1. **Problem Statement:** Find the total charge contained in the rectangular parallelepiped volume defined by $0 < x < 2$, $0 < y < 3$, $0 < z < 5$ where the electric field is given by $$\vec{E} = \frac{1}{\epsilon_0} \left[ 4 a y \hat{a}_x + 2 (x^2 + z^2) a y \hat{a}_y + 4 y z a^2 \hat{a}_z \right] \text{ V/m}.$$ 2. **Relevant Formula:** The total charge $Q$ inside a volume $V$ can be found using Gauss's law in differential form: $$\nabla \cdot \vec{E} = \frac{\rho}{\epsilon_0}$$ where $\rho$ is the volume charge density. The total charge is then: $$Q = \int_V \rho \, dV = \epsilon_0 \int_V (\nabla \cdot \vec{E}) \, dV.$$ 3. **Calculate the divergence $\nabla \cdot \vec{E}$:** Given $$\vec{E} = \frac{1}{\epsilon_0} \left[4 a y \hat{a}_x + 2 (x^2 + z^2) a y \hat{a}_y + 4 y z a^2 \hat{a}_z \right],$$ we compute $$\nabla \cdot \vec{E} = \frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z}.$$ Calculate each partial derivative: - $E_x = \frac{4 a y}{\epsilon_0}$, so $$\frac{\partial E_x}{\partial x} = 0$$ (no $x$ dependence). - $E_y = \frac{2 a y (x^2 + z^2)}{\epsilon_0}$, so $$\frac{\partial E_y}{\partial y} = \frac{2 a (x^2 + z^2)}{\epsilon_0}.$$ - $E_z = \frac{4 a^2 y z}{\epsilon_0}$, so $$\frac{\partial E_z}{\partial z} = \frac{4 a^2 y}{\epsilon_0}.$$ Thus, $$\nabla \cdot \vec{E} = 0 + \frac{2 a (x^2 + z^2)}{\epsilon_0} + \frac{4 a^2 y}{\epsilon_0} = \frac{2 a (x^2 + z^2) + 4 a^2 y}{\epsilon_0}.$$ 4. **Find the charge density $\rho$:** From Gauss's law, $$\rho = \epsilon_0 (\nabla \cdot \vec{E}) = 2 a (x^2 + z^2) + 4 a^2 y.$$ 5. **Calculate total charge $Q$ by integrating $\rho$ over the volume:** $$Q = \int_0^2 \int_0^3 \int_0^5 \left[ 2 a (x^2 + z^2) + 4 a^2 y \right] dz \, dy \, dx.$$ Break the integral into two parts: $$Q = 2 a \int_0^2 \int_0^3 \int_0^5 (x^2 + z^2) dz \, dy \, dx + 4 a^2 \int_0^2 \int_0^3 \int_0^5 y \, dz \, dy \, dx.$$ 6. **Evaluate the integrals:** - First integral: $$\int_0^5 (x^2 + z^2) dz = x^2 \int_0^5 dz + \int_0^5 z^2 dz = 5 x^2 + \left[ \frac{z^3}{3} \right]_0^5 = 5 x^2 + \frac{125}{3}.$$ - Integrate over $y$: $$\int_0^3 dy = 3.$$ - Integrate over $x$: $$\int_0^2 (5 x^2 + \frac{125}{3}) dx = 5 \int_0^2 x^2 dx + \frac{125}{3} \int_0^2 dx = 5 \left[ \frac{x^3}{3} \right]_0^2 + \frac{125}{3} (2) = 5 \times \frac{8}{3} + \frac{250}{3} = \frac{40}{3} + \frac{250}{3} = \frac{290}{3}.$$ - So the first integral part is: $$2 a \times 3 \times \frac{290}{3} = 2 a \times 290 = 580 a.$$ - Second integral: $$\int_0^5 dz = 5,$$ $$\int_0^3 y dy = \left[ \frac{y^2}{2} \right]_0^3 = \frac{9}{2},$$ $$\int_0^2 dx = 2.$$ - So the second integral part is: $$4 a^2 \times 5 \times \frac{9}{2} \times 2 = 4 a^2 \times 5 \times 9 = 180 a^2.$$ 7. **Total charge:** $$Q = 580 a + 180 a^2.$$ **Final answer:** $$\boxed{Q = 580 a + 180 a^2}.$$