For a closed surface S in $R^3$, consider the flux of F, given by the usual flux integral. For what choice of S will the flux be maximal?
So, I want to apply the divergence theorem and instead look at the triple integral of the divergence of the given vector field F (over a solid that is enclosed by S.)
Since the flux integral = the divergence integral, I can aim to maximize the divergence integral.
My computation of the divergence gives me -3($x^2 + y^2 + z^2 - \frac{5}{3}$), so this would be the integrand in the triple integral.
It sounds reasonable that in order to maximize this triple integral, I should maximize the integrand. How do I do that?
My attempt was to make $x^2 + y^2 + z^2$ - $\frac{5}{3}$ < 0, since there's a factor of -3 to consider. negative * negative will give me a positive integrand - which would be good for maximizing the flux of F.
Then this tells me that I should choose a sphere of radius $\sqrt{(\frac{5}{3})}$.
I carried out my work and it looks like I got the correct answer.
But I feel like I chose my sphere ...by luck.
How do I know for sure that I've maximized the integrand, simply by making $x^2 + y^2 + z^2$ - $\frac{5}{3}$ < 0? Could I have done even better, achieving a better maximum? I simply knew that this factor had to be < 0, since it was being multiplied by -3.
Thanks,