Question

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Set up the double integral needed to evaluate
∫∫s3 xyzdσ over the surface S, the first octant part of the plane x + 2y +3z = 6.
Simplify the integrand but do not evaluate the integral.

Answer 1

`S` is a triangle with vertices where the plane
`x+2y+3z=6` has its intercepts. These occur at the points (0,0,2), (0,3,0), and (6,0,0). Parameterize
`S` by

`\vec s(u,v)=(1-v)((1-u)(0,0,2)+u(0,3,0))+v(6,0,0)`

`\vec s(u,v)=(6v,3u(1-v),2(1-u)(1-v))`

with
`0\le u\le1` and
`0\le v\le1`. The surface element is

`\mathrm d\sigma=\left\|(\partial\vec s)/(\partial u)*(\partial\vec s)/(\partial u)\right\|\,\mathrm du\,\mathrm dv=6√(14)(1-v)\,\mathrm du\,\mathrm dv`

So the integral is

`\displaystyle\iint_Sxyz\,\mathrm d\sigma=216√(14)\int_0^1\int_0^1uv(1-u)(1-v)^3\,\mathrm du\,\mathrm dv`