Question

Evaluate the surface integral. S x2yz dS, S is the part of the plane z = 1 + 2x + 3y that lies above the rectangle [0, 4] × [0, 2]

Answer 1

Parameterize $S$ by $\vec s(u,v)=(u,v,1+2u+3v)$ with $0\le u\le4$ and $0\le v\le2$. Then the surface element is

$\mathrm dS=\left\|\dfrac{\partial\vec s}{\partial u}\times\dfrac{\partial\vec s}{\partial v}\right\|\,\mathrm du\,\mathrm dv=\sqrt{14}\,\mathrm du\,\mathrm dv$

so that the integral is

$\displaystyle\iint_Sx^2yz\,\mathrm dS=\sqrt{14}\int_0^2\int_0^4u^2v(1+2u+3v)\,\mathrm du\,\mathrm dv=\boxed{\frac{1408\sqrt{14}}3}$