15.6 Polar Area
A polar box \( [\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang} \) describes the region between two fixed radii and two fixed angles.
We'd like to use Calculus to find the area enclosed by a polar box.
Formula.
In polar coordinates, the area polyruler can be written as
\[
\begin{align}
&\rad \mult \bar{\diff} \rad \mult \bar{\diff} \ang.
\end{align}
\]
Why?
We compute.
\[
\begin{align}
\rad \mult \bar{\diff} \rad \mult \bar{\diff} \ang &= \bar{\diff} \rad \mult \bigl(\rad \mult \bar{\diff} \ang\bigr)
\\
&= \biggl(\frac{x}{\rad} \mult \bar{\standard}_{x} + \frac{y}{\rad} \mult \bar{\standard}_{y}\biggr) \mult \biggl(\frac{-y}{\rad} \mult \bar{\standard}_{x} + \frac{x}{\rad} \mult \bar{\standard}_{y}\biggr)
\\
&= \frac{x^2}{\rad^2} \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} - \frac{y^2}{\rad^2} \mult \bar{\standard}_{y} \mult \bar{\standard}_{x}
\\
&= \frac{x^2}{\rad^2} \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + \frac{y^2}{\rad^2} \mult \bar{\standard}_{x} \mult \bar{\standard}_{y}
\\
&= \frac{x^2 + y^2}{\rad^2} \mult \bar{\standard}_{x} \mult \bar{\standard}_{y}
\\
&= \bar{\standard}_{x} \mult \bar{\standard}_{y}
\end{align}
\]
Sure enough, we've found the area polyruler.
This formula will let us compute the area of a polar box!
Theorem.
Consider the polar box
\[
\begin{align}
&[\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}
\end{align}
\]
where
\[
\begin{align}
\rad_1 &: \Number,
&
\rad_2 &: \Number,
&
\ang_1 &: \Number,
&
\ang_2 &: \Number
\end{align}
\]
are constants. The polar box has area
\[
\begin{align}
&\frac{{\rad_2}^2 - {\rad_1}^2}{2} \mult (\ang_2 - \ang_1).
\end{align}
\]
Why?
We can measure the area by integrating the area polyruler.
\[
\begin{align}
A &= \biggl\langle{\rad \mult \bar{\diff} \rad \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{[\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \biggl\langle{\bar{\diff} \biggl(\frac{\rad^2}{2} \mult \bar{\diff} \ang\biggr)}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{[\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{\boundary \bigl([\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}\bigr) }\biggr\rangle
\\
&= \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{\boundary [\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle - \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{[\rad_1, \rad_2]_{\rad} \mult \boundary [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\end{align}
\]
The integral on the right will be zero because \( \bar{\diff} \ang_2 = 0 \) and \( \bar{\diff} \ang_1 = 0 \). Let's continue our calculation with the integral on the left.
\[
\begin{align}
A &= \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{\boundary [\rad_1, \rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{[\rad_2]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle - \biggl\langle{\frac{\rad^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{(\rad, \ang)}{\displaystyle{\int}}}{[\rad_1]_{\rad} \mult [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \biggl\langle{\frac{{\rad_2}^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{[\ang_1, \ang_2]_{\ang}}\biggr\rangle - \biggl\langle{\frac{{\rad_1}^2}{2} \mult \bar{\diff} \ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{[\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \frac{{\rad_2}^2}{2} \mult \biggl\langle{\bar{\diff} \ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{[\ang_1, \ang_2]_{\ang}}\biggr\rangle - \frac{{\rad_1}^2}{2} \mult \biggl\langle{\bar{\diff} \ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{[\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \frac{{\rad_2}^2 - {\rad_1}^2}{2} \mult \biggl\langle{\bar{\diff} \ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{[\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \frac{{\rad_2}^2 - {\rad_1}^2}{2} \mult \biggl\langle{\ang}\mathrel{\overset{\ang}{\displaystyle{\int}}}{\boundary [\ang_1, \ang_2]_{\ang}}\biggr\rangle
\\
&= \frac{{\rad_2}^2 - {\rad_1}^2}{2} \mult (\ang_2 - \ang_1)
\end{align}
\]
Our theorem gives us a formula for the area enclosed within a circle.
Formula.
The area inside a circle with radius \( R \) is
\[
\begin{align}
&\frac{R^2}{2} \mult \revolution.
\end{align}
\]
Why?
We can describe the region as the polar box \( [0, R]_{\rad} \mult [0, \revolution]_{\ang} \). We just apply our theorem!