# Question #51be1

Jan 24, 2018

$A = 2 + \frac{\pi}{4}$

#### Explanation:

Since the value of $\cos \theta$ goes from 1 to -1 and back to 1 (as you rotate counterclockwise starting from $\theta = 0$), the value of $1 - \cos \theta$ will go from 0 to 2 and back to 0

This means that the cardioid $r = 1 - \cos \theta$ should intersect the unit circle at two places, and the area we're looking for is between these two values.

So let's find these two values:

${r}_{1} = {r}_{2}$

$1 - \cos \theta = 1$

$\cos \theta = 0$

$\theta \in \left\{\frac{\pi}{2} , \frac{3 \pi}{2}\right\}$

So we're looking for the area outside of the circle but inside the cardioid, between $\theta = \frac{\pi}{2}$ and $\theta = \frac{3 \pi}{2}$

The area that we're looking for is equal to the area of the cardioid MINUS the area of the circle.

Now that we have our bounds and our equations, we can calculate the area. Remember the formula:

$A = \int \frac{1}{2} {r}^{2} d \theta$

$A = {\int}_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{1}{2} {\left(1 - \cos \theta\right)}^{2} d \theta - {\int}_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{1}{2} {\left(1\right)}^{2} d \theta$

$A = {\int}_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \frac{1}{2} \left(1 - 2 \cos \theta + {\cos}^{2} \theta\right) d \theta - {\left[\frac{1}{2} \theta\right]}_{\frac{\pi}{2}}^{\frac{3 \pi}{2}}$

$A = \frac{1}{2} {\int}_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} \left(1 - 2 \cos \theta + \frac{1 + \cos \left(2 \theta\right)}{2}\right) d \theta - \frac{1}{2} \pi$

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

$A = \frac{1}{2} \left[\pi - \left(- 4\right) + \frac{\pi}{2} + 0\right] - \frac{1}{2} \pi$

$A = \frac{1}{2} \pi + 2 + \frac{1}{4} \pi - \frac{1}{2} \pi$

$A = 2 + \frac{\pi}{4}$