Step 1

Let \(\displaystyle{v}={<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\) be the velocityfield of a fluid.Compute the flux of v a cross the surface \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={9}\) Where x>0 and the surface is oriented away from the origin.

Step 2

It is given that the velocity of the fluid is

\(\displaystyle{v}={<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\).

Now,divergence of v is

\(\displaystyle{<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\)

\(\displaystyle\div{\left({v}\right)}=\frac{\partial}{{\partial{x}}}{\left({4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}{\cos{{\left({z}\right)}}}\right)}+\frac{\partial}{{\partial{z}}}{\left({e}^{{z}}{\cos{{\left({y}\right)}}}\right)}\)

\(\displaystyle=−{\cos{{\left({z}\right)}}}+{\cos{{\left({z}\right)}}}−{e}^{{z}}{\cos{{\left({y}\right)}}}+{0}+{e}^{{z}}{\cos{{\left({y}\right)}}}\)

=0

Hence,by using the divergence theorem,then

\(\displaystyle\int\int_{{C}}{l}{o}{s}{e}{d}{f}{d}{s}=\int\int\int{\left(\div{\left({v}\right)}\right)}{d}{v}\)

=0

\(\displaystyle\int\int_{{C}}{l}{o}{s}{e}{d}{f}{d}{s}={0}\)

\(\displaystyle\int\int_{{S}}{f}{d}{s}+\int\int_{{d}}{i}{s}{k}{f}{d}{s}={0}\)

\(\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{f}{d}{s}\)

Step 3

The given disk is x = 0 and \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={9}\Rightarrow{y}^{{2}}+{z}^{{2}}={9}\).

Now,the normal to the surface is \(\displaystyle{N}={<}−{1},{0},{0}{>}\).

\(\displaystyle{f}\cdot{N}=−{4}+{x}{\cos{{\left({z}\right)}}}+{x}{e}^{{z}}{\cos{{\left({y}\right)}}}\)

At x=0,then

\(\displaystyle{f}\cdot{N}{\mid}_{{{x}={0}}}=-{4}\)

\(\displaystyle\int\int_{{d}}{i}{s}{k}{f}\cdot{d}{s}=\int\int{f}\cdot{N}{d}{A}\)

\(\displaystyle=\int\int{\left(-{4}\right)}{d}{A}\)

\(\displaystyle={\left(-{4}\right)}\int\int{d}{A}\)

\(\displaystyle={\left(-{4}\right)}\pi{\left({3}\right)}^{{2}}\)

\(\displaystyle=-{36}\pi\)

It implies that

\(\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{F}{d}{s}\)

\(\displaystyle=-{\left(-{36}\pi\right)}\)

\(\displaystyle={36}\pi\)

Let \(\displaystyle{v}={<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\) be the velocityfield of a fluid.Compute the flux of v a cross the surface \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={9}\) Where x>0 and the surface is oriented away from the origin.

Step 2

It is given that the velocity of the fluid is

\(\displaystyle{v}={<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\).

Now,divergence of v is

\(\displaystyle{<}{4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}},{y}{\cos{{\left({z}\right)}}},{e}^{{z}}{\cos{{\left({y}\right)}}}{>}\)

\(\displaystyle\div{\left({v}\right)}=\frac{\partial}{{\partial{x}}}{\left({4}−{x}{\cos{{\left({z}\right)}}}−{x}{e}^{{z}}{\cos{{\left({y}\right)}}}\right)}+\frac{\partial}{{\partial{y}}}{\left({y}{\cos{{\left({z}\right)}}}\right)}+\frac{\partial}{{\partial{z}}}{\left({e}^{{z}}{\cos{{\left({y}\right)}}}\right)}\)

\(\displaystyle=−{\cos{{\left({z}\right)}}}+{\cos{{\left({z}\right)}}}−{e}^{{z}}{\cos{{\left({y}\right)}}}+{0}+{e}^{{z}}{\cos{{\left({y}\right)}}}\)

=0

Hence,by using the divergence theorem,then

\(\displaystyle\int\int_{{C}}{l}{o}{s}{e}{d}{f}{d}{s}=\int\int\int{\left(\div{\left({v}\right)}\right)}{d}{v}\)

=0

\(\displaystyle\int\int_{{C}}{l}{o}{s}{e}{d}{f}{d}{s}={0}\)

\(\displaystyle\int\int_{{S}}{f}{d}{s}+\int\int_{{d}}{i}{s}{k}{f}{d}{s}={0}\)

\(\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{f}{d}{s}\)

Step 3

The given disk is x = 0 and \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={9}\Rightarrow{y}^{{2}}+{z}^{{2}}={9}\).

Now,the normal to the surface is \(\displaystyle{N}={<}−{1},{0},{0}{>}\).

\(\displaystyle{f}\cdot{N}=−{4}+{x}{\cos{{\left({z}\right)}}}+{x}{e}^{{z}}{\cos{{\left({y}\right)}}}\)

At x=0,then

\(\displaystyle{f}\cdot{N}{\mid}_{{{x}={0}}}=-{4}\)

\(\displaystyle\int\int_{{d}}{i}{s}{k}{f}\cdot{d}{s}=\int\int{f}\cdot{N}{d}{A}\)

\(\displaystyle=\int\int{\left(-{4}\right)}{d}{A}\)

\(\displaystyle={\left(-{4}\right)}\int\int{d}{A}\)

\(\displaystyle={\left(-{4}\right)}\pi{\left({3}\right)}^{{2}}\)

\(\displaystyle=-{36}\pi\)

It implies that

\(\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{F}{d}{s}\)

\(\displaystyle=-{\left(-{36}\pi\right)}\)

\(\displaystyle={36}\pi\)