Triple Integrals in Spherical Coordinates

Balls, shells and cones with their tip at the origin are simplest in spherical coordinates: a ball of radius $a$ is just ${}0 \le \rho \le a$. The volume element needs one careful picture.

Key idea

$$dV = \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta, \qquad 0 \le \varphi \le \pi,\ 0 \le \theta \le 2\pi,$$ with $x = \rho\sin\varphi\cos\theta$, $y = \rho\sin\varphi\sin\theta$, $z = \rho\cos\varphi$ and $x^2 + y^2 + z^2 = \rho^2$.

Why it works

A small spherical box has three nearly perpendicular edges. Changing $\rho$ by $\Delta\rho$ moves you $\Delta\rho$ straight out. Changing $\varphi$ by $\Delta\varphi$ moves you along a meridian, a circle of radius $\rho$: distance $\rho\,\Delta\varphi$. Changing $\theta$ by $\Delta\theta$ moves you along a circle of latitude, whose radius is the distance from the $z$-axis, $\rho\sin\varphi$: distance $\rho\sin\varphi\,\Delta\theta$. Multiply the three edges: $\Delta V \approx \rho^2\sin\varphi\,\Delta\rho\,\Delta\varphi\,\Delta\theta$. Near the poles ($\sin\varphi \approx 0$) the boxes are thin slivers, which the factor $\sin\varphi$ records.

The three directions of a spherical box: out along $\rho$, down along the purple $\varphi$ arc (radius $\rho$), and around along the circle of radius $r = \rho\sin\varphi$. Change $\varphi$ toward ${}0$ and that last circle shrinks.

Spherical limits are read off the same way. The angle $\theta$ goes around the $z$-axis; $\varphi$ is measured down from the positive $z$-axis, so the upper half of space is ${}0 \le \varphi \le \frac{\pi}{2}$ and the cone $z = \sqrt{x^2 + y^2}$ is $\varphi = \frac{\pi}{4}$; $\rho$ runs out from the origin to the outer surface. A sphere through the origin, such as $x^2 + y^2 + z^2 = 2z$, becomes $\rho^2 = 2\rho\cos\varphi$, that is, $\rho = 2\cos\varphi$ with ${}0 \le \varphi \le \frac{\pi}{2}$. Its volume is ${}2\pi \int_0^{\pi/2} \frac{8}{3}\cos^3\varphi\sin\varphi\,d\varphi = \frac{4\pi}{3}$, the volume of a ball of radius ${}1$.

Worked examples

Worked example: Volume of a ball

Find the volume of the ball $\rho \le a$.

$\int_0^{2\pi}\int_0^{\pi}\int_0^{a} \rho^2\sin\varphi\,d\rho\,d\varphi\,d\theta$: the three integrals separate.
$\int_0^a \rho^2\,d\rho = \frac{a^3}{3}$, $\int_0^{\pi}\sin\varphi\,d\varphi = 2$, $\int_0^{2\pi}d\theta = 2\pi$.
$V = \frac{a^3}{3} \cdot 2 \cdot 2\pi = \frac{4}{3}\pi a^3$.

Answer: $\frac{4}{3}\pi a^3$

Worked example: An ice cream cone

Find the volume inside the sphere $\rho = 2$ and the cone $\varphi \le \frac{\pi}{3}$.

Only the upper limit of $\varphi$ changes: $\int_0^{\pi/3}\sin\varphi\,d\varphi = 1 - \cos\frac{\pi}{3} = \frac{1}{2}$.
$V = \frac{8}{3} \cdot \frac{1}{2} \cdot 2\pi = \frac{8\pi}{3}$.

Answer: $\frac{8\pi}{3}$

Worked example: A moment

Evaluate $\iiint_H z\,dV$ over the upper half-ball $\rho \le a$, $z \ge 0$.

$z = \rho\cos\varphi$, and the upper half is ${}0 \le \varphi \le \frac{\pi}{2}$.
$$\int_0^{2\pi}\int_0^{\pi/2}\int_0^a \rho^3\cos\varphi\sin\varphi\,d\rho\,d\varphi\,d\theta = \frac{a^4}{4} \cdot \frac{1}{2} \cdot 2\pi = \frac{\pi a^4}{4}.$$

Answer: $\frac{\pi a^4}{4}$

Common mistake

Forgetting $\rho^2\sin\varphi$, or letting $\varphi$ run to ${}2\pi$. Without the factor, the "volume" of a ball comes out as ${}2\pi^2a$, not even the right units. And $\varphi$ is measured from the north pole to the south pole, so it runs from ${}0$ to $\pi$; going to ${}2\pi$ covers the ball twice and makes $\int\sin\varphi\,d\varphi = 0$. Check every setup against $\frac{4}{3}\pi a^3$.

Triple Integrals in Spherical Coordinates

Worked examples

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