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In polar/cylindrical coordinates, does $r dr d\theta=r d\theta dr$? For example, I believe the integral for the area of a half-circle is given by $$\int_0^1\int_0^{\pi}{rd\theta dr}.$$

What would this integral be if the order were reversed? I find it hard to visualize taking $dr$ first.

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2 Answers 2

The order does not matter except when you evaluate it by multiple integrals you are supposed to be evaluating the inner $dx_{i}$ first, then the second integral indexed by $dx_{j}$, etc.

So for your example, the area of the half circle is the same as $$\int_{0}^{\pi}\left(\int_{0}^{R}rdr\right)d\theta=\int^{\pi}_{0}\frac{1}{2}R^{2}\theta d\theta=\frac{\pi}{2}R^{2}$$

Or $$\int^{R}_{0}\left(\int^{\pi}_{0}d\theta\right)rdr=\int^{R}_{2}\pi rdr=\frac{\pi}{2}R^{2}$$

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You have an $r$ as the upper limit of the $r$ integral, which strikes me as a bit confusing. –  Muphrid Mar 10 '13 at 21:01
    
Note that you can get appropriately sized parentheses by preceding them with \left and \right, respectively. –  joriki Mar 10 '13 at 21:08
    
I tried but did not work well. Can you help to fix it? –  Bombyx mori Mar 10 '13 at 21:10
    
I've changed (... ) with \left(... \right) –  Américo Tavares Mar 10 '13 at 21:15
    
I see. I tried $\left \right$. Thanks. –  Bombyx mori Mar 10 '13 at 21:16

The order of integration is not important when the other integration variables do not appear in the limits, so usually you can rearrange the integrals to your liking. If, however, a function of $\theta$ appeared in the $r$-integral's limits, you would be forced to do the $r$ integral first.

I'm not sure why you find it hard to do the $r$ integral first, though. Doing the $\theta$ integral first is like tracing out the whole circle at a fixed radius and then integrating over all radii. Doing the $r$ integral first just traces out a straight line that you then integrate around a full circle.

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The way you visualize it makes more sense for cylindrical, thanks! The way I've been visualizing is a little different. For example, consider a triangle bounded by $y=0,x=1,y=x$. If I integrate over $y$ first, then in my mind I'm getting rid of the $y$ dimension, so now I just have a line from $x=0$ to $x=1$. –  Caleb Jares Mar 10 '13 at 20:58

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