Answer
$ \Phi = 3a^2\pi$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = (x - z)\mathbf{i} + (y - x)\mathbf{j} + (z - y)\mathbf{k}\) across the cylindrical solid bounded by the surface \(\sigma: \{x^2 + y^2 = a^2, z = 0, z = 1\}\) is given by \[ \Phi = \iint \mathbf{F} \cdot \mathbf{n} \, ds = \iiint \nabla \cdot \mathbf{F} \, dV \] where we have used the Divergence Theorem. We can note that \[ \sigma: \left\{ \begin{array}{l} (\mathbf{r}) \, \mid \, -a \leq x \leq a, \\ -a^2 - x^2 \leq y \leq a^2 - x^2, \\ 0 \leq z \leq 1 \end{array} \right\} \] is the region bounded by \(\sigma\). Using cylindrical coordinates, this can be written as \[ \sigma: \left\{ \begin{array}{l} (r, \theta, z) \, \mid \, 0 \leq \theta \leq 2\pi, \\ 0 \leq r \leq a, \\ 0 \leq z \leq 1 \end{array} \right\} \] Since \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(x - z) + \frac{\partial}{\partial y}(y - x) + \frac{\partial}{\partial z}(z - y) = 3 \] the flux result is \[ \Phi = \iiint \nabla \cdot \mathbf{F} \, dV = 3\iiint \, dV = 3\int_0^{2\pi} d\theta \int_0^a r\, dr \int_0^1 dz = 3(2\pi)\left(\frac{a^2}{2}\right)(1) = 3a^2\pi \] Finally, \[ \text{flux} = \Phi = 3a^2\pi \]
