I have a term $\delta g\int_V dV$, where $\delta g$ is the variation of a continuous function g(x,y,z) and $V$ is an arbitrary volume of integration over some physical 3D body. Moreover, $\delta g=0$ is true on parts of the surface bounding $V$.
My question: Does $\delta g\int_V dV=\int_V \delta g dV$ hold?
My initial reasoning was that it should hold because, since $V$ and $\delta g$ are arbitrary, we can always choose $V$ such that the equality is true whether $\delta g$ is a constant or not.
Needless to say I am not a mathematician so if you are kind enough to answer, and although I usually enjoy technical proofs, please keep it as simple as you can.
Thank you!
Best regards
If I understand your question correctly, you are asking to compare something like $$ f(x) \int_0^1 1 \;dt \quad \text{and} \quad \int_0^1 f(t) dt.$$ I have chosen to use $dt$ instead of $dx$ to emphasize the fact that the thing on the left is a function of $x$, while the thing on the right is a constant. I know the original question was phrased in 3d, but I don't think the dimension is important here.
They are not equal, unless $f(x)$ is actually a constant function $c$.