Question

Calculate the area of the surface S. S is the portion of the cone (x^2/4)+(y^2/4)=(z^2/9) that lies between z=4 and z=5

Answer 1

Parameterize
`S` by

`\vec r(u,v)=\frac23u\cos v\,\vec\imath+\frac23u\sin v\,\vec\jmath+u\,\vec k`

with
`4\le u\le5` and
`0\le v\le2\pi`. Take the normal vector to
`S` to be

`\vec r_u*\vec r_v=-\frac23u\cos v\,\vec\imath-\frac23u\sin v\,\vec\jmath+\frac49u\,\vec k`

(orientation doesn't matter here)

Then the area of
`S` is

`\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u*\vec r_v\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\frac{2√(13)}9\int_0^(2\pi)\int_4^5u\,\mathrm du\,\mathrm dv`

`=\displaystyle\frac{4√(13)\,\pi}9\int_4^5u\,\mathrm du=\boxed{2√(13)\,\pi}`