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I'm trying to calculate the integral of $(x^2 + y^2) \;dx\;dy$ where it is bounded by the positive region where the four curves intersect.

$x^2 - y^2 =2$

$x^2 - y^2 =8$

$x^2 + y^2 =8$

$x^2 + y^2 =16$

So I know I need to do a change of variables and polar looks like a good choice to me. This is where I get lost. So I use $x=r\cos t$, $y=r\sin t$ to change to polar, calculate the Jacobian from $x(r,t)$ and $y(r,t)$ Then do the integral $(x(r,t)^2+y(r,t)^2)$ (jacobian) $drdt$ ? And how do I figure out the limits of integration?

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1 Answer 1

I had a very similar problem when I took calculus. Try this:

Let $ u = x^2 - y^2 $ $ v = x^2 + y^2 $

Now your region of integration is v from 8 to 16 and u from 2 to 8. Now evaluate your Jacobian and throw it into the integral. The integrand in your problem...check it out... is now just u! Sweet. That's what we want. Your Jacobian is the only tricky part, but you'll get it.

When dealing with problems like this, a really good question to ask yourself is, "how can I make my limits of integration easier?" For instance, sometimes converting to polar makes the limits really easy, but keep your mind open to new substitutions, especially when you don't have to integrate from zero to r or some other variable or function. The best possible case is when you can get both of your regions in terms of real values, as we did by letting $ u = x^2 - y^2 $ $ v = x^2 + y^2 $ Now the only tricky part is the Jacobian. If you want a hint with that, let me know.

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I'll give you a hint with the Jacobian. Now we need to find x and y in terms of u and v, right? Here's the hint: find u+v and v-u. Then you get some nice cancellation. Get everything in terms of u and v and you're set. –  N.G. Jan 11 '12 at 18:53
    
I just noticed and changed it. Thanks, D. Mitra! –  N.G. Jan 11 '12 at 18:56
    
I was thinking u=x^2+y^2 and v=x^2-y^2 the first time I tried this. It turned into a big mess when I tried to get x and y in terms of u and v. I'll give it another shot maybe I made a mistake somewhere. –  user1543 Jan 11 '12 at 18:58
    
I'm still pretty stuck at getting x and y in terms of u and v. I take the u and v equations that you gave and solve for x and y, right? What x and y should I get? I keep getting equation for y that are imaginary and that doesn't seem right to me. –  user1543 Jan 11 '12 at 23:51
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