area between polar equation $r = \sin\theta$ and $r = \cos\theta$

Question

Below is the exact question and answer from my textbook:
Find the area of the region enclosed between the two curves $C_{1}$ and $C_{2}$ where $C_{1}$ has the polar equation $r = \sin\theta$ and $C_{2}$ has the polar equation $r = \cos\theta$.

answer is
$\frac{\pi}{8} - \frac{1}{16}$

I spend some time figuring this out...
At first I need find intersection (ie: $\sin\theta = \cos\theta$) between this 2 equations but this
obviously didn't made any sense.
But then how would i find lower and upper limit [a,b]
using the formula for area = $\int_a^b \frac{1}{2}(\sin\theta-\cos\theta)^2 \,d\theta$

i assume the textbook is asking area for this:

Answer 1

The sine and the cosine are equal when $\theta=\pi/4$. The two curves are actually circles with radii 1/2 and center $(0,1/2)$ and $(1/2,0)$ for the $\sin$ and $\cos$ respectively. You can thus find the area by computing the following integral

$$\int_0^{\pi/4} (\sin\theta)^2 d\theta$$

in which I multiplied by $2$ exploiting symmetry.

Answer 2

I just added this plot to show what is Raskolnikov trying to tell you. You can find the area. Note the integral he wrote above.