x; x=xi+yj+zk; r^2= x^2+y^2+z^2; D = ball of radius a and ceter at the origin. I

Question

x; x=xi+yj+zk; r^2= x^2+y^2+z^2; D = ball of radius a and ceter at the origin.

I figured out the right side of the equation is 4a5 . with n=x/a for x that exist in a body of D ....

Work:

double int r2x (dot) n ds= double integral (r^2x^2)/a ds then double integral r^4/a ds = a^3 double integral ds and the surface area of a sphere is 4pia^5....

I can't figure out the left side of teh equation....I know you have to eventually switch to polar coordinates to make is easier?? and take the partial derifivaties of F....

Explanation / Answer

The x-component of F is

Fx=r^2x=(x^2+y^2+z^2)x=x^3+xy^2+xz^2

So dFx/dx=3x^2+y^2+z^2

By symmetry dFy/dy=3y^2+x^2+z^2, dFz/dz=3z^2+x^2+y^2

So

divF=dFx/dx+dFy/dy+dFz/dz=5(x^2+y^2+z^2)

We use spherical coordinates

On the sphere , is between 0 and a, is between 0 and , is between 0 and 2

x^2+y^2+z^2=2

So divF dV=

52 *2 sin() ddd=

54 sin() ddd=

(note we multiplied by 2 sin() because that''s what you do always when you transform to spherical)

The bounds for integral are is between 0 and a, is between 0 and , is between 0 and 2

I let you show this is indeed 4a5

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