Volume integral of a gradient

Question

Assuming we have a scalar function f which tends to 0 at the boundary of a space. Is it true that the volume integral of the gradient of f will also tend to 0?

Answer 1

Provided the function is smooth enough, I'd say yes: the volume integral of its gradient is a vector that has three components; to calculate, say, the first component, you can integrate the derivative with respect to $x$ over $x$ first and, taking into account the boundary conditions, obtain zero.

EDIT (7/9/2014): I guess some extra conditions may be necessary to ensure that the relevant multiple integrals can be computed using iterated integrals (http://en.wikipedia.org/wiki/Fubini%27s_theorem ).

Answer 2

You know the Gauss' divergence theorem says that $\int_V \nabla \cdot \vec{A} \rm dV = \int_S \vec{A} \cdot d \vec{S}$

Now taking $\vec{A}= f \hat{c}$, where $\hat{c}$ is a constant vector, you can prove:

$\int_V (\nabla f)\rm dV = \int_S f d \vec{S}$

Now telling f tends zero on the surface will not be sufficient to say that left hand side is zero. You have to say how it tends to zero on the surface.

say your f=$\frac{1}{r}$, then obviously as r$\rightarrow \infty$, f$\rightarrow 0$

but$ \int_S f d \vec{S}$=$ \int_S \frac{1}{r}\hat{r} r^2 \sin (\theta) d\theta d\phi=r \int_S \hat{r} \sin (\theta) d\theta d\phi $ =$r \times 0$

now see you can't say $r \times 0=0$ when r$\rightarrow \infty$. (Also I have assumed perfectly spherical surface so that you can take out r out of the integral. But for any arbitrary surface r=$r(\theta)$. )

so if f decreases faster than $\frac{1}{r^2}$ then only you can say $\int_V (\nabla f)\rm dV $=0