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When evaluating double integral using polar co-ordinates, does the order of $dr ~ d\theta$ make any difference?

Suppose,

$$\int_0^{\pi/4}\int_0^{\sin\theta} r^2 dr d\theta$$ $$\int_0^{\pi/4}\int_0^{\sin\theta} r^2 d\theta dr$$

Do the above question yield different answers? If the process for solving the above is different can you please explain in general?

Sorry I have no formatting knowledge, so any light here would be helpful for any future questions.

adi rohan
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  • Remark about formatting: I fixed your integrals, you can edit your question to see what I did. Math here is generally written in LaTeX, you can learn it from many resources on the Internet. – Alfonso Fernandez Mar 20 '13 at 16:31
  • In this case, it makes no sense to try to integrate w.r.t. $\theta$ first because the delimiters on $dr$ depend on $\theta$. While it does not make a strict difference, if you want to integrate w.r.t. $\theta$ first, you're going to have to change $\sin\theta$ into something else. – Ian Coley Mar 20 '13 at 16:32
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    If you want to change the order of integration, then the limits of integration have to be changed accordingly. See this problem for instance. – Mhenni Benghorbal Mar 20 '13 at 16:32

1 Answers1

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The first integral makes perfect sense; the second one is nonsensical.

I write my integrals with the $d$ first to eliminate any confusion as to what variable goes with what integral:

$$\int_0^{\pi/4} d\theta \: \int_0^{\sin{\theta}} dr \,r^2$$

Evaluate right to left. Clearly, $r$ depends on $\theta$, and integrate over $\theta$. If you wish to switch the order of integration, you must, as Mhenni points out, redefine your integration region:

$$\int_0^{1/\sqrt{2}} dr \, r^2 \int_{\arcsin{r}}^{\pi/4} d\theta$$

Ron Gordon
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