# Thread: Polar Coordinates

1. ## Polar Coordinates

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 $x^2+y^2 = 36$ and $x^2-6x+y^2 = 0$

After plugging in $x=rcos\theta$ and $y=rsin\theta$ I ended up with $r^2 = 36$ and $r^2=6rcos\theta$. I simplified them so $r=6$ and $r=6cos\theta$

$\int_{0}^{\pi/2}\int_{6cos\theta}^{6} r^2drd\theta \rightarrow \int_{0}^{\pi/2}\frac{r^3} {3}\mid_{6cos\theta}^{6}d\theta \rightarrow$ $\int_{0}^{\pi/2}72 - 72cos^3\theta d\theta \rightarrow 72\int_{0}^{\pi/2} 1-cos^3\theta d\theta \rightarrow 72 - (72sin\theta - 24sin\theta) \mid_{0}^{\pi/2}$

After plugging in $\pi/2$ for $\theta$ I ended up with 24. Have I done this right then? Thanks for all the help!

2. 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 $x^2+y^2 = 36$ and $x^2-6x+y^2 = 0$

After plugging in $x=rcos\theta$ and $y=rsin\theta$ I ended up with $r^2 = 36$ and $r^2=6rcos\theta$. I simplified them so $r=6$ and $r=6cos\theta$

$\int_{0}^{\pi/2}\int_{6cos\theta}^{6} r^2drd\theta \rightarrow \int_{0}^{\pi/2}\frac{r^3} {3}\mid_{6cos\theta}^{6}d\theta \rightarrow$ $\int_{0}^{\pi/2}72 - 72cos^3\theta d\theta \rightarrow 72\int_{0}^{\pi/2} 1-cos^3\theta d\theta \rightarrow 72 - (72sin\theta - 24sin\theta) \mid_{0}^{\pi/2}$

After plugging in $\pi/2$ for $\theta$ I ended up with 24. Have I done this right then? Thanks for all the help!
No need for an integral, you can just use elementary geometry.

First, it helps to write the circles in general form...

$x^2 + y^2 = 36$, which has radius 6 units centred at the origin, and $x^2 - 6x + y^2 = 0$.

The second can be rewritten as

$x^2 - 6x + (-3)^2 + y^2 = (-3)^2$

$(x - 3)^2 + y^2 = 9$.

So it has radius 3 units centred at (3, 0).

By drawing a graph of the situation, you have the required region as a quarter of the larger circle minus half of the smaller circle.

So $A = \frac{1}{4}\cdot \pi \cdot 6^2 - \frac{1}{2} \cdot \pi \cdot 3^2$

$= 9\pi - \frac{9\pi}{2}$

$= \frac{9\pi}{2}\,\textrm{units}^2$