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i have this double integral: $$ I=\int \int_{R} (x+y),\;\; R=\left \{ (x,y):\frac{x^{2}}{3} \leq y\leq 3,\; -1\leq x\leq 3\right \} $$

and this is the domain of integration NOT in polar coordinates:enter image description here

i don't see any radial simmetry, so how can i switch to polar coordinates?

EDIT: the question is: What's the best way to handle such problems, when you're asked to switch in polar coordinates but with a "not-radial like" domain?

Psykomantis
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    That doesn't look like one I'd want to switch to polar coordinates. It's got none of the features that would make polar coordinates desirable. – Dan Uznanski Jan 09 '15 at 14:58
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    The question is rather: Why should you switch to polar coordinates? That does not seem to be giving anything here. Just do it via iterated integration instead. – mickep Jan 09 '15 at 14:58
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    I need to do that integral and i have to represent the domain with polar coordinates, my question is, how to switch in polar coordinates with difficult domains. Thanks. – Psykomantis Jan 09 '15 at 15:20
  • "need to"? What silly person decided that? Well, if you really are stuck with it: you either have four pieces (one, plus one for each corner as you sweep around), or you can displace the whole thing so -3, 1 is your new center and have a single piece from there, though that still doesn't make this thing all that sensible in polar. – Dan Uznanski Jan 09 '15 at 15:25
  • This is easy to do by hand in rectangular coordinates. I got $\frac{928}{45}$. If you really have to do polar maybe you could first do a substitution that moves the upper left corner of your domain to the origin. Then switching to polar should be easier. – Wintermute Jan 09 '15 at 15:35
  • my prof is a silly person =) anyway i will try moving the corner! but i think it would be still difficult to represent – Psykomantis Jan 09 '15 at 15:52

1 Answers1

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If you want to write this in polar coordinates, you can draw line segments from the origin to the points $(3,3), (-1,3), \text{ and }(-1,\frac{1}{3})$, thus dividing the region into 4 subregions.

Then you can write the integral as the sum of 4 integrals in polar coordinates using $\hspace{.4 in}x=r\cos\theta \text{ and } y=r\sin\theta$ to get

$\displaystyle\int_0^{\frac{\pi}{4}}\int_0^{3\sec\theta\tan\theta}r(\cos\theta+\sin\theta)r\;dr d\theta+\int_{\frac{\pi}{4}}^{\cos^{-1}(-\frac{1}{\sqrt{10}})}\int_0^{3\csc\theta}r(\cos\theta+\sin\theta)r\;dr d\theta$

$\displaystyle+\int_{\cos^{-1}(-\frac{1}{\sqrt{10}})}^{\cos^{-1}(-\frac{3}{\sqrt{10}})}\int_0^{-\sec\theta}r(\cos\theta+\sin\theta)r\;dr d\theta+\int_{\cos^{-1}(-\frac{3}{\sqrt{10}})}^{\pi}\int_0^{3\sec\theta\tan\theta}r(\cos\theta+\sin\theta)r\;dr d\theta$

user84413
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  • oh god, can you explain at least the first integral? or the one that you prefere, i understand your subregions but how can you find the limits of the regions in polar? – Psykomantis Jan 09 '15 at 16:28
  • Sure, for the first integral I used that $\theta$ varies from 0 to $\frac{\pi}{4}$, since $(3,3)$ has polar coordinates $(3\sqrt{2},\frac{\pi}{4})$, and that $r$ varies from 0 to $3\sec\theta\tan\theta$ since $y=\frac{x^2}{3}\implies r\sin\theta=\frac{r^2\cos^{2}\theta}{3}\implies r=3\sec\theta\tan\theta$. – user84413 Jan 09 '15 at 19:10