Answer
area of intersection $ = \frac{\pi }{8} - \frac{1}{4}$
Work Step by Step
First, we find the intersection points by solving the equation: $r = \sin \theta = \cos \theta $.
For the interval $0 \le \theta \le \pi $, the solution is $\theta = \frac{\pi }{4}$.
Let $A$ denote the region bounded by the circle $r = \sin \theta $ and the ray $\theta = \frac{\pi }{4}$ as is shown in the figure attached. So, the area of $A$ is
area of $A$ $ = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} {\sin ^2}\theta {\rm{d}}\theta $
By symmetry, the area of intersection of the circles $r = \sin \theta $ and $r = \cos \theta $ is twice the area of $A$. Therefore,
area of intersection $ = 2\cdot\frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} {\sin ^2}\theta {\rm{d}}\theta $
Since ${\sin ^2}\theta = \frac{1}{2}\left( {1 - \cos 2\theta } \right)$, the integral becomes
area of intersection $ = \frac{1}{2}\cdot\mathop \smallint \limits_0^{\pi /4} \left( {1 - \cos 2\theta } \right){\rm{d}}\theta $
area of intersection $ = \frac{1}{2}\left( {\theta - \frac{1}{2}\sin 2\theta } \right)|_0^{\pi /4}$
area of intersection $ = \frac{1}{2}\left( {\frac{\pi }{4} - \frac{1}{2}} \right) = \frac{\pi }{8} - \frac{1}{4}$
