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

$\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}=2(v-1)\,\vec\imath+2(v-1)\,\vec\jmath+(v-1)\,\vec k$

with magnitude

$\left\|\dfrac{\partial\vec r}{\partial u}\times\dfrac{\partial\vec r}{\partial v}\right\|=\sqrt{(2(v-1))^2+(2(v-1))^2+(v-1)^2}=3(1-v)$

Then the integral is

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

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

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