How to find the volume of a cone inside a cylinder using integration? 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 $h$ ).


*

*The Volume of cylinder $\Omega$ = the Volume of cylinder $\Sigma$ = _____.

*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!)

*What is the Volume of that part of cylinder $\Omega$ that lies outside of the cone ?

*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?
 A: 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.
A: First convert the equation of the slant into cylindrical coordinates.
Start with the observation that axes are arbitrary. Next we need to find the equation of the line in a cylinder with the height on the Z axis. $${Z=mX +H}$$ Where H is the Z intercept (the apex of the cone)
Change the X axis to ${\rho}$ then the slope of the line is $${(H/R)}$$ Where H is the apex of the cone and R is the radius of the base. (the limit of the base on ${\rho}$)
THen the equation of the slant is $${Z=(H/R)\rho + H}$$ IF we assume the slope is negative then the equation becomes $${Z=H-(H/R)\rho}$$
Now we can integrate $${\int_0^R\int_0^{2\pi}\int^H_{(H/R)p}(\rho )d\rho d\theta dz}$$
Integrating over z will give you the equation of the line but the ${\rho}$ keeps you from dragging it out of the integral with respect to ${\rho}$ So now we have $${\int_0^R \int_0^{2\pi} \rho[H-(H/R)\rho]d\rho d\theta}$$ ... $${\int_0^R\int_0^{2\pi}H\rho-(H\rho^2)/R}$$.. $${\int_0^{2\pi} (HR^2)/2-(HR^3)/3R}$$... $${\int_0^{2\pi}[(3HR^2)/6-(2HR^2)/6 = \int_0^{2\pi} HR^2/6}$$... finally $${(2\pi HR^2)/6 = (R^2\pi H)/3}$$
Drawing out your problem will help you to apply it to your problem.
When you have the volume of the cone subtract it from the volume of the cylinder.
