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

 A: 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. 

A: 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.
A: For the curve $r=\sin(\theta)$
$$
\left.\begin{array}{}
x=r\cos(\theta)=\sin(\theta)\cos(\theta)\\
y=r\sin(\theta)=\sin^2(\theta)
\end{array}\right\}\quad x^2+\left(y-\tfrac12\right)^2=\tfrac14
$$
For the curve $r=\cos(\theta)$
$$
\left.\begin{array}{}
x=r\cos(\theta)=\cos^2(\theta)\\
y=r\sin(\theta)=\sin(\theta)\cos(\theta)
\end{array}\right\}\quad\left(x-\tfrac12\right)^2+y^2=\tfrac14
$$
So the area is the intersection of the two circles:

The area of the purple lune is $\frac14$ of the area of a radius $\frac12$ circle minus the area of a $\frac12,\frac12,\frac1{\sqrt2}$ right triangle, that is
$$
\frac14\cdot\frac\pi4-\frac12\cdot\frac12\cdot\frac12
$$
Since the area of the green lune is the same, the area of the given intersection is
$$
\frac\pi8-\frac14
$$
A: @robjohn's calculation states "pi/4 x 1/4 - 1/2 x 1/2 x 1/2 = pi/8 -1/4" .
This is false, it equals pi/16 -1/8.
Here's a simpler argument that requires NO CALCULUS. Take @robjohn's diagram. Notice that the full square has area 1/2 x 1/2 = 1/4.
Then, notice that one quarter of one of these circles has area pi/16 (since pi/4 divided by 4 is pi/16).
Take the following difference: 1/4 - pi/16 = (4-pi)/16 call this D
Now, the area of interest is the total area of the square (1/4) minus TWO of the D areas, i.e. 1/4 -2D = 1/4 -2( (4-pi)/16 ) = pi/8 - 1/4
