Answer
See proof
Work Step by Step
Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ be a radial vector. The flux of the vector field $\vec{F}(x, y, z) = \vec{r}$ across the surface $\sigma$ of the solid $G$ is given by: \[ \Phi = \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS = \iiint_G \nabla \cdot \vec{F} \, dV, \] where we have used the Divergence Theorem. Since \[ \nabla \cdot \vec{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3, \] the flux result is: \[ \Phi = \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS = 3\iiint_G \, dV = 3 \text{ volume of } G. \] Thus, the volume of $G$ is: \[ \text{volume of } G = \frac{1}{3} \iint_{\sigma} \vec{F} \cdot \vec{n} \, dS. \] QED
