$$\def\ans#1{\bbox[border:1px solid green,6pt]{#1}}$$

# Triple Integrals in Cylindrical Coordinates

## Introduction

Now we finally get to use what we covered about cylindrical coordinates.

Recall that

\begin{align*} x &= r \cos \theta\\ y &= r \sin \theta\\ z &= z\\ \\ r &= \sqrt{x^2+y^2}\\ \theta &= \tan^{-1} \dfrac{y}{x}\\ z &= z \end{align*}

Therefore, a triple integral in rectangular coordinates can be rewritten (similar to a double integral in polar coordinates):

$\iiint_D f(x,y,z)\ dV = \iiint_D f(r \cos \theta, r \sin \theta, z)\ dz\ r\ dr\ d\theta$

Note carefully the $$r$$ in the midst of the differentials.

As with double integrals and polar coordinates, we'll tend to use cylindrical coordinates when we encounter a triple integral with $$x^2+y^2$$ somewhere.

## Examples

Convert the following integral to cylindrical coordinates and evaluate.

$\int_0^{2\sqrt{2}} \int_{-\sqrt{8-x^2}}^{\sqrt{8-x^2}} \int_{-1}^2 \sqrt{1+x^2+y^2}\ dz\ dy\ dx$

#### Solution

Note that the limits on $$z$$ will remain the same, $$r$$ will go from 0 to $$\sqrt{8} = 2\sqrt{2}$$, and $$\theta$$ will go from $$-\pi/2$$ to $$\pi/2$$.

\begin{align} \int_0^{2\sqrt{2}} \int_{-\sqrt{8-x^2}}^{\sqrt{8-x^2}} \int_{-1}^2 &\sqrt{1+x^2+y^2}\ dz\ dy\ dx\\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\sqrt{2}} \int_{-1}^2 \sqrt{1+r^2} \cdot r\ dz\ dr\ d\theta\\ &= \int_{-\pi/2}^{\pi/2} \int_0^{2\sqrt{2}} 3\sqrt{1+r^2} \cdot r\ dr\ d\theta\\ &= \int_{-\pi/2}^{\pi/2} (1+r^2)^{3/2} \bigg|_0^{2\sqrt{2}}\ d\theta\\ &= \int_{-\pi/2}^{\pi/2} 26\ d\theta = \ans{26\pi} \end{align}

Find the volume of the solid $$D$$ bounded by the paraboloid $$z=4-x^2-y^2$$ and the plane $$z=0$$.

#### Solution

\begin{align} V &= \iiint_D r\ dz\ dr\ d\theta\\ &= \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\\ &= \int_0^{2\pi} 2r^2-\dfrac{1}{4}r^4 \bigg|_0^2 \ d\theta\\ &= \int_0^{2\pi} 4 \ d\theta = \ans{8\pi} \end{align}

#### Try it yourself:

(click on a problem to show/hide its answer)