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*\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*\vec r_v\|=√(\sin^2v+(-\cos v)^2+u^2)=√(1+u^2)`

So the integral is

`\displaystyle\iint_Sy\,\mathrm dS=\int_0^\pi\int_0^6u\sin v√(1+u^2)\,\mathrm du\,\mathrm dv`

`=\displaystyle\left(\int_0^\pi\sin v\,\mathrm dv\right)\left(\int_0^6u√(1+u^2)\,\mathrm du\right)`

`=\frac23(37^(3/2)-1)`