Area between two polar areas I could use some help with this problem.
Let a be a constant. Find the area that stays inside both the circle $r = a$, and the cardioid $r = a(1-\sin\theta)$.
I tried to find the point of intersect by making the r's equal but I didn't really understand how it is supposed to be done when the constant is involved.
 A: No problem, the constant cancels. From $a=a(1-\sin\theta)$ we conclude that $1=1-\sin\theta$, so $\theta=0$ or $\theta=\pi$. Note that the cardioid is "inside" the circle from $\theta=0$ to $\theta=\pi$, and outside the circle from $\pi$ to $2\pi$.
Draw a picture. For the part from $0$ to $\pi$, you will need to calculate an integral, For the rest, you don't even need to integrate.
A: $$a(1-\sin\theta) = a \Leftrightarrow 1-\sin\theta = 1 \Leftrightarrow \theta \in \pi\mathbb Z$$
Thus $(r,\theta) \in A \Leftrightarrow (r<a$ and $-\pi<\theta<0)$ or $(r < a(1-\sin\theta)$ and $0<\theta<\pi)$. This means get the area of the cardiodid for $-\pi<\theta<0$ and add half a circle, i.e. $\pi \frac{a^2}2$ to it.
$$\begin{align*}
A & = \frac\pi2 a^2 + \int_{-\pi}^0 \int_0^{a(1-\sin\theta)} r \ \mathrm dr \ \mathrm d\theta \\
& = \frac\pi2 a^2 + \frac12 \int_{-\pi}^0 a^2 (1-\sin\theta)^2\ \mathrm d\theta \\
& = \frac{a^2}2 \left(\pi + \int_{-\pi}^0 1 -2\sin\theta + \sin^2 \theta \ \mathrm d\theta\right) \\
& = \frac{a^2}2 \left( 2\pi + 4 + \int_{-\pi}^0 \sin^2\theta\ \mathrm d\theta \right) \\
& = a^2 (\frac54 \pi + 2)
\end{align*}$$
A: See the figure below. Just add the partial areas as shown :

A: Plotting the curves is helpful, it gives an idea that constant $a$ is not necessary for $\theta$ limits of area integration in this case, confirming the algebraic solution 
$ a = a (1- \sin \theta), \theta = (0,  \pi) , ( 0< \theta < \pi). $
It is required to integrate only for brown common area of cardoid , remaining area is obtained by symmetry and the semi-circle.

