I found the intersecting point to be at $\pi/2$ and ${3\pi}/2$
$a=a(1-\cos(\theta) )$
$\cos (\theta)=0$
$\theta= \pi/2$, $3\pi/2$
I'm confused as to what angles to use for integration. As for $r=a$, I'm assuming I'm supposed to draw a circle, and for $r=a(1-cos (\theta))$ I draw a cardioid.
As well, I'm supposed to get the answer $A=a^2 (2-\pi/4)$
Draw a picture. Our region is the part of the circle to the "right" of the cardioid. I would probably find the area of the top half of the region, and multiply by $2$.
Thus we confine attention to $\theta$ going from $0$ to $\pi/2$. The area inside the quarter-circle is $\pi/4$. The area of the part of the cardioid from $\theta=0$ to $\theta=\pi/2$ is $\int_0^{\pi/2}\frac{1}{2}a^2(1-\cos\theta)^2\,d\theta$. Calculate this integral, subtract from $\pi/4$, and multiply the result by $2$.
Remark: It is almost advantageous not to know the area of a circle! The area of our quarter-circle is $\int_0^{\pi/2}\frac{1}{2}a^2 \,d\theta$. Thus half the desired area is $$\int_0^{\pi/2} \frac{1}{2}a^2 \left(1-(1-\cos\theta)^2\right)\,d\theta.$$ By symmetry, the whole area is twice that. That gets rid of the $\frac{1}{2}$ in the integrand, and the whole area is $$\int_0^{\pi/2} a^2(2\cos\theta-\cos^2\theta)\,d\theta.$$