How can I show that volume of cone of heigh $h$ and base radius $a$ is equal to $\frac{\pi a^2h}{3}$ using the following formula:
$$V=\frac{1}{3}\cdot \int x\,\mathrm{d}y\,\mathrm{d}z + y\,\mathrm{d}x\,\mathrm{d}z + z \,\mathrm{d}x\,\mathrm{d}y$$
Let $S$ be a cone:
$$ V=\frac{1}{3}\cdot \int_S x\,\mathrm{d}y\,\mathrm{d}z + y\,\mathrm{d}x\,\mathrm{d}z + z \,\mathrm{d}x\,\mathrm{d}y= \frac13\int_S \langle x,y,z\rangle.n \,\mathrm{d}S $$
I can't use the divergence theorem as I don't have the equation of the cone with me.
I clearly have parametrise the surface of cone so as to solve this surface integral, but I am unable to find a nice parametrisation.
Can we take $z=h$ and so $\mathrm{d}z=0$?
This does give me the desired result as follow but I don't know if this is logical.
Let $R:x^2+y^2=a^2$
$$ V=\frac13\cdot \int_S x\,\mathrm{d}y\,\mathrm{d}z + y\,\mathrm{d}x\,\mathrm{d}z + z \,\mathrm{d}x\,\mathrm{d}y = \frac{1}{3}\cdot \int_R h \,\mathrm{d}x\,\mathrm{d}y = \frac{h}3\cdot \pi a^2 $$
But can we do this?