Triple Integral converting from cylindrical to spherical

  • Thread starter mpn17
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  • #1
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Homework Statement



Convert the following integral to an equivalent integral in spherical coordinates.
Do NOT evaluate the integral.

∫∫∫ r^3 dz dr dtheta

limits of integration
pi/4<theta<pi/2
0<r<2
0<z<√(2r-r^2)



Homework Equations



z=pcos(theta)
r^2=x^2 +y^2
p^2=x^2 +y^2 +z^2
theta=theta



The Attempt at a Solution



My problem is not converting the actual integrand but finding the new limits of integration. I realize that the theta limits will stay the same, but I am not sure how to find the rho or phi limits.

I have found through conversions that the new integrand will be

∫∫∫(p^5)sin(phi)^4 dp dphi dtheta
 

Answers and Replies

  • #2
1,796
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That's kina' messy and just cus' he said not doesn't mean you don't have too. First need to get straight the cylindrical and spherical coordinate system. But for starters, just get the answer and then work towards that. You have:

[tex]\int_{\pi/4}^{\pi/2}\int_0^2\int_{0}^{\sqrt{2r-r^2}} f(z,r,\theta) rdzdrd\theta[/tex]

[tex]\int_{\pi/4}^{\pi/2}\int_0^2\int_{0}^{\sqrt{2r-r^2}} \left(r^2\right) rdzdrd\theta\approx 2.15898[/tex]

See, just did it. So anything I come up with in spherical coordinates, has to equal that (numerically in Mathematica or whatever).

Now, if I have a function:

[tex]f(z,r,\theta)=r^2[/tex]

How can I write that in terms of a function in spherical coordinates:

[tex]f(z,r,\theta)=r^2=h(\rho,\theta,\phi)[/tex]

Keep in mind the angle [itex]\phi[/itex] is mearured down from the y-axis where it is zero, to the x-axis where it is [itex]\pi/2[/itex]

So draw some triangles, z, rho, phi and all, and can you compute:

[tex]f(z,r,\theta)=r^2=h(\rho,\theta,\phi)=k(\rho,\phi)[/tex]

I'll get you started. Surely, we can write between the two coordinate systems along a typical triangle in both spaces:

[tex]\rho^2=r^2+z^2[/tex]

for the angle [itex]\pi/2-\phi[/itex]

so from just that relation, can you come up with an expression for [itex]r^2[/itex] in terms of [itex]\rho[/itex] and [itex]\phi[/itex]?
 
  • #3
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After converting r^3 from cylindrical to spherical I attained the value of (p^5)*sin(phi)^4, I still dont understand how to get the new limits of integration though
 
  • #4
LCKurtz
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Normally you would use a combination of algebra and a picture to figure out the limits. I would suggest that you first plot [itex]z =\sqrt{2r-r^2}[/itex] in a z-r plane for r from 0 to 2. Your cylindrical equation doesn't depend on [itex]\theta[/itex] so the surface is that graph rotated through [itex]\frac \pi 4[/itex]. That should help you figure out the [itex]\phi[/itex] limits.

To figure out the [itex]\rho[/itex] limits you need to express your equation in spherical coordinates and solve for [itex]\rho[/itex]. You know the spherical coordinates for z and remember [itex]r =\rho\sin\phi[/itex]. Start by squaring your equation and substituting.
 

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