Sunday

November 23, 2014

November 23, 2014

Posted by **drwls** on Sunday, August 5, 2007 at 2:15pm.

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!!!!!!

Direct computation:

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:

12 pi/5

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

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