Given the following triple integral:
$-\iiint_D (x^2 + y^2 + z^2 -4) \,dV$
How can I find the closed surface out of which the above integral is maximal? I am using the divergence theorem to calculate flux through a surface.
2026-04-11 03:27:51.1775878071
Find maximizing region for triple integral
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The integral is maximized for the largest set $\subseteq \mathbb{R}^3$ for which $f(x,y,z) = -(x^2+y^2+z^2-4)$ remains positive, i.e., when $$ x^2+y^2+z^2\le 4 $$ which is indeed the closed ball of radius $2$ centered at the origin.