\( \int _ { 0 } ^ { 1 } \int _ { 0 } ^ { 2 \pi } \int _ { 0 } ^ { 2 n } 15 x ^ { 2 } z ^ { 2 } d z d y d x \)

To evaluate the integral \[ \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{2n} 15 x^2 z^2 \, dz \, dy \, dx, \] we can break it down into three separate integrals, since the integrand \(15 x^2 z^2\) does not depend on \(y\): \[ \int_{0}^{1} \int_{0}^{2\pi} \int_{0}^{2n} 15 x^2 z^2 \, dz \, dy \, dx = \int_{0}^{1} 15 x^2 \left( \int_{0}^{2\pi} dy \right) \left( \int_{0}^{2n} z^2 \, dz \right) dx. \] 1. **Evaluate the integral with respect to \(y\)**: \[ \int_{0}^{2\pi} dy = 2\pi. \] 2. **Evaluate the integral with respect to \(z\)**: \[ \int_{0}^{2n} z^2 \, dz. \] Using the power rule for integration, we have: \[ \int z^2 \, dz = \frac{z^3}{3} + C. \] Thus, \[ \int_{0}^{2n} z^2 \, dz = \left[ \frac{z^3}{3} \right]_{0}^{2n} = \frac{(2n)^3}{3} - 0 = \frac{8n^3}{3}. \] 3. **Combine the results**: Now we can substitute back into the integral: \[ \int_{0}^{1} 15 x^2 (2\pi) \left( \frac{8n^3}{3} \right) dx = 15 \cdot 2\pi \cdot \frac{8n^3}{3} \int_{0}^{1} x^2 \, dx. \] 4. **Evaluate the integral with respect to \(x\)**: \[ \int_{0}^{1} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{1} = \frac{1^3}{3} - 0 = \frac{1}{3}. \] 5. **Combine everything**: Now we can substitute this back into our expression: \[ 15 \cdot 2\pi \cdot \frac{8n^3}{3} \cdot \frac{1}{3} = 15 \cdot 2\pi \cdot \frac{8n^3}{9}. \] Calculating this gives: \[ \frac{240\pi n^3}{9} = \frac{80\pi n^3}{3}. \] Thus, the final result of the integral is: \[ \boxed{\frac{80\pi n^3}{3}}. \]