Question

Evaluate the surface integral. S x2z2 dS S is the part of the cone z2 = x2 + y2 that lies between the planes z = 4 and z = 5

Answer 1

The surface $S$ can be parameterized in cylindrical coordinates using

$\mathbf r(u,v)=(x(u,v),y(u,v),z(u,v))=(u\cos v,u\sin v,u)$

where $4\le u\le5$ and $0\le v\le2\pi$. The surface integral is then equivalent to

$\displaystyle\iint_Sx^2z^2\,\mathrm dS=\int_{v=0}^{v=2\pi}\int_{u=4}^{u=5}(u\cos v)^2u^2\left\|\mathbf r_u\times\mathbf r_v\right\|\,\mathrm du\,\mathrm dv$
$=\displaystyle\sqrt2\int_{v=0}^{v=2\pi}\int_{u=4}^{u=5}u^5\cos^2v\,\mathrm du\,\mathrm dv$
$=\dfrac{3843\pi}{\sqrt2}$