
Originally Posted by
Spudwad
So I am just fiddling around with converting iterated integral into polar coordinates and I am just a bit stumped with this problem. I think I have figured it out, but I just wanted to make sure:
Using polar coordinates, evaluate the integral which gives the area which lies in the first quadrant between the circles $\displaystyle x^2+y^2 = 36$ and $\displaystyle x^2-6x+y^2 = 0$
After plugging in $\displaystyle x=rcos\theta$ and $\displaystyle y=rsin\theta$ I ended up with $\displaystyle r^2 = 36$ and $\displaystyle r^2=6rcos\theta$. I simplified them so $\displaystyle r=6$ and $\displaystyle r=6cos\theta$
$\displaystyle \int_{0}^{\pi/2}\int_{6cos\theta}^{6} r^2drd\theta \rightarrow \int_{0}^{\pi/2}\frac{r^3} {3}\mid_{6cos\theta}^{6}d\theta \rightarrow $ $\displaystyle \int_{0}^{\pi/2}72 - 72cos^3\theta d\theta \rightarrow 72\int_{0}^{\pi/2} 1-cos^3\theta d\theta \rightarrow 72 - (72sin\theta - 24sin\theta) \mid_{0}^{\pi/2}$
After plugging in $\displaystyle \pi/2$ for $\displaystyle \theta$ I ended up with 24. Have I done this right then? Thanks for all the help!