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

$\vec r(u,v)=\dfrac23u\cos v\,\vec\imath+\dfrac23u\sin v\,\vec\jmath+u\,\vec k$

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

$\vec r_u\times\vec r_v=-\dfrac23u\cos v\,\vec\imath-\dfrac23u\sin v\,\vec\jmath+\dfrac49u\,\vec k$

(orientation doesn't matter here)

Then the area of $S$ is

$\displaystyle\iint_S\mathrm dA=\iint_S\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$=\displaystyle\frac{2\sqrt{13}}9\int_0^{2\pi}\int_4^5u\,\mathrm du\,\mathrm dv$

$=\displaystyle\frac{4\sqrt{13}\,\pi}9\int_4^5u\,\mathrm du=\boxed{2\sqrt{13}\,\pi}$