[Math] How to find the volume of a cone inside a cylinder using integration

multivariable-calculus

A cone of radius $r$ and height $h$ sits inside a cylinder, $C$, of radius $r$ and height $2h$ in such a way that the axis of the cone and the axis of cylinder $C$ coincide (call this the $z$-axis). The vertex of the cone lies exactly at the center of the bottom circular base of cylinder $C$ (so that the top circular face of the cone to the top circular face of cylinder $C$ forms an upper cylinder (call it $\Sigma$) of radius $r$ and height $h$, and the top circular face of the cone to the bottom circular base of cylinder $C$ forms a lower cylinder (call it $\Omega$) also of radius $r$ and height $h$ ).

  1. The Volume of cylinder $\Omega$ = the Volume of cylinder $\Sigma$ = _____.
  2. The volume of the cone = ______ (re-derive the formula for the Volume of a cone of radius $r$ and height $h$ if you don’t remember it!)
  3. What is the Volume of that part of cylinder $\Omega$ that lies outside of the cone ?
  4. Suppose you forgot the formula for the Volume of a cone of radius $r$ and height $h$. Using Integration, find the Volume of that part of cylinder $\Omega$ that lies outside of the cone by “summing up” the Volumes of infinitesimal washers perpendicular to the $z$-axis which lie inside cylinder $\Omega$ and outside of the cone.

My professor is not the best at explaining and I'm sort of struggling with this problem. Where do I start?

Best Answer

For the volume of the cone, fix a height $z$ between $0$ and $h$, and consider the radius $r(z)$ of the circular slice of the cone at that height. If you "flatten" the picture into two dimensions, you get two similar right triangles :

One has height $h$ and base $r$. The other has height $h-z$ and base $r(z)$. Taking ratios of these, you see that $$ \frac{h}{r} = \frac{h-z}{r(z)} $$ So $$ r(z) = \frac{r(h-z)}{h} $$ Now the area of that slice that you chose is $\pi r(z)^2$, so the area of the "washer" outside this slice is $$ \pi r^2 - \pi r(z)^2 $$ when you "add" up these washers, you get $$ \int_0^h \left [\pi r^2 - \pi \frac{r^2(h-z)^2}{h^2}\right] dz $$ which gives the volume of the region outside of the cone.

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