Volume of a Cone -- Stuck On My Approach I'd like to calculate the volume of a right circular cone via my way.
If I have a right-triangle with base $D$ and height $H$ then its area is $\frac{1}{2}HD$. Now if we imagine rotating this shape through space about the $z$ axis then we know it will trace out the cone I desire. Appropriate integration will give me this volume:
$$\int \frac{1}{2}HD $$
The hard part, obviously, is filling in the details. So I suppose the best place to start is looking at $dx dy dz$, $$\int \int \int \frac{1}{2}HD dx dy dz$$ I'm imagining rotation about the $z$ axis such that $dz =0$ so we can simplify our integration a little: $$\int \int \frac{1}{2}HD dx dy$$
Now how about finding $dx$ and $dy$... If we imagine the following scenario:

then we can obviously approximate $dx = -y d\theta$ and $dy=xd\theta$ but in our case $y=H$ and $D=x$ such that $$\int_0^{2\pi}\int_0^{2\pi} \frac{1}{2}HD (-H d\theta) ( D d\theta)$$
$$\int_0^{2\pi}\int_0^{2\pi} -\frac{1}{2}H^2 D^2 d\theta d\theta $$
but this doesn't seem to work. Can you tell me where I went wrong and how I can fix this so that it will work how I intend?
 A: Unless you are pursuing this multiple integral approach for a reason, why not just use the method of disks to compute the volume of the cone? (Video #28 here is a great resource.)
The triangle passes through the point $(0,H)$ and $(D,0)$; the line between these points is $y=-\frac{H}{D}x+H$, or equivalently, $x=-\frac{D}{H}y+D$ (this is the form we will need in a moment).
Then, using the method of disks, the volume of the cone is given by $$V=\int_0^H \pi\cdot(\text{radius})^2\,dy=\int_0^H \pi \left(-\frac{D}{H}y+D\right)^2\,dy=\frac{1}{3}\pi D^2H,$$ which is the well-known formula for the volume of a right circular cone of base radius $D$ and height $H$.
A: 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 $${\int_0^R\int_0^{2\pi}\int^H_{(H/R)p}(\rho )d\rho d\theta dz}$$ (don't forget the jacobian)
Integrating over z will give you the equation of the slant but the ${\rho}$ keeps you from dragging it out of the integral with respect to ${\rho}$ So now we have $${\int_0^R \int_0^{2\pi} \rho[H-(H/R)\rho]d\rho d\theta}$$ ... $${\int_0^R\int_0^{2\pi}H\rho-(H\rho^2)/R d\rho d\theta}$$.. $${\int_0^{2\pi} (HR^2)/2-(HR^3)/3R d\theta}$$... $${\int_0^{2\pi}[(3HR^2)/6-(2HR^2)/6 = \int_0^{2\pi}( HR^2/6) d\theta}$$... finally $${(2\pi HR^2)/6 = ( \pi R^2 H)/3}$$
A: It may be easier to see ( at least it was for me) if you inscribe the cone in a unit- sphere since the height is equal to the radius, the triangle integrates over ${2\pi}$ much nicer.
