# evaluating a double integral using polar coordinates

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• Jul 28th 2013, 02:46 PM
mrb165
evaluating a double integral using polar coordinates
Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x (D is under the double integral)

so I decided it would be easier to use polar coordinates for this problem. My limits for theta are 0 to Pi/2 and I used pythag to find the first limit of r which is 2, My professor gave us a hint and told us for x^2 + y^2 = 2x , r = 2cos(theta). Is there a way to actually find the other limit or is this just some kind of rule?

I also set up the double integral

Pi/2 2
∫ . . ∫ rcos(theta) *r drd(theta)
0 . . 2cos(theta)

I integrated the r but I think I integrated incorrectly, I got this

Pi/2 . . . . . . . . . . . . . . . . . . 2
∫ (r^2/2)cos(theta) *(r^2/2)| d(theta)
0 . . . . . . . . . . . . . . . . . . . . 2cos(theta)

I'm not sure how I should go about integrating the first r in front of the cosine.
• Jul 28th 2013, 09:20 PM
chiro
Re: evaluating a double integral using polar coordinates
Hey mrb165.

I think that if you are integrating between x^2 + y^2 = 4 and x^2 + y^2 = 2x then shouldn't the limit for r be 2cos(theta) to 2?

Doing 0 to 2cos(theta) should (I think) find the area between the first curve and the x/y axes.
• Jul 29th 2013, 05:41 PM
HallsofIvy
Re: evaluating a double integral using polar coordinates
I find it very difficult to read your math "expressions". You say you want to "Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x".

Okay, in the first quadrant $\theta$ goes from 0 to $\pi/2$. Converting the first circle to polar coordinates again is easy- $r^2= 4$ means that r= 2 (r is never negative).
The other circle is not much harder: $x^2+ y^2$ is again $r^2$ and $2x= 2r cos(\theta)$ so the equation is $r^2= 2rcos(\theta)$ or ( if r is not 0, $r= 2cos(\theta)$. So the double integral is $\int_{\theta= 0}^{\pi/2}\int_{r= 2}^{2cos(\theta)} (r cos(\theta)) r dr d\theta$
• Jul 29th 2013, 07:27 PM
Prove It
Re: evaluating a double integral using polar coordinates
Quote:

Originally Posted by HallsofIvy
I find it very difficult to read your math "expressions". You say you want to "Evaluate the integral given that ∫ ∫ x dA, where D is the region in the first quadrant that lies between the circles x^2 + y^2 = 4 and x^2 + y^2 = 2x".

Okay, in the first quadrant $\theta$ goes from 0 to $\pi/2$. Converting the first circle to polar coordinates again is easy- $r^2= 4$ means that r= 2 (r is never negative).
The other circle is not much harder: $x^2+ y^2$ is again $r^2$ and $2x= 2r cos(\theta)$ so the equation is $r^2= 2rcos(\theta)$ or ( if r is not 0, $r= 2cos(\theta)$. So the double integral is $\int_{\theta= 0}^{\pi/2}\int_{r= 2}^{2cos(\theta)} (r cos(\theta)) r dr d\theta$

I think your r limits should be around the other way...