Evaluate the volume by solving this triple integral: \( \int_{0}^{1} \int_{0}^{4} \int_{0}^{2}(x y z) d z d y d x \)

To evaluate the triple integral \( \int_{0}^{1} \int_{0}^{4} \int_{0}^{2}(x y z) \, dz \, dy \, dx \), we will proceed step-by-step, integrating with respect to \( z \), then \( y \), and finally \( x \). 1. **Integrate with respect to \( z \)**: \[ \int_{0}^{2} (x y z) \, dz = xy \left[ \frac{z^2}{2} \right]_{0}^{2} = xy \left( \frac{2^2}{2} - \frac{0^2}{2} \right) = xy \left( \frac{4}{2} \right) = 2xy. \] 2. **Now, substitute this result into the next integral and integrate with respect to \( y \)**: \[ \int_{0}^{4} (2xy) \, dy = 2x \left[ \frac{y^2}{2} \right]_{0}^{4} = 2x \left( \frac{4^2}{2} - 0 \right) = 2x \left( \frac{16}{2} \right) = 2x \cdot 8 = 16x. \] 3. **Finally, substitute into the last integral and integrate with respect to \( x \)**: \[ \int_{0}^{1} (16x) \, dx = 16 \left[ \frac{x^2}{2} \right]_{0}^{1} = 16 \left( \frac{1^2}{2} - 0 \right) = 16 \cdot \frac{1}{2} = 8. \] Thus, the volume evaluated by the triple integral is \( \boxed{8} \).