Answer
$\Phi = \frac{4608}{35}$
Work Step by Step
The flux of the vector field \(\mathbf{F}(x, y, z) = x^3\mathbf{i} + x^2y\mathbf{j} + xy\mathbf{k}\) across the surface of the solid bounded by \(z = 4 - x^2\) and the planes \(y + z = 5\), \(z = 0\), and \(y = 0\) is given by: \[ \Phi = \iiint \nabla \cdot \mathbf{F} \, dV \] Where \(\nabla \cdot \mathbf{F}\) is the divergence of the vector field. We have already calculated that \(\nabla \cdot \mathbf{F} = 4x^2\). The region bounded by the surface is defined as: \[ \begin{align*} \mathcal{D} : \{ (x, y, z) &\mid -2 \leq x \leq 2, 0 \leq y \leq 4 - x^2, 0 \leq z \leq 5 - y \} \\ G : \{ (x, z) &\mid -2 \leq x \leq 2, 0 \leq z \leq 4 - x^2 \} \end{align*} \] Now, let's calculate the flux: \[ \begin{align*} \Phi &= \int_{-2}^{2} \int_{0}^{4-x^2} \int_{0}^{5-y} 4x^2 \, dz \, dy \, dx \\ &= \int_{-2}^{2} \int_{0}^{4-x^2} \left[ 4x^2(5-y) \right] \, dy \, dx \\ &= \int_{-2}^{2} \left[ \frac{4}{3}x^2(5-y)^2 \right]_{0}^{4-x^2} \, dx \\ &= \int_{-2}^{2} \left[ \frac{4}{3}x^2(5-(4-x^2))^2 - 0 \right] \, dx \\ &= \int_{-2}^{2} \left[ \frac{4}{3}x^2(1+x^2)^2 \right] \, dx \\ &= \int_{-2}^{2} \left[ \frac{4}{3}x^2(1+2x^2+x^4) \right] \, dx \\ &= \int_{-2}^{2} \left[ \frac{4}{3}x^2 + \frac{8}{3}x^4 + \frac{4}{3}x^6 \right] \, dx \\ &= \left[ \frac{4}{9}x^3 + \frac{8}{15}x^5 + \frac{4}{21}x^7 \right]_{-2}^{2} \\ &= \left[ \frac{256}{9} + \frac{1024}{15} + \frac{256}{21} \right] - \left[ -\frac{256}{9} + \frac{1024}{15} - \frac{256}{21} \right] \\ &= \frac{4608}{35} \end{align*} \] So, the flux of the vector field across the given surface is \(\Phi = \frac{4608}{35}\).
