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?
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 $H$.