Question

Evaluate the surface of the integral rr

Answer 1

The part of the plane $z=18-3x-9y$ in the first octant is one of the faces of a tetrahedron whose vertices correspond exactly to the intercepts of the plane. These are

$3x+9\cdot0+0=18\implies x=6\implies(6,0,0)$
$3\cdot0+9y+0=18\implies y=2\implies(0,2,0)$
$3\cdot0+9\cdot0+z=18\implies z=18\implies(0,0,18)$

We can parameterize this surface with the vector-valued function

$\mathbf s(u,v)=(6(1-u)(1-v),2u(1-v),18v)$


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

$\mathrm dS=\|\mathbf s_u\times\mathbf s_v\|\,\mathrm du\,\mathrm dv=12\sqrt{91}(1-v)\,\mathrm du\,\mathrm dv$

so the surface integral becomes

$\displaystyle\iint_{\mathcal S}e^z\,\mathrm dS=12\sqrt{91}\int_{v=0}^{v=1}\int_{u=0}^{u=1}(1-v)e^{18v}\,\mathrm du\,\mathrm dv$
$=\displaystyle12\sqrt{91}\int_{v=0}^{v=1}(1-v)e^{18v}\,\mathrm dv$
$=\dfrac{12\sqrt{91}}{324}e^{18v}(19-18v)\bigg|_{v=0}^{v=1}$
$=\dfrac{\sqrt{91}}{27}(e^{18}-228)$