# 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.

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
$$\displaystyle\int\int_{{S}}{f}{d}{s}=-\int\int_{{S}}{F}{d}{s}$$
$$\displaystyle=-{\left(-{36}\pi\right)}$$
$$\displaystyle={36}\pi$$

### Relevant Questions

Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({\cos{{\left({z}\right)}}}+{x}{y}^{{2}}\right)}{i}+{x}{e}^{{-{{z}}}}{j}+{\left({\sin{{\left({y}\right)}}}+{x}^{{2}}{z}\right)}{k}$$
S is the surface of the solid bounded by the paraboloid $$\displaystyle{z}={x}^{{2}}+{y}^{{2}}$$ and the plane z = 9.
Let C be the ellipse contained in the xy plane whose equation is $$\displaystyle{4}{x}^{{2}}+{y}^{{2}}={4}$$, oriented clockwise. The force field F described by $$\displaystyle{F}{\left({x},{y},{z}\right)}={x}^{{2}}{i}+{2}{x}{j}+{z}^{{2}}{k}$$, moves a particle along C in the same direction as the curve orientation, performing a W job. C as the surface boundary S: $$\displaystyle{z}={4}-{4}{x}^{{2}}-{y}^{{2}},{z}\ge{0}$$ (with ascending orientation, that is, the component in the z direction equal to 1) and assuming $$\displaystyle\pi={3.14}$$, we can state what:
a) It is not necessary to apply Stokes' Theorem, as C is a closed curve and therefore W = 0.
b) Inverting the orientation of the surface S, we can apply Stokes' Theorem and conclude that W = 12.56.
c) We can apply Stokes' Theorem and conclude that W = 6.28
d) We can apply Stokes' Theorem and conclude that W = 12.56.
Use the Divergence Theorem to calculate the surface integral F · dS, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={3}{x}{y}^{{2}}{i}+{x}{e}^{{z}}{j}+{z}^{{3}}{k}$$,
S is the surface of the solid bounded by the cylinder $$\displaystyle{y}^{{2}}+{z}^{{2}}={9}$$ and the planes x = −3 and x = 1.
Divergence Theorem for more general regions Use the Divergence Theorem to compute the net outward flux of the following vector fields across the boundary of the given regions D.
$$\displaystyle{F}={\left\langle{z}-{x},{x}-{y},{2}{y}-{z}\right\rangle}$$, D is the region between the spheres of radius 2 and 4 centered at the origin.
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{5}{\cos{{\left({y}\right)}}},{8}{\sin{{\left({y}\right)}}}\right\rangle}$$.Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
(Use symbolic notation and fractions where needed.)
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$ ?
Let $$\displaystyle{F}{\left({x},{y}\right)}={\left\langle{4}{\cos{{\left({y}\right)}}},{2}{\sin{{\left({y}\right)}}}\right\rangle}$$. Compute the flux $$\displaystyle\oint{F}\cdot{n}{d}{s}$$ of F across the boundary of the rectangle $$\displaystyle{0}\le{x}\le{5},{0}\le{y}\le\frac{\pi}{{2}}$$ using the vector form of Green's Theorem.
$$\displaystyle\oint{F}\cdot{n}{d}{s}=$$?
Use the Divergence Theorem to calculate the surface integral $$\displaystyle\int\int_{{S}}{F}·{d}{S}$$, that is, calculate the flux of F across S.
$$\displaystyle{F}{\left({x},{y},{z}\right)}={\left({x}^{{3}}+{y}^{{3}}\right)}{i}+{\left({y}^{{3}}+{z}^{{3}}\right)}{j}+{\left({z}^{{3}}+{x}^{{3}}\right)}{k}$$, S is the sphere with center the origin and radius 2.
$$\displaystyle{F}={\left\langle{x}{\sin{{y}}},-{\cos{{y}}},{z}{\sin{{y}}}\right\rangle}$$ , S is the boundary of the region bounded by the planes x = 1, y = 0, $$\displaystyle{y}=\frac{\pi}{{2}},{z}={0}$$, and z = x.
Consider the vector field $$\displaystyle{F}={\left\langle{5}{z},{x},{5}{y}\right\rangle}$$ and the surface which is the part of the elliptic paraboloid $$\displaystyle{z}={x}^{{2}}+{5}{y}^{{2}}$$ that lies below the plane z = 5. Calculate curl(F) and then apply Stokes' Theorem to compute the exact magnitude of the flux of curl(F) through the surface using line integral. You do not need to cinfirm your answer by evaluating the double integral of curl(F) over the surface(the right-hand side of Stokes' Theorem).