Step 1

Given:

\(\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{z}{\sin{{y}}}\right\rangle}\), the surface S is the boundary of the region bounded by the planes x = 1, \(\displaystyle{y}=\frac{\pi}{{2}},{z}={0}\), and z = x.

Step 2

Using Divergence theorem

\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}=\int\int_{{S}}{F}\cdot{n}{d}{s}\)

Compute The divergence of the field

\(\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{\sin{{y}}}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}{\sin{{y}}}\right)}\)

\(\displaystyle={\sin{{y}}}\frac{\partial}{{\partial{x}}}{\left({x}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+{\sin{{y}}}\frac{\partial}{{\partial{z}}}{\left({z}\right)}\)

\(\displaystyle={\sin{{y}}}+{\sin{{y}}}+{\sin{{y}}}\)

\(\displaystyle={3}{\sin{{y}}}\)

Step 3

Therefore, the outward flux is

\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\int_{{0}}^{{x}}}{3}{\sin{{y}}}{\left.{d}{z}\right.}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\sin{{y}}}{{\left[{z}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{x}{\sin{{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[\frac{{x}^{{2}}}{{2}}\right]}_{{0}}^{{1}}}{\sin{{y}}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{{\left[-{\cos{{y}}}\right]}_{{0}}^{{\frac{\pi}{{2}}}}}\)

\(\displaystyle=\frac{{3}}{{2}}{\left(-{\cos{{\left(\frac{\pi}{{2}}\right)}}}-{\left(-{\cos{{\left({0}\right)}}}\right)}\right)}\)

\(\displaystyle=\frac{{3}}{{2}}{\left({0}-{\left(-{1}\right)}\right)}\)

\(\displaystyle=\frac{{3}}{{2}}\)

Step 4

Thus, the outward flux of the given field is \(\displaystyle\frac{{3}}{{2}}\).

Given:

\(\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{z}{\sin{{y}}}\right\rangle}\), the surface S is the boundary of the region bounded by the planes x = 1, \(\displaystyle{y}=\frac{\pi}{{2}},{z}={0}\), and z = x.

Step 2

Using Divergence theorem

\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}=\int\int_{{S}}{F}\cdot{n}{d}{s}\)

Compute The divergence of the field

\(\displaystyle\nabla\cdot{F}=\frac{\partial}{{\partial{x}}}{\left({x}{\sin{{y}}}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+\frac{\partial}{{\partial{z}}}{\left({z}{\sin{{y}}}\right)}\)

\(\displaystyle={\sin{{y}}}\frac{\partial}{{\partial{x}}}{\left({x}\right)}+\frac{\partial}{{\partial{y}}}{\left(-{\cos{{y}}}\right)}+{\sin{{y}}}\frac{\partial}{{\partial{z}}}{\left({z}\right)}\)

\(\displaystyle={\sin{{y}}}+{\sin{{y}}}+{\sin{{y}}}\)

\(\displaystyle={3}{\sin{{y}}}\)

Step 3

Therefore, the outward flux is

\(\displaystyle\int\int_{{D}}\int\nabla\cdot{F}{d}{v}={\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\int_{{0}}^{{x}}}{3}{\sin{{y}}}{\left.{d}{z}\right.}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{\sin{{y}}}{{\left[{z}\right]}_{{0}}^{{x}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\int_{{0}}^{{1}}}{x}{\sin{{y}}}{\left.{d}{x}\right.}{\left.{d}{y}\right.}\)

\(\displaystyle={3}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{{\left[\frac{{x}^{{2}}}{{2}}\right]}_{{0}}^{{1}}}{\sin{{y}}}{\left.{d}{y}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{\int_{{0}}^{{\frac{\pi}{{2}}}}}{\left[{\sin{{y}}}{\left.{d}{y}\right.}\right.}\)

\(\displaystyle=\frac{{3}}{{2}}{{\left[-{\cos{{y}}}\right]}_{{0}}^{{\frac{\pi}{{2}}}}}\)

\(\displaystyle=\frac{{3}}{{2}}{\left(-{\cos{{\left(\frac{\pi}{{2}}\right)}}}-{\left(-{\cos{{\left({0}\right)}}}\right)}\right)}\)

\(\displaystyle=\frac{{3}}{{2}}{\left({0}-{\left(-{1}\right)}\right)}\)

\(\displaystyle=\frac{{3}}{{2}}\)

Step 4

Thus, the outward flux of the given field is \(\displaystyle\frac{{3}}{{2}}\).