# Integration using polar coordinates

I have to solve this exercise I will describe but I am facing some problems.

$$\int\int_\tau(x+y)\;dx\;dy$$ where $$(x-2)^2 + y^2 \leq 4$$ and $$y\geq0$$

So I am trying to find this integration inside half of a circle.

$\theta$ is from 0 to $\pi/2$ and $r\leq 4 \cos \theta$

$$\int_0^{\pi/2}\int_{4\cos\theta}^0r^2(\cos \theta+\sin\theta)\;dr\;d\theta$$

Something is wrong with the plane of integration cause I am getting wierd results.

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You appear to have inverted the limits on the inner integral. – Brian M. Scott May 10 '12 at 19:19
@BrianM.Scott Yes. I did that because for $\theta=0 r=4$ and for $\theta=\pi/2 r=0$. So you say that the limits are correct but the second one is inverted? – pikapaki May 10 '12 at 19:25
For any fixed value of $\theta$, the inner end of the slice through the area is at $r=0$, and the outer end is at $r=4\cos\theta$, so $r$ ranges from a low of $0$ to a high of $4\cos\theta$. – Brian M. Scott May 10 '12 at 19:32
I may be missing something completely, but I get the integral $\int_{r=0}^2 \int_{\theta=-\pi}^{\pi} (2+r \cos \theta + r \sin \theta) r \; dr d\theta$? – copper.hat May 10 '12 at 19:43
@BrianM.Scott can you please show what you mean with a graph? Also you may want to turn your comments into an answer in order to accept it. Thanks again. – pikapaki May 10 '12 at 19:44

## 2 Answers

Okay, here's a (fairly crude) picture:

It shows a ray from the origin at an angle $\theta$ from the positive $x$-axis. When you calculate the inner integral, intuitively speaking you're 'adding up' the values of the function $x+y$ along that ray, so you want $r$ to range from the smallest radius along that ray to the largest one. The smallest one is at $O$, the origin, and is $0$; the largest is at $P$, where the ray intersects the circle $r=4\cos\theta$, and is $4\cos\theta$. Thus, you should have $$\int_0^{\pi/2}\int_0^{4\cos\theta}r^2(\cos\theta+\sin\theta)\,drd\theta\;.$$

What you did with the inner integral is like trying to find the area under the curve $y=x^2$ between $x=0$ and $x=1$ by calculating $$\int_1^0 x^2\,dx\;.$$

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Thanks a lot for this. That ray thing helped a lot. – pikapaki May 10 '12 at 20:25
+1 I deleted my answer because it was equivalent and yours is better explained. – Américo Tavares May 10 '12 at 21:02

Let $\phi(r,\theta) = (2+r\cos\theta, r\sin\theta)$, then $|J_{\phi}(r,\theta)| = |r|$, and $\phi([0,2],[0,\pi]) = \tau$. Use the change of variables formula to get $$\int_{\tau} f = \int_{\phi^{-1}(\tau)} f \circ \phi |J_{\phi}|.$$ From this you get $$\int_{\tau} x+y \; dx dy = \int_{r=0}^2 \int_{\theta=0}^{\pi} (2+r\cos\theta + r\sin\theta)r \; d\theta dr$$ There is a minor technicality with $\phi$ not being invertible at $r=0$ (ie, $(x,y) = (2,0)$), but this is easy to deal with (the measure of $\{(2,0)\}$ is zero, so we may integrate over $\tau \setminus \{(2,0)\}$ instead; then $\phi$ is invertible at all points of interest).

Here's my crude picture:

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Interesting. I wanted to do something like that or just translate the circle to the 0,0. – pikapaki May 10 '12 at 21:36
Same thing, basically. Generally, exploiting symmetry is a good thing. In this case, the resulting integral is much simpler to evaluate. – copper.hat May 10 '12 at 21:37