Let v = << 4 - x cos z - xe^z cos y, y cos z, e^z cos y>> be the velocity field of a fluid. Compute the flux of v across the surface x^2+y^2+z^2=9 where x > 0 and the surface is oriented away from the origin.

Let v = << 4 - x cos z - xe^z cos y, y cos z, e^z cos y>> be the velocity field of a fluid. Compute the flux of v across the surface x^2+y^2+z^2=9 where x > 0 and the surface is oriented away from the origin.

Question
Let \(\displaystyle{v}={\left\langle{4}-{x}{\cos{{z}}}-{x}{e}^{{z}}{\cos{{y}}},{y}{\cos{{z}}},{e}^{{z}}{\cos{{y}}}\right\rangle}\) be the velocity field of a fluid. Compute the flux of v across the surface \(\displaystyle{x}^{{2}}+{y}^{{2}}+{z}^{{2}}={9}\) where x > 0 and the surface is oriented away from the origin.

Answers (1)

2021-01-08
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
\(\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{F}{d}{s}\)
\(\displaystyle=-{\left(-{36}\pi\right)}\)
\(\displaystyle={36}\pi\)
0

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