1. calc help

Using polar coordinates evaluate the integral which gives the area which lies in the first quadrant between the circles x^2+y^2=144 and x^2-12x+y^2=0.

2. Hello, karnold9!

Using polar coordinates evaluate the integral which gives the area which lies
in the first quadrant between the circles $x^2+y^2\:=\:144\text{ and }x^2-12x+y^2\:=\:0$
Convert the equations to polar form:

. $\begin{array}{cccc}x^2 + y^2\:=\:144 & \Rightarrow & r \:=\:12 & \text{Circle: center (0,0), radius 12}\\
x^2-12x + y^2 \:=\:0 & \Rightarrow & r \:=\:12\cos\theta & \text{Circle: center (6,0), radius 6}\end{array}$

$A \;=\;\frac{1}{2}\int^{\frac{\pi}{2}}_0\left[12^2 - (12\cos\theta)^2\right] \,d\theta \;=\;72\int^{\frac{\pi}{2}}_0\left[1 - \cos^2\!\theta\right]\,d\theta \;=\;72\int^{\frac{\pi}{2}}_0\sin^2\theta\,d\theta
$

. . Got it?