Answer
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {xy + {{\rm{e}}^z}} \right){\rm{d}}S = 3{{\rm{e}}^3} - 6{{\rm{e}}^2} + 3{\rm{e}} + 1$
Work Step by Step
1. Find the equation of the plane determined by the three points: $P = \left( {0,0,3} \right)$, $Q = \left( {1,0,2} \right)$, and $R = \left( {0,4,1} \right)$.
Since the vectors $\overrightarrow {PQ} $ and $\overrightarrow {PR} $ lie in the plane, their cross product is normal to the plane:
$\overrightarrow {PQ} = \left( {1,0,2} \right) - \left( {0,0,3} \right) = \left( {1,0, - 1} \right)$
$\overrightarrow {PR} = \left( {0,4,1} \right) - \left( {0,0,3} \right) = \left( {0,4, - 2} \right)$
${\bf{n}} = \overrightarrow {PQ} \times \overrightarrow {PR} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - 1}\\
0&4&{ - 2}
\end{array}} \right| = 4{\bf{i}} + 2{\bf{j}} + 4{\bf{k}}$
Now, we choose $P = \left( {0,0,3} \right)$ and compute
${\bf{n}}\cdot\left( {x - 0,y - 0,z - 3} \right) = 0$
$\left( {4,2,4} \right)\cdot\left( {x,y,z - 3} \right) = 0$
$4x + 2y + 4\left( {z - 3} \right) = 0$
We conclude that the equation of the plane is $2x + y + 2z = 6$.
2. Parametrize the plane and evaluate $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {xy + {{\rm{e}}^z}} \right){\rm{d}}S$
Using $2x + y + 2z = 6$, we can parametrize the plane as
$G\left( {x,y} \right) = \left( {x,y,\frac{1}{2}\left( {6 - 2x - y} \right)} \right)$
So,
${{\bf{T}}_x} = \frac{{\partial G}}{{\partial x}} = \left( {1,0, - 1} \right)$
${{\bf{T}}_y} = \frac{{\partial G}}{{\partial y}} = \left( {0,1, - \frac{1}{2}} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - 1}\\
0&1&{ - \frac{1}{2}}
\end{array}} \right| = {\bf{i}} + \frac{1}{2}{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {1 + {{\left( {\frac{1}{2}} \right)}^2} + 1} = \sqrt {\frac{9}{4}} = \frac{3}{2}$
Write $f\left( {x,y,z} \right) = xy + {{\rm{e}}^z}$, we obtain $f\left( {G\left( {x,y} \right)} \right) = xy + {{\rm{e}}^{\left( {6 - 2x - y} \right)/2}}$.
From Figure 18, we obtain the domain ${\cal D}$ as the projection of the plane on the $xy$-plane:
${\cal D}$ is bounded by the positive axes and the line $y - 0 = - 4\left( {x - 1} \right)$.
So, the domain description is ${\cal D} = \left\{ {\left( {x,y} \right):0 \le x \le 1,0 \le y \le - 4x + 4} \right\}$.
By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {xy + {{\rm{e}}^z}} \right){\rm{d}}S = \frac{3}{2}\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^{ - 4x + 4} \left( {xy + {{\rm{e}}^{\left( {6 - 2x - y} \right)/2}}} \right){\rm{d}}y{\rm{d}}x$
$ = \frac{3}{2}\mathop \smallint \limits_{x = 0}^1 \left( {\frac{1}{2}x{y^2} - 2{{\rm{e}}^{\left( {6 - 2x - y} \right)/2}}} \right)|_0^{ - 4x + 4}{\rm{d}}x$
$ = \frac{3}{2}\mathop \smallint \limits_{x = 0}^1 \left[ {\frac{1}{2}x{{\left( { - 4x + 4} \right)}^2} - 2{{\rm{e}}^{1 + x}} + 2{{\rm{e}}^{3 - x}}} \right]{\rm{d}}x$
$ = \frac{3}{2}\mathop \smallint \limits_{x = 0}^1 \left( {8{x^3} - 16{x^2} + 8x - 2{{\rm{e}}^{1 + x}} + 2{{\rm{e}}^{3 - x}}} \right){\rm{d}}x$
$ = \frac{3}{2}\left( {2{x^4} - \frac{{16}}{3}{x^3} + 4{x^2} - 2{{\rm{e}}^{1 + x}} - 2{{\rm{e}}^{3 - x}}} \right)|_0^1$
$ = \frac{3}{2}\left( {2 - \frac{{16}}{3} + 4 - 2{{\rm{e}}^2} - 2{{\rm{e}}^2} + 2{\rm{e}} + 2{{\rm{e}}^3}} \right)$
$ = \frac{3}{2}\left( {\frac{2}{3} - 4{{\rm{e}}^2} + 2{\rm{e}} + 2{{\rm{e}}^3}} \right)$
$ = 1 - 6{{\rm{e}}^2} + 3{\rm{e}} + 3{{\rm{e}}^3}$
So, $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} \left( {xy + {{\rm{e}}^z}} \right){\rm{d}}S = 3{{\rm{e}}^3} - 6{{\rm{e}}^2} + 3{\rm{e}} + 1$.
