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
\(0\le \theta <2\pi \)?
\[ (r,\theta ,z) = \left (\answer {\sqrt {2}}, \answer {7\pi /4},\answer {5}\right ) \]
1 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*}
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*}