What are the Limits for Computing the Volume of a Cylinder Inside a Sphere?

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Discussion Overview

The discussion revolves around computing the volume of a cylinder that is situated inside a sphere, specifically focusing on the limits of integration in cylindrical coordinates. Participants explore the appropriate limits for the integral based on the geometric constraints of the cylinder and sphere.

Discussion Character

  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents the initial limits for the integral based on the cylinder defined by x^2+y^2=4 and the sphere defined by x^2+y^2+z^2=9, suggesting limits of 0 to 2π for θ, 0 to 2 for r, and 0 to √(9-r^2) for z.
  • Another participant questions the choice of the lower limit for z, suggesting that the limits may not be correctly defined solely by z=√(9-r^2).
  • A subsequent reply acknowledges the need to consider the sphere's bounds above and below the z-axis, proposing that the limits for z should be from -√(9-r^2) to √(9-r^2).
  • A later reply confirms the revised limits for z as appropriate.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the initial limits for z, but there is agreement on the revised limits after discussion. The conversation reflects a progression from uncertainty to a more refined understanding of the limits involved.

Contextual Notes

The discussion highlights potential ambiguities in defining limits for integration in cylindrical coordinates, particularly regarding the vertical bounds set by the sphere. The assumptions about the orientation of the cylinder and sphere are not fully resolved.

boneill3
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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
 
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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?
 
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
 
That's right indeed. :smile:
 
Thanks for your help!
 

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