Question

–/1 points scalc8 16.7.005.my notes question part points submissions used evaluate the surface integral. s (x + y + z) ds, s is the parallelogram with parametric equations x = u + v, y = u − v, z = 1 + 2u + v, 0 ≤ u ≤ 4, 0 ≤ v ≤ 1.

Answer 1

You're conveniently given the parameterization of the surface
`\mathcal S`,

`\mathbf s(u,v)=\langle u+v,u-v,1+2u+v\rangle`

where
`0\le u\le4` and
`0\le v\le1`, so all we need to set up and compute the integral is to compute the surface element,
`\mathrm dS`:

`\mathrm dS=\|\mathbf s_u*\mathbf s_v\|\,\mathrm du\,\mathrm dv`

`\mathrm dS=√(14)\,\mathrm du\,\mathrm dv`

So the surface integral is

`\displaystyle\iint_(\mathcal S)x+y+z\,\mathrm dS=√(14)\int_(v=0)^(v=1)\int_(u=0)^(u=4)(u+v)+(u-v)+(1+2u+v)\,\mathrm du\,\mathrm dv`

`=\displaystyle√(14)\int_(v=0)^(v=1)\int_(u=0)^(u=4)1+4u+v\,\mathrm du\,\mathrm dv=38√(14)`