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Triple Integral

  • Thread starter eok20
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Hi, I need to find the volume of the solid that lies above the cone with equation (in spherical coordinates) [tex] \phi = \frac{\Pi}{3} [/tex] and inside the torus with equation [tex] \rho = 4\sin\phi [/tex]. I thought that the bounds are: [tex] 0\leq\rho\leq4\sin\phi[/tex], [tex]\frac{\Pi}{3}\leq\phi\leq\frac{\Pi}{2}[/tex], and [tex]0\leq\theta\leq2\Pi[/tex] but when I evaluated the integral (using Mathematica) of [tex]\rho^2\sin\phi[/tex] (the Jacobian) using these bounds I got the wrong answer. Any help would be greatly appreciated.

Thanks.
 
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[tex]\frac{\Pi}{3}\leq\phi\leq\frac{\Pi}{2}[/tex]

Between Pi/3 and Pi/2 you ahve the volume outside the cone, if you want the volume inside the cone (like water ina coneshaped cup) then you'd want to go from phi = 0 to Pi/3.
 

saltydog

Science Advisor
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The attached plot shows the cross-section of the donut intersecting with the cone. This is what I come up with:

[tex]\int_0^{2\pi}\int_0^{\pi/3}\int_0^{4Sin(\phi)}\rho^2 Sin(\phi)d\rho d\phi d\theta[/tex]

Mathematica reports approx 19.99. What did you get?
 

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Thanks a lot for both of your help. After fixing the bounds of phi i got the correct answer of around 19.99
 

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