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.

Tobias Ali 2021-01-07 Answered
Let v=4xcoszxezcosy,ycosz,ezcosy be the velocity field of a fluid. Compute the flux of v across the surface x2+y2+z2=9 where x > 0 and the surface is oriented away from the origin.
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2abehn
Answered 2021-01-08 Author has 88 answers

Step 1
Let v=<4xcos(z)xezcos(y),ycos(z),ezcos(y)> be the velocityfield of a fluid.Compute the flux of v a cross the surface x2+y2+z2=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=<4xcos(z)xezcos(y),ycos(z),ezcos(y)>.
Now,divergence of v is
<4xcos(z)xezcos(y),ycos(z),ezcos(y)>
÷(v)=x(4xcos(z)xezcos(y))+y(ycos(z))+z(ezcos(y))
=cos(z)+cos(z)ezcos(y)+0+ezcos(y)
=0
Hence,by using the divergence theorem,then
Closedfds=(÷(v))dv
=0
Closedfds=0
Sfds+diskfds=0
Sfds=Sfds
Step 3
The given disk is x = 0 and x2+y2+z2=9y2+z2=9.
Now,the normal to the surface is N=<1,0,0>.
fN=4+xcos(z)+xezcos(y)
At x=0,then
fNx=0=4
diskfds=fNdA
<

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