Help me setup and solve double integral in polar coord.

  • #1
RJLiberator
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Homework Statement


The region between sphere x^2+y^2+z^2=3 and the upper sheet of the hyperboloid z^2=x^2+y2+1.

Homework Equations




The Attempt at a Solution



Curve of intersection: We set the two equations equal to each other and find x^2+y^2=1, a circle of radius 1 is the curve of intersection.
I then drew the graph of the two elements here.

Now, I need to set up the actual double integral.
I know my bounds: theta from 0 to 2pi and r from 0 to 1.

I know that using polar coordinates I have to use r*dr*d(theta).

My problem is, I don't know how to set up the rest of the integral. I think it has something to do with f(r,theta)-g(r,theta) but I've been trying various things and nothing seems to be clear/working out.

I think I should do (3-x^2-y^2) - (x^2+y^2+1) as that seems to be the correct order, but how do I switch this to polar coordinates :/.
 

Answers and Replies

  • #2
RJLiberator
Gold Member
1,095
63
After re-looking at some conversion formula's, I may have solved it:

So we do (3-x^2-y^-2)-(x^2+y^2+1) which is upper-lower and we get 2-2x^2-2y^2.
We then convert to polar coordinates and simplify the sin^2(theta)+cos^2(theta) so that we get: 2-2r^2
We can now use this in the integral with the bounds that I posted above, we multiply both by r: 2r-2r^3 and then we simply integrate.
The answer should be V = pi.

:)
 

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