Question

Compute the scalar surface integral (z y) dS where T is the triangle (including its interior) with vertices (0, 0, 2), (0, 1, 0) and (1, 0, 0). g

Answer 1

Parameterize
`T` by

`\vec r(u,v)=(1-v)((1-u)(2\,\vec k)+u\,\vec\jmath)+v\,\vec\imath=v\,\vec\imath+u(1-v)\,\vec\jmath+2(1-u)(1-v)\,\vec k`

with
`0\le u\le1` and
`0\le v\le1`. Take the normal vector to
`T` to be

`(\partial\vec r)/(\partial u)*(\partial\vec r)/(\partial v)=2(v-1)\,\vec\imath+2(v-1)\,\vec\jmath+(v-1)\,\vec k`

with magnitude

`\left\|(\partial\vec r)/(\partial u)*(\partial\vec r)/(\partial v)\right\|=√((2(v-1))^2+(2(v-1))^2+(v-1)^2)=3(1-v)`

Then the integral is

`\displaystyle\iint_Tyz\,\mathrm dS=\int_0^1\int_0^12u(1-u)(1-v)^2(3(1-v))\,\mathrm du\,\mathrm dv`

`=\displaystyle6\int_0^1\int_0^1u(1-u)(1-v)^3\,\mathrm du\,\mathrm dv`

`=\displaystyle6\left(\int_0^1u(1-u)\,\mathrm du\right)\left(\int_0^1(1-v)^3\,\mathrm dv\right)=\boxed{\frac14}`