Question

Evaluate the surface integral. s x ds, s is the part of the plane 18x + 9y + z = 18 that lies in the first octant.

Answer 1

In the first octant, the given plane forms a triangle with vertices corresponding to the plane's intercepts along each axis.

$(x,0,0)\implies 18x+9\cdot0+0=18\implies x=1$
$(0,y,0)\implies 18\cdot0+9y+0=18\implies y=2$
$(0,0,z)\implies 18\cdot0+9\cdot0+z=18\implies z=18$

Now that we know the vertices of the surface $\mathcal S$, we can parameterize it by

$\mathbf s(u,v)=\langle(1-u)(1-v),2u(1-v),18v\rangle$

where $0\le u\le1$ and $0\le v\le1$. The surface element is

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=2\sqrt{406}(1-v)\,\mathrm du\,\mathrm dv$

With respect to our parameterization, we have $x(u,v)=(1-u)(1-v)$, so the surface integral is

$\displaystyle\iint_{\mathcal S}x\,\mathrm dS=2\sqrt{406}\int_{u=0}^{u=1}\int_{v=0}^{v=1}(1-u)(1-v)^2\,\mathrm dv\,\mathrm du=\frac{\sqrt{406}}3$