# 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 $v=⟨4-x\mathrm{cos}z-x{e}^{z}\mathrm{cos}y,y\mathrm{cos}z,{e}^{z}\mathrm{cos}y⟩$ 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.
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Step 1
Let $v=<4-x\mathrm{cos}\left(z\right)-x{e}^{z}\mathrm{cos}\left(y\right),y\mathrm{cos}\left(z\right),{e}^{z}\mathrm{cos}\left(y\right)>$ be the velocityfield of a fluid.Compute the flux of v a cross the surface ${x}^{2}+{y}^{2}+{z}^{2}=9$ Where x>0 and the surface is oriented away from the origin.
Step 2
It is given that the velocity of the fluid is
$v=<4-x\mathrm{cos}\left(z\right)-x{e}^{z}\mathrm{cos}\left(y\right),y\mathrm{cos}\left(z\right),{e}^{z}\mathrm{cos}\left(y\right)>$.
Now,divergence of v is
$<4-x\mathrm{cos}\left(z\right)-x{e}^{z}\mathrm{cos}\left(y\right),y\mathrm{cos}\left(z\right),{e}^{z}\mathrm{cos}\left(y\right)>$
$÷\left(v\right)=\frac{\partial }{\partial x}\left(4-x\mathrm{cos}\left(z\right)-x{e}^{z}\mathrm{cos}\left(y\right)\right)+\frac{\partial }{\partial y}\left(y\mathrm{cos}\left(z\right)\right)+\frac{\partial }{\partial z}\left({e}^{z}\mathrm{cos}\left(y\right)\right)$
$=-\mathrm{cos}\left(z\right)+\mathrm{cos}\left(z\right)-{e}^{z}\mathrm{cos}\left(y\right)+0+{e}^{z}\mathrm{cos}\left(y\right)$
=0
Hence,by using the divergence theorem,then
$\int {\int }_{Closed}fds=\int \int \int \left(÷\left(v\right)\right)dv$
=0
$\int {\int }_{Closed}fds=0$
$\int {\int }_{S}fds+\int {\int }_{d}iskfds=0$
$\int {\int }_{S}fds=-\int {\int }_{S}fds$
Step 3
The given disk is x = 0 and ${x}^{2}+{y}^{2}+{z}^{2}=9⇒{y}^{2}+{z}^{2}=9$.
Now,the normal to the surface is $N=<-1,0,0>$.
$f\cdot N=-4+x\mathrm{cos}\left(z\right)+x{e}^{z}\mathrm{cos}\left(y\right)$
At x=0,then
$f\cdot N{\mid }_{x=0}=-4$
$\int {\int }_{disk}f\cdot ds=\int \int f\cdot NdA$
$<$