# Thread: Area with Double Integral

1. ## Area with Double Integral

Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles x^2 + y^2 = 100 and x^2 - 10x +y^2 = 0

2. If we convert the second circle to polar we have

$x^{2}+y^{2}-10x=0$

$r^{2}-10rcos{\theta}=0$

$r=10cos{\theta}$

The equation of the first circle is $r=10$

The area between them is:

$\frac{1}{2}\int_{0}^{\frac{\pi}{2}}[10^{2}-(10cos{\theta})^{2}]d{\theta}$