi d\theta$, so $zdV=(\rho \cos \varphi )({\rho }^2\sin \varphi )d\rho d\varphi d\theta ={\rho }^3\sin \varphi \cos \varphi d\rho d\varphi d\theta .$ $I={\int }_0^{2\pi }{\int }_{\pi /6}^{\pi /3}{\int }_0^2{\rho }^3\sin \varphi \cos \varphi d\rho d\varphi d\theta .$ $\rho$-integral: ${\int }_0^2{\rho }^{3}d\rho =\frac{2^4}{4}=4$. $\varphi$-integral: use $\sin \varphi \cos \varphi =\frac{1}{2}\sin 2\varphi$: ${\int }_{\pi /6}^{\pi /3}\frac{1}{2}\sin 2\varphi d\varphi =\frac{1}{2}[-\frac{\cos 2\varphi }{2}{]}_{\pi /6}^{\pi /3}=-\frac{1}{4}(\cos \frac{2\pi }{3}-\cos \frac{\pi }{3}).$ Now $\cos \frac{2\pi }{3}=-\frac{1}{2}$ and $\cos \frac{\pi }{3}=\frac{1}{2}$, so the bracket is $-\frac{1}{2}-\frac{1}{2}=-1$, giving $-\frac{1}{4}\cdot (-1)=\frac{1}{4}$. $\theta$-integral: ${\int }_0^{2\pi }d\theta =2\pi$. $I=2\pi \cdot \frac{1}{4}\cdot 4=\boxed{2\pi }.$

WHY spherical: ball region ($\rho \le a$) plus the constraint $z\ge 0$ becomes $\varphi \in [0,\frac{\pi }{2}]$ (upper half). The integrand $z=\rho \cos \varphi$ is clean in spherical.

Denominator (volume) is Exercise 2.1: $V=\frac{8\pi }{3}$.

Numerator ${\iiint }_{E}zdV={\int }_0^{2\pi }{\int }_0^2{\int }_r^{2}zrdzdrd\theta .$ Inner ($z$): ${\int }_r^{2}zrdz=r\cdot [\frac{z^2}{2}{]}_r^2=r\cdot \frac{4-r^2}{2}=2r-\frac{r^3}{2}.$ Middle ($r$): ${\int }_0^2(2r-\frac{r^3}{2})dr=[r^2-\frac{r^4}{8}{]}_0^2=4-2=2.$ Outer ($\theta$): $\times 2\pi$ $\Rightarrow \iiint zdV=4\pi$.

$\bar{z}=\frac{4\pi }{8\pi /3}=4\pi \cdot \frac{3}{8\pi }=\boxed{\frac{3}{2}}.$ Sanity check: the cone spans $z\in [0,2]$ but is fatter near the top (radius grows with $z$), so the balance point sits above the midpoint $1$. Indeed $\bar{z}=1.5>1$. ✓

WHY spherical is the only sane choice: $(x^2+y^2+z^2{)}^{3/2}=({\rho }^2{)}^{3/2}={\rho }^3$. That exponent ${\rho }^3$ is exactly what the ${\rho }^{2}d\rho$ from the Jacobian is built to handle.

$I={\int }_0^{2\pi }{\int }_0^{\pi }{\int }_0^1{e}^{{\rho }^3}{\rho }^2\sin \varphi d\rho d\varphi d\theta .$ Angular part factors out: ${\int }_0^{2\pi }d\theta \cdot {\int }_0^{\pi }\sin \varphi d\varphi =2\pi \cdot 2=4\pi .$ Radial part — the trick is the substitution $u={\rho }^3$, $du=3{\rho }^{2}d\rho$, so ${\rho }^{2}d\rho =\frac{1}{3}du$: ${\int }_0^1{e}^{{\rho }^3}{\rho }^{2}d\rho =\frac{1}{3}{\int }_0^1{e}^{u}du=\frac{1}{3}(e-1).$ WHY this substitution: the leftover ${\rho }^2$ from the Jacobian is precisely $\frac{1}{3}du$ — the integral was designed to be doable only because of the ${\rho }^2$ factor. Without spherical's Jacobian there is no clean $u$-substitution. $I=4\pi \cdot \frac{1}{3}(e-1)=\boxed{\frac{4\pi }{3}(e-1)}.$

WHAT the region is — an ice-cream cone: a spherical cap sitting on a $45^{\circ }$ cone. Both the cone and the sphere are coordinate surfaces in spherical, so use it.

Find $\varphi$ for the cone. The cone $z=\sqrt{x^2+y^2}$ means $z=r$, i.e. angle from the $+z$ axis is $45^{\circ }=\frac{\pi }{4}$. Check with spherical: $z=\rho \cos \varphi$ and $r=\rho \sin \varphi$, so $z=r\Rightarrow \cos \varphi =\sin \varphi \Rightarrow \varphi =\frac{\pi }{4}$. The solid is inside the cone (near the axis), so $\varphi :0\to \frac{\pi }{4}$.

Find $\rho$ for the sphere: $x^2+y^2+z^2=8\Rightarrow {\rho }^2=8\Rightarrow \rho =2\sqrt{2}$. So $\rho :0\to 2\sqrt{2}$.

$V={\int }_0^{2\pi }{\int }_0^{\pi /4}{\int }_0^{2\sqrt{2}}{\rho }^2\sin \varphi d\rho d\varphi d\theta .$

• ${\int }_0^{2\sqrt{2}}{\rho }^{2}d\rho =\frac{(2\sqrt{2}{)}^3}{3}=\frac{16\sqrt{2}}{3}$ (since $(2\sqrt{2}{)}^3=8\cdot 2\sqrt{2}=16\sqrt{2}$).
• ${\int }_0^{\pi /4}\sin \varphi d\varphi =[-\cos \varphi {]}_0^{\pi /4}=1-\frac{\sqrt{2}}{2}$.
• ${\int }_0^{2\pi }d\theta =2\pi$.

Which $dV$ for a sphere-shaped integrand $(x^2+y^2+z^2{)}^p$? ::: Spherical, because $(x^2+y^2+z^2{)}^p={\rho }^{2p}$ pairs with the ${\rho }^{2}d\rho$ Jacobian to give a clean radial integral. Where does a paraboloid's shadow radius come from? ::: Intersect the paraboloid with the capping plane and solve, e.g. $r^2=4\Rightarrow r=2$. Why does the cone's center of mass sit above its midpoint? ::: The solid is fatter (more volume) near the top, so the balance point shifts upward: $\bar{z}=\frac{3}{2}>1$.