If I have a function v equal to some integral then the differential is just what ever I'm taking the integral of. But what if my function of V is equal to a volume integral? When I say dV is that just what I'm integrating again?

Question

anonymous · Answer

Sorry, to clarify: the differential of V.

turingtest · Answer

yes but you are phrasing this very odd

turingtest · Answer

$$V=\iiint\limits_E dV=\iiint dxdydz$$

anonymous · Answer

function v and function of V are these meant to be different?

turingtest · Answer

integrating a function over a volume in $\mathbb R^3$ would be$$\iiint_E f(x,y,z)dxdydz\neq V$$unless f(x,y,z)=1
not sure if that answers your question at all

anonymous · Answer

, sorry v and V are supposed to be the same symbol. and I'm declaring V (not to be confused with volume but actually a voltage) to be some volume integral of a function of charge density (normally called rho(r) a scalar function of position as expressed by some vector r from the origin).
so I'm thinking that dV is then just the integrand.