Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
1  
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
show 2 more comments

2 Answers

up vote 1 down vote accepted

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

enter image description here

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 $$\int_1^0 x^2\,dx\;.$$

share|improve this answer
    
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
add comment

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: enter image description here

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.