We integrate over regions in spherical coordinates.
Another way to generalize polar coordinates to three dimensions is with spherical
coordinates.
An ordered triple consisting of a radius, an angle, and a height
\((\rho ,\theta , \varphi )\) can be graphed as
\begin{align*} x &= \rho \cdot \cos (\theta )\sin (\varphi )\\ y &= \rho \cdot \sin (\theta )\sin (\varphi )\\ z &= \rho \cdot \cos (\varphi ) \end{align*}
meaning:
Coordinates of this type are called
spherical coordinates.
Consider the point
\((\rho ,\theta ,\phi )=(2,-\pi /4,\pi /4)\) in spherical coordinates. What is this point when expressed in
\((x,y,z)\)-coordinates?
\[ (x,y,z) = \left (\answer {2\cos (-\pi /4)\sin (\pi /4)}, \answer {2 \sin (-\pi /4)\sin (\pi /4)},\answer {2\cos (\pi /4)}\right ) \]
1 Triple integrals in spherical 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 spherical coordinates and
let’s express \(\d V\) in terms of \(\d \rho \), \(\d \phi \), and \(\d \theta \). To do this, consider the diagram below:
Here we see
\begin{align*} \d V &= (\rho \sin (\varphi )\d \theta )\cdot (\rho \d \varphi )\cdot (\d \rho )\\ &= \rho ^2 \sin (\varphi )\d \rho \d \varphi \d \theta . \end{align*}
We may now state at theorem:
Fubini Let
\(F:\R ^3\to \R \) be continuous on the region
\[ R=\{(\rho ,\theta ,\phi ):\text {$\alpha \leq \theta \leq \beta $, $a\leq \phi \leq b$, $G_1(\theta ,\phi )\le \rho \le G_2(\theta ,\phi )$}\} \]
Then:
\begin{align*} \iiint _R &F(\rho ,\theta ,\phi )\d V\\ &= \int _\alpha ^\beta \int _{a}^{b} \int _{G_1(\theta ,\phi )}^{G_2(\theta ,\phi )}F(\rho ,\theta ,\phi ) \rho ^2 \sin (\phi )\d \rho \d \varphi \d \theta . \end{align*}
Write down a triple integral in spherical coordinates that will compute the volume
of a sphere of radius
\(a\).
\[ \iiint _R \d V = \int _{\answer {0}}^{\answer {2\pi }} \int _{\answer {0}}^{\answer {\pi }} \int _{\answer {0}}^{\answer {a}} \rho ^2 \sin (\varphi )\d \rho \d \varphi \d \theta \]