Question

Find ∬xz ds, where s is the triangle with vertices (1,0,0), (0,1,0), and (0,1,1).

Answer 1

Parameterize the triangle
`\mathcal S` by the vector-valued function
`\mathbf s(u,v)` with


`\mathbf s(u,v)=(1-u)((1-v)(1,0,0)+v(0,1,0))+u(0,1,1)`

`\implies\mathbf s(u,v)=(\underbrace{(1-u)(1-v)}_(x(u,v)),\underbrace{v(1-u)+u}_(y(u,v)),\underbrace{u}_(z(u,v)))`

where
`0\le u\le1` and
`0\le v\le1`. Then the integral becomes


`\displaystyle\iint_(\mathcal S)xz\,\mathrm dS=\int_(v=0)^(v=1)\int_(u=0)^(u=1)x(u,v)z(u,v)\left\|(\partial\mathbf s)/(\partial u)*(\partial\mathbf s)/(\partial v)\right\|\,\mathrm du\,\mathrm dv`

`=\displaystyle\sqrt2\int_(v=0)^(v=1)\int_(u=0)^(u=1)(1-u)^2u(1-v)\,\mathrm du\,\mathrm dv=\frac1{12\sqrt2}`