Volume of a cone using cylindrical coordinates and integration

[jolt527]

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:

$$\int_{0}^{2\pi}\int_{0}^{h}\int_{0}^{R}\rho d\rho dz d\phi$$

...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?

$$V_{cone} = \frac{1}{3}\pi r^2 h$$

Thank you!

 

[HallsofIvy]

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.