14 pts ) The two parts below are not related. (a) Find the volume of the solid in the first octant bounded by the parabolic cylinder \( z=12-3 x^{2} \) and the plane \( y=3 \).

To find the volume of the solid in the first octant bounded by the parabolic cylinder \( z=12-3x^2 \) and the plane \( y=3 \), we need to integrate the volume of the solid over the region defined by the parabolic cylinder and the plane. The volume of the solid can be calculated using the following formula: \[ V = \int_{a}^{b} \int_{c}^{d} \int_{e}^{f} \rho(x,y,z) \, dz \, dy \, dx \] where \( \rho(x,y,z) \) is the density function, and \( a, b, c, d, e, f \) are the limits of integration. In this case, the density function is 1 (since it's a solid with no density variation), and the limits of integration are: - For \( x \): from 0 to \( \sqrt{\frac{12}{3}} = 2 \) - For \( y \): from 0 to 3 - For \( z \): from \( 12-3x^2 \) to 0 Therefore, the volume of the solid can be calculated as: \[ V = \int_{0}^{2} \int_{0}^{3} \int_{12-3x^2}^{0} 1 \, dz \, dy \, dx \] Let's calculate this integral step by step. Evaluate the integral by following steps: - step0: Evaluate using formulas and rules: \(\int_{0}^{2} \int_{0}^{3} \int_{12-3x^{2}}^{0} 1 dz dy dx\) - step1: Evaluate the inner integral: \(\int_{0}^{2} \int_{0}^{3} \left(-12+3x^{2}\right) dy dx\) - step2: Evaluate the inner integral: \(\int_{0}^{2} \left(-36+9x^{2}\right) dx\) - step3: Evaluate the integral: \(\int \left(-36+9x^{2}\right) dx\) - step4: Use properties of integrals: \(\int -36 dx+\int 9x^{2} dx\) - step5: Evaluate the integral: \(-36x+\int 9x^{2} dx\) - step6: Evaluate the integral: \(-36x+3x^{3}\) - step7: Return the limits: \(\left(-36x+3x^{3}\right)\bigg |_{0}^{2}\) - step8: Calculate the value: \(-48\)