# Using Polar coordinates, what is the area of region A?

## F: $r$ = 16 $\cos \left(\Theta\right)$ ${r}_{f}$ = 8 G: $r$ = 5 ${r}_{g}$ = 5

Jul 10, 2018

$\left(\left(\frac{5}{2}\right) \sqrt{231} + 103 {\cos}^{- 1} \left(\frac{5}{16}\right)\right)$
$= \left(\frac{5}{2} \sqrt{231} + 129.056 r a d\right)$= 167.05 areal units, nearly.
This is nearly 83.1 % of the area of the larger circle.

#### Explanation:

At the common points,

$r = 16 \cos \theta = 5$, giving $\cos \theta = \frac{5}{16}$ and

$\left(5 , \pm {\cos}^{- 1} \left(\frac{5}{16}\right)\right)$, as the points of intersection.

$= 2 \int \left[\int r \mathrm{dr}\right] d \theta$,

between the limits 5 and $16 \cos \theta$, for r,

and 0 and ${\cos}^{- 1} \left(\frac{5}{16}\right)$, for $\theta$

$= \int \left({16}^{2} {\cos}^{2} \theta - {5}^{2}\right) d \theta$,

$\theta$ from 0 to ${\cos}^{- 1} \left(\frac{5}{16}\right)$

$= \int \left(256 {\cos}^{2} \theta - 25\right) d \theta , \theta$ from 0 to cos^(-1)

(5/16)

= int ( 128 cos 2theta + 103 )] d theta , between the limits

$= \left[64 \sin 2 \theta + 103 \theta\right]$, between the limits

$= \left(128 \left(\sqrt{1 - {\left(\frac{5}{16}\right)}^{2}} \left(\frac{5}{16}\right)\right) + 103 \cos \left(- 1\right) \left(\frac{5}{16}\right)\right)$

$= \left(\left(\frac{5}{2}\right) \sqrt{231} + 103 \cos \left(- 1\right) \left(\frac{5}{16}\right)\right)$

$= \left(\frac{5}{2} \sqrt{231} + 129.056 r a d\right)$

#= 167.05 areal units, nearly.