Math 344: Calculus III

15.7 Triple Integrals in Cylindrical Coordinates

---

15.7.1 Cylindrical Coordinates

Figure 1

Like polar coordinates, cylindrical coordinates describe coordinates $x$ and $y$ in terms of the distance

$$ \require{color} \definecolor{brightblue}{rgb}{.267, .298, .812} \definecolor{darkblue}{rgb}{.08, .18, .28} \definecolor{palepink}{rgb}{1, .73, .8} \definecolor{softmagenta}{rgb}{.99,.34,.86} \def\ihat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ı}}}} \def\jhat{\mathbf{\hat{\mmlToken{mi}[mathvariant="bold"]{ȷ}}}} \def\khat{\mathbf{\hat{k}}} \newcommand{\pypx}[2][x]{\dfrac{\partial #2}{\partial #1}} \newcommand{\dydx}[2][x]{\dfrac{d #2}{d #1}} \newcommand{\deltax}[2][x]{\frac{\Delta #2}{\Delta #1}} r = \sqrt{x^2 + y^2} $$
and the angle that the line segment $\langle x, y, 0 \rangle$ makes with the positive $x$-axis

$$ \theta = \tan^{-1}\left(\dfrac{y}{x}\right) $$
The third coordinate is the rectangular distance of the point from the $xy$-plane. Like rectangular coordinates, positive values for $z$ indicate a distance above the $xy$-plane, and negative values indicate a distance below the $xy$-plane. When we choose the orientation of the three axis, we are choosing the directions above and below . The standard orientation is that above is the direction pointed by $\ihat\times\jhat$. This orientation is called the right-hand rule .

Thus our change of coordinates from rectangular coordinates to cylindrical coordinates is given as

$$ \begin{align*} x &= r\cos(\theta) \\ y &= r\sin(\theta) \\ z &= z \end{align*} $$
In figure 1, vector $\overrightarrow{OQ}$ is the projection of vector $\overrightarrow{OP}$ onto the $xy$-plane. Thus angle $\angle OQP$ is a right angle , and the length of vector $\overrightarrow{OP}$ is given by

$$ \rho = \sqrt{\left|\overrightarrow{OQ}\right|^2 + \left|\overrightarrow{PQ}\right|^2} = \sqrt{r^2 + z^2} = \sqrt{x^2 + y^2 + z^2} $$
The coordinates $r$ and $\theta$ are related to coordinates $x$ and $y$ just as polar coordinates relate to rectangular coordinates, and the third coordinate $z$ is the same in both coordinate systems.

Example 1

In figure 1, the cylindrical coordinates of vector $P = \left\langle 4, \frac{\pi}{3}, 3 \right\rangle$. What are the rectangular coordinates?

The rectangular coordinates are given by

$$ \begin{align*} x &= r\cos(\theta) = 4\cdot\cos\left(\frac{\pi}{3}\right) = 4\cdot\frac{1}{2} = 2 \\ y &= r\sin(\theta) = 4\cdot\cos\left(\frac{\pi}{3}\right) = 4\cdot\frac{\sqrt{3}}{2} = 2\sqrt{3} \\ z &= 3 \end{align*} $$
In rectangular coordinates point $P = \left\langle 2, 2\sqrt{3}, 3 \right\rangle$.

15.7.2 Why "Cylindrical" Coordinates?

The graph of the equation $r = 1$, or $r =$ constant is a right circular cylinder.

Figure 2

Example 2

Convert the rectangular coordinates for point $P = \langle -1, \sqrt{3}, 5 \rangle$ into cylindrical coordinates.

Figure 3

$$ \begin{align*} r &= \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + (\sqrt{3})^2} = \sqrt{1 + 3} = 2 \\ \text{reference angle} &= \tan^{-1}\left(\left|\frac{y}{x}\right|\right) = \tan^{-1}\left(\frac{\sqrt{3}}{1}\right) = \frac{\pi}{3} \\ \end{align*} $$
Since $P$ is in the second quadrant we have

$$ \theta = \pi - \text{reference angle} = \pi - \frac{\pi}{3} = \frac{2\pi}{3} $$

15.7.3 The Jacobian

To perform an integration in cylindrical coordinates we will need the appropriate Jacobian. In three dimensions the Jacobian is

$$ \begin{align*} J &= \text{abs}\left(\begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial z} \end{vmatrix}\right) \\ \\ &= \text{abs}\left(\begin{vmatrix} \cos(\theta) & -r\sin(\theta) & 0 \\ \sin(\theta) & r\cos(\theta) & 0 \\ 0 & 0 & 1 \end{vmatrix}\right) \\ \\ &= \text{abs}\left(0\begin{vmatrix} -r\sin(\theta) & 0 \\ r\cos(\theta) & 0 \end{vmatrix} - 0\begin{vmatrix} \cos(\theta) & 0 \\ \sin(\theta) & 0 \end{vmatrix} + 1\begin{vmatrix} \cos(\theta) & -r\sin(\theta) \\ \sin(\theta) & r\cos(\theta) \end{vmatrix}\right) \\ \\ &= \text{abs}\left(0 - 0 + r\cos^2(\theta) + r\sin^2(\theta)\right) = r \\ \end{align*} $$

15.7.4 Triple Integrals

We may use cylindrical coordinates to compute the volume under a quadratic surface described in rectangular coordinates.

Example 3

Find the volume of the solid $E$ bounded by the paraboloid $S$ defined by $x^2 + y^2 + z = 1$ and the plane $R$ defined by $z=0$.

Figure 4

If we use cylindrical coordinates we have $z = 1 - r^2$ for the equation of the surface of the paraboloid, and $z=0$ as the equation of the surface of the plane. Hence we compute the volume as follows,

$$ \begin{align*} V &= \displaystyle\iiint_E dV = \displaystyle\iiint_E dxdydz = \displaystyle\iiint_E r\,dz\,dr\,d\theta \\ \\ &= \displaystyle\int_0^{2\pi}\displaystyle\int_0^1\displaystyle\int_0^{1-r^2} r\,dz\,dr\,d\theta = \displaystyle\int_0^{2\pi}\displaystyle\int_0^1 r\,\left[\,\displaystyle\int_0^{1-r^2} \,dz\,\right]\,dr\,d\theta \\ \\ &= \displaystyle\int_0^{2\pi}\displaystyle\int_0^1 r\,\left[\,1-r^2\,\right]\,dr\,d\theta = \displaystyle\int_0^{2\pi}\,\left[\,\displaystyle\int_0^1 r-r^3\,dr\,\right]\,d\theta \\ \\ &= \displaystyle\int_0^{2\pi}\,\left[\,\frac{r^2}{2}-\frac{r^4}{4}\,\right]_0^1\,d\theta = \frac{1}{4}\displaystyle\int_0^{2\pi}\,dz = \frac{\pi}{2} \\ \end{align*} $$

We may need to derive an equation for the surface and then convert the integral to cylindrical coordinates.

Example 4

Find the volume of a right circular cone whose base has a radius of $a$ units and height $h$ units.

Recall the equation of a circular cone with radius $a$ at distance $h$ along the $z$-axis is given by

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2} - \frac{z^2}{h^2} = 0 $$

Figure 5

However the surface of the right circular cone we want is below the $xy$-plane with its base of radius $a$ at a distance $h$ below the $xy$-plane. We must shift the cone a distance of $h$ units up along the $z$-axis.

$$ \frac{x^2}{a^2} + \frac{y^2}{a^2} - \frac{(z-h)^2}{h^2} = 0 $$

Figure 6

We also must solve the equation of our surface for $z$ and ignore the surface above $z=h$.

$$ \begin{align*} \frac{(z-h)^2}{h^2} &= \frac{x^2}{a^2} + \frac{y^2}{a^2} \\ \\ \frac{z-h}{h} &= - \sqrt{\frac{x^2}{a^2} + \frac{y^2}{a^2}} \\ \\ z-h &= - \frac{h}{a}\sqrt{x^2 + y^2} \\ \\ z &= h - \frac{h}{a}\sqrt{x^2 + y^2} \\ \\ z &= h\left( 1 - \frac{r}{a}\right) \end{align*} $$
We choose the negative root because we only want the surface below the vertex. Now we can compute the volume of the right circular cone.

$$ \begin{align*} V &= \displaystyle\iiint_E\,dV = \displaystyle\iiint_E\,dx\,dy\,dz \\ \\ &= \displaystyle\int_0^{2\pi}\,\displaystyle\int_0^a\,\displaystyle\int_0^{h\left(1-\frac{r}{a}\right)}\,r\,dz\,dr\,d\theta = \displaystyle\int_0^{2\pi}\,\displaystyle\int_0^a\,r\,\left[\,\displaystyle\int_0^{h\left(1-\frac{r}{a}\right)}\,dz\,\right]\,dr\,d\theta \\ \\ &= \displaystyle\int_0^{2\pi}\,\displaystyle\int_0^a\,r\,\left[\,h\left(1-\frac{r}{a}\right)\,\right]\,dr\,d\theta = h\,\displaystyle\int_0^{2\pi}\,\left[\,\displaystyle\int_0^a\,\left(\,r-\frac{r^2}{a}\,\right)\,dr\,\right]\,d\theta \\ \\ &= h\,\displaystyle\int_0^{2\pi}\,\left[\,\frac{r^2}{2}-\frac{r^3}{3a}\,\right]_0^a\,d\theta = h\,\displaystyle\int_0^{2\pi}\,\left(\,\frac{a^2}{2}-\frac{a^3}{3a}\,\right)\,d\theta \\ \\ &= \frac{a^2h}{6}\,\displaystyle\int_0^{2\pi}\,d\theta = \frac{h}{3}\pi a^2 \end{align*} $$

15.7.5 Additional Practice

Exercise 1

Compute the value of the integral

$$ \displaystyle\iiint_E\,\dfrac{dx\,dy\,dz}{\sqrt{x^2 + y^2}}, $$
where the volume $E$ is the cylinder $x^2 + y^2=16$ bounded above by the plane $z=4$, below by the plane $z=1$.

Check your work

First draw a graph of your solid domain


$$ \begin{align*} \displaystyle\iiint_E\,\dfrac{dx\,dy\,dz}{\sqrt{x^2 + y^2}} &= \displaystyle\iiint_E\,\dfrac{r\,dz\,dr\,d\theta}{r} = \displaystyle\int_0^{2\pi}\,\displaystyle\int_0^4\,\displaystyle\int_1^4\,dz\,dr\,d\theta \\ \\ &= \left[\displaystyle\int_0^{2\pi}\,d\theta\,\right]\,\left[\,\displaystyle\int_0^4\,dr\,\right]\,\left[\,\displaystyle\int_1^4\,dz\,\right] = (2\pi)(4)(4-1) = 24\pi \\ \end{align*} $$

Exercise 2

Compute the value of the integral

$$ \displaystyle\iiint_E\,dx\,dy\,dz, $$
where the volume $E$ is the cylindrical shell of height $h\gt 0$ with base on the $xy$-plane between the cylinders $x^2 + y^2 = a^2$ and $x^2 + y^2 = b^2$, where $b\gt a\gt 0$.

Check your work

First draw a graph of your solid domain


In cylindrical coordinates the limits of integration are much simpler with a figure that is symmetric with respect to the $z$-axis.
$$ \begin{align*} \displaystyle\iiint_E\,dx\,dy\,dz &= \displaystyle\iiint_E\,r\,dz\,dr\,d\theta = \displaystyle\int_0^{2\pi}\,\displaystyle\int_a^b\,\displaystyle\int_0^h\,r\,dz\,dr\,d\theta \\ \\ &= \left[\,\displaystyle\int_0^{2\pi}\,d\theta\,\right]\,\left[\,\displaystyle\int_a^b\,r\,dr\,\right]\,\left[\,\displaystyle\int_0^h\,dz\,\right] \\ \\ &= (2\pi)\,\left[\frac{r^2}{2}\right]_a^b\,(h) = 2h\pi\,\left[\,\frac{b^2}{2} - \frac{a^2}{2}\,\right] \\ \\ &= \pi(b^2-a^2)h \end{align*} $$

Your use of this self-initiated mediated course material is subject to our Creative Commons License .

Creative Commons Attribution-NonCommercial-ShareAlike 4.0


Attribution
You must give appropriate credit, provide a link to the license, and indicate if changes were made. You may do so in any reasonable manner, but not in any way that suggests the licensor endorses you or your use.


Noncommercial
You may not use the material for commercial purposes.


Share Alike
You are free to share, copy and redistribute the material in any medium or format. If you adapt, remix, transform, or build upon the material, you must distribute your contributions under the same license as the original.