area of the region that lies inside both curves

Hi,

I am trying to solve this problem:

Find the area of the region that lies inside both curves.

$\displaystyle

r_{1}=\sin\theta, r_{2}=\cos\theta

$

Which are the circles $\displaystyle x^2+(y+\frac{1}{2})^2=\frac{1}{4}$ and $\displaystyle (x+\frac{1}{2})^2+y^2=\frac{1}{4}$ on the Cartesian coordinate system.

When I try to solve the problem in the following way, I get a different answer than in my textbook:

$\displaystyle

A_{1}=\int_0^\frac{\pi}{4}\frac{1}{2}\sin^2\theta\ d\theta=\frac{1}{4}\int_0^\frac{\pi}{4}(1-\cos2\theta)\ d\theta=\frac{1}{4}(\frac{\pi}{4}-\frac{1}{2})=\frac{\pi}{16}-\frac{1}{8}

$

$\displaystyle

A_{2}=\int_0^\frac{\pi}{4}\frac{1}{2}\cos^2\theta\ d\theta=\frac{1}{4}\int_0^\frac{\pi}{4}(1+\cos2\th eta)\ d\theta=\frac{1}{4}(\frac{\pi}{4}+\frac{1}{2})=\fr ac{\pi}{16}+\frac{1}{8}

$

$\displaystyle

A=A_{1}+A_{2}=(\frac{\pi}{16}-\frac{1}{8})+(\frac{\pi}{16}+\frac{1}{8})=\frac{\p i}{8}

$

However, if I do it by symmetry, it gives me a different answer (the correct answer in my textbook):

$\displaystyle

A=2\int_0^\frac{\pi}{4}\frac{1}{2}\sin^2\theta\ d\theta=\frac{1}{2}\int_0^\frac{\pi}{4}(1-\cos2\theta)\ d\theta=\frac{1}{2}(\frac{\pi}{4}-\frac{1}{4})=\frac{\pi}{8}-\frac{1}{4}

$

Can anyone explain this to me? I've been trying to figure it out, but I can't seem to get it.