Triple Integrals in Cylindrical Coordinates

Solids with a vertical axis of symmetry (cans, pipes, bowls, cones) are described simply in cylindrical coordinates: polar coordinates for the floor plus the height $z$. Combining the polar area element with a height step gives the volume element.

Key idea

$$\iiint_E f\,dV = \int_\alpha^\beta\int_{r_1(\theta)}^{r_2(\theta)}\int_{u_1(r, \theta)}^{u_2(r, \theta)} f(r\cos\theta, r\sin\theta, z)\,r\,dz\,dr\,d\theta,$$ so $dV = r\,dz\,dr\,d\theta$.

Why it works

A small cylindrical box has a polar rectangle as its base, with area about $r\,\Delta r\,\Delta\theta$, and height $\Delta z$. Its volume is base times height: $r\,\Delta z\,\Delta r\,\Delta\theta$. As in the plane, boxes far from the axis are bigger, and the factor $r$ keeps count. In practice: describe the top and bottom surfaces as functions of $r$ (and $\theta$), the shadow in polar coordinates, and integrate $z$ first.

Cylindrical coordinates: $r$ is the distance from the $z$-axis, $\theta$ the angle around it, $z$ the height. A small box with these sides has volume about $r\,\Delta r\,\Delta\theta\,\Delta z$.

To set up the limits, describe the solid in three stages. First, its shadow on the $xy$-plane, in polar form: this gives the limits for $\theta$ and $r$. Second, for a point $(r, \theta)$ in the shadow, the bottom and top surfaces above it: these give the limits for $z$, written in terms of $r$ and $\theta$. Third, multiply the integrand by $r$. For example, the solid inside $x^2 + y^2 = 1$ between $z = 0$ and the plane $z = 2 + x$ has shadow ${}0 \le r \le 1$, ${}0 \le \theta \le 2\pi$, and ${}0 \le z \le 2 + r\cos\theta$. Its volume is $\int_0^{2\pi} \int_0^1 (2 + r\cos\theta)\,r\,dr\,d\theta = 2\pi$.

Worked examples

Worked example: Under a paraboloid

Find the volume under $z = 4 - x^2 - y^2$ and above the $xy$-plane.

In cylindrical form the top is $z = 4 - r^2$, which meets the floor at $r = 2$.
$$V = \int_0^{2\pi}\int_0^2\int_0^{4 - r^2} r\,dz\,dr\,d\theta = \int_0^{2\pi}\int_0^2 (4r - r^3)\,dr\,d\theta.$$
$= 2\pi\left[2r^2 - \frac{r^4}{4}\right]_0^2 = 2\pi(8 - 4) = 8\pi$.

Answer: ${}8\pi$

Worked example: A weighted cylinder

Evaluate $\iiint_E (x^2 + y^2)\,dV$ over the cylinder $x^2 + y^2 \le 4$, ${}0 \le z \le 3$.

$x^2 + y^2 = r^2$, so the integrand times $r$ is $r^3$.
$$\int_0^{2\pi}\int_0^2\int_0^3 r^3\,dz\,dr\,d\theta = 2\pi \cdot 4 \cdot 3 = 24\pi.$$

Answer: ${}24\pi$

Worked example: Between a cone and a sphere

Find the volume above the cone $z = \sqrt{x^2 + y^2}$ and below the sphere $x^2 + y^2 + z^2 = 2$.

Cylindrically: $r \le z \le \sqrt{2 - r^2}$. The surfaces meet where $r^2 = 2 - r^2$, at $r = 1$.
$$V = 2\pi\int_0^1 \left(r\sqrt{2 - r^2} - r^2\right) dr = 2\pi\left[-\frac{(2 - r^2)^{3/2}}{3} - \frac{r^3}{3}\right]_0^1.$$
$$= 2\pi\left(-\frac{1}{3} - \frac{1}{3} + \frac{2\sqrt{2}}{3}\right) = \frac{4\pi}{3}\left(\sqrt{2} - 1\right).$$

Answer: $\frac{4\pi}{3}\left(\sqrt{2} - 1\right)$

Common mistake

Dropping the $r$, or writing the top surface in $x$ and $y$. Every integrand in cylindrical coordinates carries the factor $r$ from $dV$. And the limits for $z$ must be written in $r$ (and $\theta$): $z = 4 - x^2 - y^2$ becomes $z = 4 - r^2$. Check with a cylinder of radius $R$ and height $H$: the integral of ${}1$ must give $\pi R^2H$.

Triple Integrals in Cylindrical Coordinates

Worked examples

Practice

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