# Question 75f31

Dec 19, 2017

$96 \pi$

#### Explanation:

The graph of the polar curve would be as shown below:

The area enclosed by the curve would be twice the area enclosed by the curve lying over x-axis. The formula for the area enccloseed by a polar curve is given by the formula ${\int}_{\alpha}^{\beta} \frac{1}{2} {r}^{2} d \theta$.
In the present case $\alpha = 0$ which comes by solving $r = 16 = 8 \left(1 + \cos \theta\right)$ and $\beta = \pi$, which is obtained by solving $r = 0 = 8 \left(1 + \cos \theta\right)$

Accordingly, the desired area would be $2 {\int}_{0}^{\pi} \frac{1}{2} {r}^{2} d \theta$

=${\int}_{0}^{\pi} 64 {\left(1 + \cos \theta\right)}^{2} d \theta$

=$64 {\int}_{0}^{\pi} \left(1 + 2 \cos \theta + {\cos}^{2} \theta\right) d \theta$

=64int_0^pi (1+2 cos theta +1/2 (1 +cos 2theta)d theta#

=$64 {\left[\frac{3}{2} \theta + 2 \sin \theta + \frac{1}{4} \sin 2 \theta\right]}_{0}^{\pi}$

=$96 \pi$