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.

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
${\displaystyle <4-x\cos \left(z\right)-x{e}^z\cos \left(y\right),y\cos \left(z\right),e^z\cos \left(y\right)>}$
${\displaystyle ÷\left(v\right)=\frac{\partial }{\partial x}(4-x\cos \left(z\right)-x{e}^z\cos \left(y\right))+\frac{\partial }{\partial y}\left(y\cos \left(z\right)\right)+\frac{\partial }{\partial z}\left(e^z\cos \left(y\right)\right)}$
${\displaystyle =-\cos \left(z\right)+\cos \left(z\right)-e^z\cos \left(y\right)+0+e^z\cos \left(y\right)}$
=0
Hence,by using the divergence theorem,then
${\displaystyle \int {\int }_{Closed}fds=\int \int \int (÷\left(v\right))dv}$
=0
${\displaystyle \int {\int }_{Closed}fds=0}$
${\displaystyle \int {\int }_{S}fds+\int {\int }_{d}iskfds=0}$
${\displaystyle \int {\int }_{S}fds=-\int {\int }_{S}fds}$
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
${\displaystyle f\cdot N{\mid }_{x=0}=-4}$
${\displaystyle \int {\int }_{disk}f\cdot ds=\int \int f\cdot NdA}$