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Question is as follows Evaluate the integral.

$$\iint_S \cfrac{dxdy}{(1+x^{2}+y^{2})^2}$$

$S= (x,y)\in \mathbb{R}^{2}$ such that $ x=0$ or $x>0$

Little lost here $y=-2, x=-1$ and we divide by $0$ in fact we have a whole line of solutions where we cannot evaluate our integral what am i missing? we have covered pretty much everything in a 3rd level calculus class so feel free to take any interpretation you with but it must be solvable by hand.

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How is this a surface integral? It looks like a double improper integral. –  Ron Gordon Mar 22 '13 at 2:14
    
its defiantly improper the book i use has HORIBBLE notation ( i copied it) i haven integrated something maybe 3 times in the last 3 years just a basic push in the right direction and im sure i can make sense of where i should be going –  Faust7 Mar 22 '13 at 2:17
    
Of course $x>0$ or $x=0$ is equivalent to $x\geq 0$. Is this really what you meant? –  1015 Mar 22 '13 at 2:36
    
my understanding of latex is limited so you get a wide range of different ways to same the same thing =) –  Faust7 Mar 22 '13 at 2:36
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2 Answers 2

up vote 3 down vote accepted

Go to polar coordinates $$ x = r\cos \phi \\ y = r\sin \phi $$ Since $x \ge 0$ then $-\frac \pi 2 \le \phi \le \frac \pi 2$ $$ \int\int_S \frac {dx\,dy}{\left( 1+ x^2+y^2 \right)^2} = \int_{-\frac \pi 2}^{\frac \pi 2} \int_0^\infty \frac {rdr\,d\phi}{\left( 1+r^2\right)^2} = \frac \pi 2 \int_0^\infty \frac {d \left(r^2+1 \right)}{\left( 1+r^2\right)^2} = -\frac \pi 2 \left .\left(\frac 1{r^2+1} \right) \right|_0^\infty = \\ = -\frac \pi 2 \left( 0 - 1\right) = \frac \pi 2 $$

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both of you that's a very clever way around the improper integral! i have seen this type of change in coordinates to stay in the positive plane before but i never knew what i was doing! i wish i understood this in first year! –  Faust7 Mar 22 '13 at 2:36
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The integral is equivalent to

$$\int_0^{\infty} dr\,\frac{r}{(1+r^2)^2} \int_{-\pi/2}^{\pi/2} d\theta = \frac{\pi}{2} \int_0^{\infty} du \, \frac{1}{(1+u)^2}=\frac{\pi}{2}$$

Note the integral over $\theta$ corresponds to the right half-plane.

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