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#1 2007-02-03 23:35:00

asv
Member
Registered: 2006-01-18
Posts: 5

Triple integral from hell...

Hi all,

I have to evaluate the volume integral ∫∫∫ ( 2 - y²/x² - x²/y² ) dV over the circular cylinder bounded by the planes z=0 and z=1 and the curved surface x²+y²=a² for 0≤z≤1.

The approach I've tried was to integrate w.r.t z first, leaving ∫∫ ( 2 - y²/x² - x²/y² ) dxdy, but that didn't end up getting anywhere because the limits for x and y left things that seemed fairly unintegratable.

Then I tried going polar, giving me ∫∫∫ ( 2 - tan²θ - cot²θ ) dθdrdz = ∫∫ ( 2 - tan²θ - cot²θ ) drdθ. This seems to come out as [4θ-tanθ+cotθ] evaluated between 2pi and 0 (?), but cotθ isn't defined at these values :S

So yeah, I'm kinda stuck dizzy - any help would be greatly appreciated smile

asv smile

Last edited by asv (2007-02-03 23:36:11)

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#2 2007-02-04 16:45:01

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Triple integral from hell...

One thing you are forgetting is that you must multiply by r when switching to polar coordinates.  If you know what a jacobian is, this is it.  Otherwise, just remember the rule that you have to multiply by r.

But that alone does not look like it will allow you to integrate.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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