# How to compute this integral by converting it into polar coordinate?

The problem is to convert this integral from $dxdy$ form into $rdrd\theta$ form , please upload the graph also. I have the problem especially on the domain of the $\theta$ when convert from Cartesian coordinate into polar coordinate, thanks in advance for your help. $$\int ^2 _{-2} \int ^3 _{-3} (x^2+y^2)\,dx\,dy$$

I just want to learn how to convert this into polar coordinate and learn how to compute this in polar coordinate as a example

The answer is $$\int_{-\arctan \frac{3}{2}}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r^2 r\,dr\,d\theta,$$ please explain

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I would suggest not using polar coordinates for this... –  David Mitra Dec 11 '11 at 17:22
I second David's remark. This is trivial to do in $x,y$ coordinates and the region is perfectly adapted to that. –  user20266 Dec 11 '11 at 17:33
I think this is course homework. If the borders are explicitly known then I wouldn't use polar coordinates either, but in this case it isn't too bad to carry the (unnecessarily tedious) calculation out with all the details. –  David Heider Dec 11 '11 at 17:36
@DavidHeider - So just assume i just have a problem with this problem in a homework with many question –  Victor Dec 11 '11 at 17:38
@Victor: This sounds pretty much like a homework exercise to me. But I may be mistaken. In this case I apologize. But why do you use such a poorly chosen example to learn polar coordinates? Why don't you use polar coordinates and calculate e.g. the area of an ellipsis, or a ball with a function defined in it? –  David Heider Dec 11 '11 at 18:05

The reason people are suggesting not doing this in polar coordinates is that although the integrand is nice in those coordinates, the region of integration is not. Converting the integral $\int ^2 _{-2} \int ^3 _{-3} (x^2+y^2)dxdy=\int \int r^2 r\;dr\;d\theta$ gives you something easy, but expressing the region of integration is tougher. I would break it into four pieces, with one being $\int_{-\arctan \frac{3}{2}}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r^2 r\;dr\;d\theta$. This represents the triangle $(0,0), (2,-3), (2,3)$

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May i ask you how do you express that into polar coordinate? –  Victor Dec 11 '11 at 18:00
Use the transformations I have included in my edited article. –  David Heider Dec 11 '11 at 18:07
@Victor: I don't understand what needs expressing in polar coordinates. The triangle is represented by its sides: the two rays at angles $\pm \arctan \frac{3}{2}$ and the line $r \cos \theta=2$, which is $x=2$. The $r$ integral is easy. I didn't try the $\theta$ one. –  Ross Millikan Dec 11 '11 at 19:38

Check Wikipedia for the inverse transformation of polar coordinates. Note, that $x^2+y^2=r^2$, and that $rdrd\theta = dxdy$. By using the inverse transformation, you can easily find the upper and lower bound for the integral for both $r,\theta$. I would like to refer you to Wikipedia for that as it is rewarding to carry this calculation out on one's own.

The inverse transformations are given by:

$r=\sqrt{x^2+y^2}, \theta = \arctan\left(y/x\right)$ (Watch the signs!!!).

Now simply find the maximal and minimal values of both$r,\theta$. However, I'd advise you not to use polar coordinates. The region is well-adapted, actually, it is a square. But since you want to perform this tedious calculation...

Try to make use of the inverse transformation (Frankly, I don't know the result in polar coordinates) and try to calculate this integral in polar coordinates. Then however, calculate it in Cartesian coordinates. If you reach the same result you're done.

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May you show me how to compute it in polar coordinate, actually i am not a college student yet, so i would like to learn it for my own interest –  Victor Dec 11 '11 at 17:36
As I said before, check Wikipedia for the inverse transformation. If you learn it for your own interest, then you should have a high level of frustration tolerance. OK, I'll edit my answer. –  David Heider Dec 11 '11 at 17:56