Answer
(a) $Area\left( S \right) \approx 1.0781$
(b) $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S \approx 0.0981$
Work Step by Step
(a) The surface $z = \ln \left( {5 - {x^2} - {y^2}} \right)$ can be parametrized by $G\left( {x,y} \right) = \left( {x,y,\ln \left( {5 - {x^2} - {y^2}} \right)} \right)$. So,
${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0, - \dfrac{{2x}}{{5 - {x^2} - {y^2}}}} \right)$
${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1, - \dfrac{{2y}}{{5 - {x^2} - {y^2}}}} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{ - \dfrac{{2x}}{{5 - {x^2} - {y^2}}}}\\
0&1&{ - \dfrac{{2y}}{{5 - {x^2} - {y^2}}}}
\end{array}} \right|$
$ = \dfrac{{2x}}{{5 - {x^2} - {y^2}}}{\bf{i}} + \dfrac{{2y}}{{5 - {x^2} - {y^2}}}{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( {\dfrac{{2x}}{{5 - {x^2} - {y^2}}}} \right)}^2} + {{\left( {\dfrac{{2y}}{{5 - {x^2} - {y^2}}}} \right)}^2} + 1} $
$ = \sqrt {\dfrac{{4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}}}{{{{\left( {5 - {x^2} - {y^2}} \right)}^2}}}} = \dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}$
By Theorem 1, the surface area of $S$ is
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$Area\left( S \right) = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}{\rm{d}}y{\rm{d}}x$
Using a computer algebra system, we evaluate the integral and obtain
$Area\left( S \right) \approx 1.0781$
(b) By Eq. (7) of Theorem 1, we have
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} f\left( {x,y,z} \right){\rm{d}}S = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} f\left( {G\left( {x,y} \right)} \right)||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
where $f\left( {x,y,z} \right) = {x^2}{y^3}$.
We have $f\left( {G\left( {x,y} \right)} \right) = {x^2}{y^3}$. So,
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = 0}^1 \left( {{x^2}{y^3}} \right)\left( {\dfrac{{\sqrt {4{x^2} + 4{y^2} + {{\left( {5 - {x^2} - {y^2}} \right)}^2}} }}{{5 - {x^2} - {y^2}}}} \right){\rm{d}}x{\rm{d}}y$
Using a computer algebra system, we evaluate the integral and obtain
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {x^2}{y^3}{\rm{d}}S \approx 0.0981$