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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
up vote 3 down vote accepted

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

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