Answer
$\Phi = 24$
Work Step by Step
Step 1: Let \(\vec{F}(x, y, z) = xy\hat{i} + yz\hat{j} + xz\hat{k}\) and \(\sigma\) be the surface bounded by the cube delimited by the planes \(x = 0\), \(y = 0\), \(z = 0\), \(x = 2\), \(y = 2\), and \(z = 2\). On the face \(x = 0\), \(\vec{n} = -\hat{i}\), resulting in: \[ \vec{n}(x=0) = \iint_{\sigma} (xy\hat{j}) \cdot (-\hat{i}) \,dS = 0 \] Step 2: On the face \(x = 1\), \(\vec{F}(1, y, z) = 2y\hat{i} + yz\hat{j} + 2z\hat{k}\). The flux across this face is: \[ \begin{align*} \vec{n}(x=1) &= \iint_{\sigma} (2y\hat{i} + yz\hat{j} + 2z\hat{k}) \cdot \hat{i} \,dS \\ &= \iint_{\sigma} (2y) \,dS \\ &= \int_0^2 \int_0^2 (2y) \,dzdy \\ &= 8 \end{align*} \] Analogously, the flux across the faces \(y = 0\) and \(z = 0\) is zero, and across the faces \(x = 2\), \(y = 2\), and \(z = 2\) is also 8. The total flux across the closed surface \(\sigma\) is: \[ \Phi = 0 + 0 + 0 + 8 + 8 + 8 = 24 \] Using the Divergence Theorem: We can also calculate the flux using the Divergence Theorem. Let \(G\) be the region enclosed by the cube. Then: \[ \begin{align*} \Phi &= \iint_{\sigma} \vec{F} \cdot \vec{n} \,dS \\ &= \iiint_G \nabla \cdot \vec{F} \,dV \end{align*} \] Calculating the divergence: \[ \begin{align*} \nabla \cdot \vec{F} &= \frac{\partial}{\partial x}(xy) + \frac{\partial}{\partial y}(yz) + \frac{\partial}{\partial z}(xz) \\ &= y + z + x \end{align*} \] So, \[ \begin{align*} \Phi &= \iiint_G (y + z + x) \,dV \\ &= \int_0^2 \int_0^2 \int_0^2 (y + z + x) \,dzdydx \\ &= \int_0^2 \int_0^2 \left(\frac{y}{2} + z + x\right) \,dydx \\ &= \int_0^2 \left(2 + 2x\right) \,dx \\ &= 8 + 16 \\ &= 24 \end{align*} \] So, the flux \(\Phi = 24\).
