Volume of a cone using cylindrical coordinates and integration

  • Thread starter jolt527
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  • #1
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Hi all! I was trying to figure out how to find the volume of a cone with radius R and height h using integration with cylindrical coordinates. I first tried to set the the integral as:

[tex]\int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{R}\rho d\rho dz d\phi[/tex]

...but I think that this is setting up the integral for a cylinder and not a cone. Any suggestions so I end up with the correct volume?

[tex]V_{cone} = \frac{1}{3}\pi r^2 h[/tex]

Thank you!
 

Answers and Replies

  • #2
HallsofIvy
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Draw a picture. Assuming the cone has base of radius R and height h, in the xz-plane, it will be a triangle with vertices at (-R, 0), (R, 0), and (0, h). The slant side will be a line from (R,0) to (0,h) and that has equation z= h- (h/R)x= h(1- x/R). Rotating that around the z-axis to get a cone, "x" becomes "r": z= h(1- r/R). Your z integral should go from 0 to that.
 

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