# area of the region that lies inside both curves

• September 5th 2008, 11:39 PM
dammitpoo
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.

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

Which are the circles $x^2+(y+\frac{1}{2})^2=\frac{1}{4}$ and $(x+\frac{1}{2})^2+y^2=\frac{1}{4}$ 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:

$
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}
$

$
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}
$

$
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):

$
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.
• September 6th 2008, 01:58 AM
Laurent
I think for $A_2$ you would need to integrate from $\frac{\pi}{4}$ to $\frac{\pi}{2}$.

Laurent.
• September 6th 2008, 02:20 AM
dammitpoo
I see it now. Thanks. =)