Let
\[V\]
be a region enclosed by a surface \[S\]
.If
\[\mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k}\]
, and let \[f=f(x,y,z)\]
be continuous with first and second partial derivatives. Then\[\int \int \int_V \frac{\nabla^2 f}{r} dV +A = \in \int_S (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) \cdot \mathbf{n} dS ( \left\{ \begin{array}{cc} 0 & (0,0,0) \notin S \\ 4 \pi & (0,0,0) \in S \end{array} \right. \]
where
\[A = \left\{ \begin{array}{cc} 0 & (0,0,0) \notin S \\ 4 \pi f(0,0,0) & (0,0,0) \in S \end{array} \right.
\]
Proof
Apply the Divergence Theorem
\[\int \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV = \int \int_S \mathbf{F} \cdot \mathbf{n} dS\]
with \[\mathbf{F} = \frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})\]
to give\[\begin{equation} \begin{aligned} \int \int_S (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) \cdot \mathbf{n} dS &=
\int \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{\nabla} f}{r} - f \mathbf{\nabla}(\frac{1}{r})) dV \\ &=
\int \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{\nabla} f}{r}) - \mathbf{\nabla} \cdot (f \mathbf{\nabla}(\frac{1}{r})) dV \\ &=
\end{aligned} \end{equation}\]
Use the identities
\[\mathbf{\nabla} (f \mathbf{F}) = \mathbf{F} \cdot (\mathbf{\nabla} f) + f \mathbf{\nabla} \cdot \mathbf{F}\]
\[\mathbf{\nabla} r^n = n r^{n-2} \mathbf{r}\]
The volume integral becomes
\[\begin{equation} \begin{aligned}\int \int \int \int_V \mathbf{\nabla} \cdot \mathbf{F} dV &= \int \int \int_V (\mathbf{\nabla} f) \cdot (\mathbf{\nabla} (\frac{1}{r})) +\frac{1}{r} \nabla^2 f - \cdot (\mathbf{\nabla} (\frac{1}{r})) \cdot (\mathbf{\nabla } f)-f \nabla^2 (\frac{1}{r}) dV \\ &= \int \int \int_V \frac{1}{r} \nabla^2 f 0 f \mathbf{\nabla} \cdot ( \mathbf{\nabla} (\frac{1}{r})) dV \\ &= \int \int \int_V \frac{ \nabla^2 f} {r} dV + \int \int \int_V f \mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} dV\end{aligned} \end{equation} \]
\[\mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} =0\]
everywhere except the origin so the second integral is zero if the origin is outside \[S\]
.If the origin is inside
\[S\]
, since \[\mathbf{\nabla} \cdot \frac{\mathbf{r}}{r^3} =0\]
except the origin we can write\[\int \int \int_V f \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) dV = f(0,0,0) \int \int \int_V \mathbf{\nabla} \cdot (\frac{\mathbf{r}}{r^3}) dV = 4 \pi f(0,0,0) \]
Using Gauss's Theorem.