Question

Use a surface integral to find the general formula for the surface area of a cone with height latex: h and base radius latex: a(excluding the base).

Answer 1

We can parameterize this part of a cone by

$\mathbf s(u,v)=\left\langle u\cos v,u\sin v,\dfrac hau\right\rangle$

with $0\le u\le a$ and $0\le v\le2\pi$. Then

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+\dfrac{h^2}{a^2}}u\,\mathrm du\,\mathrm dv$

The area of this surface (call it $\mathcal S$) is then

$\displaystyle\iint_{\mathcal S}\mathrm dS=\sqrt{1+\frac{h^2}{a^2}}\int_{v=0}^{v=2\pi}\int_{u=0}^{u=a}u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+\frac{h^2}{a^2}}\pi=a\sqrt{a^2+h^2}\pi$