Answer
See proof
Work Step by Step
Step 1 Let \(\vec{F}\) be the vector field \[ \vec{F} = \vec{F}(x, y, z) = x\hat{i} + y\hat{j} + z\hat{k} = \vec{r} \] and \(\sigma\) the surface that encloses the box bounded by the planes \(x=y=z=0\) and \(x=y=z=1\). Let \(\sigma_1\): \[ \{ (x, y, z) \mid 0 < x \leq 1, 0 < y \leq 1, z = 0 \} \] \(\sigma_1\) with unit normal \(\hat{n} = -\hat{k}\), the face on the plane \(z=0\), then the flux across this face results in: \[ \Phi = \iint_{\sigma_1} \vec{F} \cdot \hat{n} \, ds = \iint_{\sigma_1} (x\hat{i} + y\hat{j}) \cdot (-\hat{k}) \, ds = 0 \] Analogously, on the face \(z=1\): \(\sigma_2\): \[ \{ (x, y) \mid 0 \leq x \leq 1, 0 \leq y \leq 1 \} \] with unit normal \(\hat{k}\), the flux results in: \[ \Phi = \iint_{\sigma_2} \vec{F} \cdot \hat{n} \, ds = \iint_{\sigma_2} (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{k} \, ds = 1 \] Step 2 Across the faces \(x=y=0\): \(\sigma_3\): \[ \Phi = \iint_{\sigma_3} (y\hat{j} + z\hat{k}) \cdot (-\hat{i}) \, ds = 0 \] and across \(x=y=1\): \(\sigma_4\): \[ \Phi = \iint_{\sigma_4} (x\hat{i} + z\hat{k}) \cdot (-\hat{j}) \, ds = 0 \] and across \(x=z=1\): \(\sigma_5\): \[ \Phi = \iint_{\sigma_5} (y\hat{j} + z\hat{k} + x\hat{i}) \cdot \hat{i} \, ds = 1 \] Finally, \[ \Phi = 0 + 0 + 0 + 1 + 1 + 1 = 3 \] Now, using the Divergence Theorem, the flux is \[ \int_{\vec{F}} \cdot \hat{n} \, d\sigma = \iiint_{G} \nabla \cdot \vec{F} \, dV = 3 \iiint_{G} \, dV = 3(\text{Volume of } G) \] where we have used that \[ \vec{F} = \vec{F}_1 + \vec{F}_2 + \vec{F}_3 \Rightarrow \nabla \cdot \vec{F} = \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 3 \] Since \(G\) is given by \[ G: \{ (x, y, z) \mid 0 \leq x \leq 1, 0 \leq y \leq 1, 0 \leq z \leq 1 \} \] then \[ \Phi = 3(1 - 0)(1 - 0)(1 - 0) = 3 \]
