Question

Use the Divergence Theorem to evaluate the outward flux ∫∫F * ndS of the given vector field across the surface F(x,y,z)=x^3i + (y^3 + xz)j + (z^3 + z^2)K.

x^2 + y^2 + z^2 = a^2, a is greater than zero.

Answer 1

Hmm I don't really remember how to do these :\
I can try to work through it though.

Answer 2

\[\Large \vec F(x,y,z)=\left<x^3,\;y^3+xz,\; z^3+z^2\right>\]The divergence of F will be,\[\Large \nabla\cdot\vec F=\frac{\partial(x^3)}{\partial x}+\frac{\partial(y^3+xz)}{\partial y}+\frac{\partial(z^3+z^2)}{\partial z}\]Which gives us,\[\Large \nabla\cdot\vec{F}=3x^2+3y^2+3z^2+2z\]

And the divergence theorem tells us somethingggggg likeeeeee...\[\Large \Phi\qquad=\qquad\iint\vec F\cdot \vec n\;dS \qquad=\qquad \iiint_G \nabla\cdot\vec F\;dV\]

So we have,\[\Large \iiint_G 3x^2+3y^2+3z^2+2z\;dV\]
Where G is the spherical region that they gave us.
From here we could probably convert to Spherical Coordinates I guess?

Answer 3

\[\Large \iiint_G \color{royalblue}{3x^2+3y^2+3z^2+2z}\;dV\]Spherical coordinates gives us ummmmm,\[\Large \iiint_G \color{royalblue}{(3r^2+2r \cos \theta)}\;r^2 \sin \theta\;dr\;d \theta\;d \phi\]

I dunno if I did any of that correctly or not :3 it's been too long...
Ok that's at least something to chew on.

Answer 4

Awesome thank you.