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Volume, Stewart ed 5, ch 6.2 # 48

  • Thread starter rocomath
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  • #1
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Check plz :)

A frustum of a right circular cone with height h, lower base radius R, and top radius r.

I chose to lay it upright, revolving it wrt the y-axis.

I have coordinates B(R,0) & b(r,h), thus [tex]y=\frac{h}{r-R}(x-R) \rightarrow x=\frac{y(r-R)}{h}+R[/tex]

[tex]V=\pi\int_0^h\left[\frac{y(r-R)}{h}+R\right]^2dy[/tex]

Is my set up right? I can integrate it myself, no prob there.

Thanks!
 

Answers and Replies

  • #2
Dick
Science Advisor
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Looks right to me. You could just integrate and check it by looking up the volume of a cone frustrum.
 
  • #3
dynamicsolo
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Another way of checking the volume of a frustum is that it is a cone minus a cone taken off the apex of the original cone. Since a cone is a self-similar figure, the ratio of the top radius of the frustum to the base radius tells you the ratio of the height of the cone to be removed to the height of the original cone. As the cone is a three-dimensional figure, the volume of the removed cone will have a ratio to the original cone which is the cube of the ratio of the removed height to the original cone's height. It sounds more complicated in words than what you have to do geometrically. (Self-similarity is a wonderful thing...)

This should give you another way to check your result.
 

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