Answer
The total charge is $4 \times {10^{ - 6}}$ coulombs.
Work Step by Step
We have a charge density of $\delta \left( {x,y} \right) = {10^{ - 6}}xy$ and the region $0 \le x \le 10$, $0 \le y \le 4{{\rm{e}}^{ - {x^2}/2}}$.
Using Eq. (1), the total charge is given by
${\rm{total{\ }charge}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} \delta \left( {x,y} \right){\rm{d}}A = \mathop \smallint \limits_{x = 0}^{10} \mathop \smallint \limits_{y = 0}^{4{{\rm{e}}^{ - {x^2}/2}}} {10^{ - 6}}xy{\rm{d}}y{\rm{d}}x$
$ = \frac{{{{10}^{ - 6}}}}{2}\mathop \smallint \limits_{x = 0}^{10} x\left( {{y^2}|_0^{4{{\rm{e}}^{ - {x^2}/2}}}} \right){\rm{d}}x$
$ = \frac{{{{10}^{ - 6}}}}{2}\mathop \smallint \limits_{x = 0}^{10} x\left( {16{{\rm{e}}^{ - {x^2}}}} \right){\rm{d}}x$
$ = 8 \times {10^{ - 6}}\mathop \smallint \limits_{x = 0}^{10} x{{\rm{e}}^{ - {x^2}}}{\rm{d}}x$
$ = - 4 \times {10^{ - 6}}\left( {{{\rm{e}}^{ - {x^2}}}|_0^{10}} \right) = - 4 \times {10^{ - 6}}\left( {{{\rm{e}}^{ - 100}} - 1} \right) \simeq 4 \times {10^{ - 6}}$
So, the total charge is $4 \times {10^{ - 6}}$ coulombs.
