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,\frac hau\right\rangle`

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


`\mathrm dS=\|\mathbf s_u*\mathbf s_v\|\,\mathrm du\,\mathrm dv=\sqrt{1+(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+(h^2)/(a^2)}\int_(v=0)^(v=2\pi)\int_(u=0)^(u=a)u\,\mathrm du\,\mathrm dv=a^2\sqrt{1+(h^2)/(a^2)}\pi=a√(a^2+h^2)\pi`