# In polar/cylindrical coordinates, does $r dr d\theta=r d\theta dr$?

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

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

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