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?