Cylindrical coordinates

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We integrate over regions in cylindrical coordinates.

The first way we will generalize polar coordinates to three dimensions is with cylindrical coordinates.

An ordered triple consisting of a radius, an angle, and a height $(r,\theta ,z)$ can be graphed as $\begin{align*} x &= r\cdot \cos(\theta)\\ y &= r\cdot \sin(\theta)\\ z &= z \end{align*}$ meaning:

Coordinates of this type are called cylindrical coordinates.

Consider the point $(2, \pi /3,5)$ in cylindrical coordinates. What is this point when expressed in $(x,y,z)$ -coordinates? $(x,y,z) = \left (\answer {2\cos (\pi /3)}, \answer {2 \sin (\pi /3)},\answer {5}\right )$

Consider the point $(1, -1,5)$ in $(x,y,z)$ -coordinates. What is this point when expressed in cylindrical coordinates where $0\le \theta <2\pi$ ? $(r,\theta ,z) = \left (\answer {\sqrt {2}}, \answer {7\pi /4},\answer {5}\right )$

Triple integrals in cylindrical coordinates

If you want to evaluate this integral $\iiint _R F \d V,$ you have to change $R$ to a region defined in $(x,y,z)$ -coordinates, and change $\d V$ to some combination of $\d x\d y\d z$ leaving you with some iterated integral: $\int _a^b\int _c^d\int _p^q F(x,y,z) \d y \d x\d z$ Now consider representing a region $R$ in cylindrical coordinates and let’s express $\d V$ in terms of $\d r$ , $\d \theta$ , and $\d z$ . To do this, consider the diagram below:

Here we see $\begin{align*} \d V &= \d r \cdot (r \d \theta)\cdot \d z \\ &= \d z\ r \d r \d \theta. \end{align*}$

Recalling that the determinate of a $3\times 3$ matrix gives the volume of a parallelepiped, we could also deduce the correct form for $\d V$ by setting $\begin{align*} x(r,\theta,z) &= r \cos(\theta)\\ y(r,\theta,z) &= r \sin(\theta)\\ z(r,\theta,z) &= z \end{align*}$ and computing: $\begin{align*} \d V &= \left| \det \begin{bmatrix} \pp[x]{r} \d r & \pp[y]{r} \d r & \pp[z]{r} \d r\\ \pp[x]{\theta} \d \theta & \pp[y]{\theta} \d \theta & \pp[z]{\theta} \d \theta\\ \pp[x]{z} \d z & \pp[y]{z} \d z & \pp[z]{z} \d z \end{bmatrix} \right|\\ &= \left| \det \begin{bmatrix} \answer[given]{\cos(\theta)} \d r & \answer[given]{\sin(\theta)} \d r & 0 \\ \answer[given]{-r\sin(\theta)} \d \theta & \answer[given]{r\cos(\theta)} \d \theta & 0\\ 0 & 0 & \d z \end{bmatrix} \right|\\ &= r \d z\d r \d \theta \end{align*}$

Fubini Let $F:\R ^3\to \R$ be continuous on the region $R=\{(r,\theta ,z):\alpha \leq \theta \leq \beta , g_1(\theta )\leq r\leq g_2(\theta ), G_1(x,y)\le z\le G_2(x,y)\}$ Then: $\iiint _R F(r,\theta ,z)\d V = \int _\alpha ^\beta \int _{g_1(\theta )}^{g_2(\theta )} \int _{G_1(r\cos (\theta ),r\sin (\theta ))}^{G_2(r\cos (\theta ),r\sin (\theta ))}F(r,\theta ,z) \d z \ r\d r\d \theta .$

Write down a triple integral in cylindrical coordinates that will compute the volume of a cylinder of radius $a$ and height $h$ . $\iiint _R \d V = \int _{\answer {0}}^{\answer {2\pi }} \int _{\answer {0}}^{\answer {a}} \int _{\answer {0}}^{\answer {h}} \d z \ r \d r \d \theta$

Find the volume under $z= \sqrt {4-r^2}$ above the quarter circle inside $x^2 + y^2 = 4$ in the first quadrant.

In this case $R=\{(r,\theta ,z):\answer [given]{0}\leq \theta \leq \answer [given]{\pi /2}, \answer [given]{0}\leq r\leq \answer [given]{2}, \answer [given]{0}\le z\le \sqrt {4-r^2}\}$ So our integral is $\begin{align*} \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \int_{\answer[given]{0}}^{\answer[given]{2}} \int_{\answer[given]{0}}^{\answer[given]{\sqrt{4-r^2}}} r \d z \d r \d \theta &= \int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \int_{\answer[given]{0}}^{\answer[given]{2}} \answer[given]{\sqrt{4-r^2}} r \d r \d \theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \eval{\answer[given]{\frac{-(4-r^2)^{3/2}}{3}}}_{\answer[given]{0}}^{\answer[given]{2}} \d \theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{\pi/2}} \answer[given]{\frac{8}{3}} \d \theta\\ &=\answer[given]{4\pi/3}. \end{align*}$

Find the volume of the object defined as the intersection of the cylinder $x^2+y^2=1$ and the sphere $x^2+y^2+z^2=4$ .

First note that we can express our region as $\begin{align*} R=\{(r,\theta,z):&\answer[given]{0}\leq\theta\leq\answer[given]{2\pi}, \\ &\answer[given]{0}\leq r\leq \answer[given]{1}, \\ &-\sqrt{4-r^2}\le z\le \answer[given]{\sqrt{4-r^2}}\} \end{align*}$ Now write with me, $\begin{align*} \int_{\answer[given]{0}}^{\answer[given]{2\pi}} \int_{\answer[given]{0}}^{\answer[given]{1}} \int_{\answer[given]{-\sqrt{4-r^2}}}^{\answer[given]{\sqrt{4-r^2}}} r \d z \d r \d \theta &= \int_{\answer[given]{0}}^{\answer[given]{2\pi}} \int_{\answer[given]{0}}^{\answer[given]{1}} \answer[given]{2\sqrt{4-r^2}} r \d r \d \theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{2\pi}} \eval{\answer[given]{\frac{-2(4-r^2)^{3/2}}{3}}}_{\answer[given]{0}}^{\answer[given]{1}} \d \theta\\ &=\int_{\answer[given]{0}}^{\answer[given]{2\pi}} \answer[given]{\frac{16}{3}-2\sqrt{3}} \d \theta\\ &=\answer[given]{2\pi\left(\frac{16}{3}-2\sqrt{3}\right)}. \end{align*}$