Question

Evaluate the surface integral. s y ds, s is the helicoid with vector equation r(u, v) = u cos(v), u sin(v), v , 0 ≤ u ≤ 6, 0 ≤ v ≤ π.

Answer 1

Compute the surface element:

$\mathrm dS=\|\vec r_u\times\vec r_v\|\,\mathrm du\,\mathrm dv$

$\vec r(u,v)=(u\cos v,u\sin v,v)\implies\begin{cases}\vec r_u=(\cos v,\sin v,0)\\\vec r_v=(-u\sin v,u\cos v,1)\end{cases}$

$\|\vec r_u\times\vec r_v\|=\sqrt{\sin^2v+(-\cos v)^2+u^2}=\sqrt{1+u^2}$

So the integral is

$\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v\sqrt{1+u^2}\,\mathrm du\,\mathrm dv$

$=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u\sqrt{1+u^2}\,\mathrm du\right)$

$=\dfrac23(37^{3/2}-1)$