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

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

\[\begin{align} V &= \iiint_D r\ dz\ dr\ d\theta\\ &= \int_0^{2\pi} \int_0^2 \int_0^{4-r^2} r\ dz\ dr\ d\theta\\ &= \int_0^{2\pi} \int_0^2 4r-r^3\ dr\ d\theta\\ &= \int_0^{2\pi} 2r^2-\dfrac{1}{4}r^4 \bigg|_0^2 \ d\theta\\ &= \int_0^{2\pi} 4 \ d\theta = \ans{8\pi} \end{align}\]