areacalculuscirclesintegrationpolar coordinates
I am not sure of my answer. In the figure, $r=10 \cos\theta$ is a circle that doesn't look like a circle.
The area of $r=5$ is $\pi r^2 = 25 \pi$.
You remove the area from $-\pi/3$ to $\pi/3$ of $10 \cos\theta$ from $25\pi$.
That is remove $$\frac 12 \int (10 \cos\theta)^2\,d\theta = 74.0105$$
Required area = $25 \pi – 74.0105 = 4.5293$
HINT...You need to establish where the curves intersect in order to determine the limits, so solve $3\cos \theta=1+\cos \theta$.
So your limits are $\pm\frac {\pi}{3}$
If $$r_1(\theta) = 3a \cos \theta$$ is the equation of the circle, and $$r_2\theta) = a(1+\cos \theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(\theta) = r_2(\theta)$, or $3a \cos \theta = a(1+\cos \theta)$, or $$2 \cos \theta = 1.$$ This gives us the intersection points $$\theta = \pm \frac{\pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $\theta \in [-\pi/3, \pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = \int_{\theta = -\pi/3}^{\pi/3} \frac{1}{2}\left(r_1^2(\theta) - r_2^2(\theta)\right) \, d\theta.$$ Note this is not the same as $$\int_{\theta = -\pi/3}^{\pi/2} \frac{1}{2}\left(r_1(\theta) - r_2(\theta)\right)^2 \, d\theta.$$ The second integral is incorrect for the area enclosed.
Best Answer
The area you want is the red-only region on the right.
We quickly get the expression
$$\begin{align} Area &= \frac 12\int_{-\pi/3}^{\pi/3}[(10\cos\theta)^2-5^2]\,d\theta \\[2 ex] \end{align}$$
I'm sure you can finish from here. This can also be done by simple geometry, which I recommend you do as a check.