posted by drwls on .
We are not going to do that work for you, but will be glad to help you.
The unit sphere is the sphere with radius 1 centered in the origin. They give you the coordinates of a vector. Compute the dot product of the vector and the local surface area normal and integrate it over the spherical surface area. Then do it again with Gauss' theorem, which involves a volume integral of the divergence of the vector. They should be equal.
We will be glad to critique your work. There is quite a bit involved.
Compute the flux of the vector field F = (3xy2; 3x2y; z3) out of the unit sphere both
directly and using Gauss' Theorem.
Show all work please!!!!!!
The outward normal on the unit sphere is, of course:
n = (x,y,z)
Take inner product with F:
F dot n =
(3xy^2, 3x^2y, z^3) dot (x,y,z) =
6 x^2y^2 + z^4
Now switch to spherical coordinates:
x = sin(theta) cos(phi)
y = sin(theta) sin(phi)
z = cos(theta)
F dot n =
6 sin^4(theta) cos^2(phi)sin^2(phi)+ cos^4(theta)
Surface element on unit sphere in spherical coordinates is:
sin(theta) d theta d phi
So we have to integrate:
[6 sin^4(theta) cos^2(phi)sin^2(phi)+ cos^4(theta)] sin(theta) d theta d phi
from phi = 0 to 2 pi and theta = 0 to pi.
The integrals are high school level, the result is:
Now using Gauss Theorem, compute the divergence of F:
div F = d(3xy^2)/dx + d(3x^2y)/dy +
d(z^3)/dz = 3(x^2 + y^2 + z^2) = 3 r^2
We must integrate this over the volume of the unit sphere. The integrand does not depend on the angles so, we can write the integral as:
Integral from 0 to 1 of
3 r^2* 4 pi r^2 dr
= 12 pi/5