Cylindrical coordinate limits

  • Thread starter boneill3
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  • #1
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Main Question or Discussion Point

Hi Guys,

I have been given the coordinates of a cylinder inside a sphere and want to convert to Cylindrical coordinates to compute the volume of the cylinder.

Can you please check the limits and integral I have?

The cylinder is x^2+y^2= 4

sphere = x^2+y^2+z^2= 9

As its a cylinder we have

Limits are 0<= theta <= 2\pi 0<= r <= 2 and

Inside a sphere with limits

sphere = x^2+y^2+z^2= 9

z = sqrt{9-r^2}

So would my integral be:


\int{{0}{2\pi} \int{0}{2} \int{0}{sqrt{9-r^2}} r dz dr d(theta)


regards
 

Answers and Replies

  • #2
arildno
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Why should the lower limit of z be 0?

Inside a sphere with limits

sphere = x^2+y^2+z^2= 9

z = sqrt{9-r^2}
Are you sure about z=sqrt(9-r^2)) is the only limit set upon z by the above equation?
 
  • #3
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Sorry I want to compute the solid bounded above and below by the sphere and inside the cylinder.

I see your point the sphere can be either side of the z axis .

it should be:

int{-sqrt{9-r^2}} {sqrt{9-r^2}} r d(theta)


Is that alright
 
  • #4
arildno
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That's right indeed. :smile:
 
  • #5
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Thanks for your help!
 

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