Problem
\( D \) is the portion of the solid cone \( z=\sqrt{x^{2}+y^{2}} \) between the planes \( z=2 \) and \( z=4 \). Use the Divergence Theorem to find the flux of the vector field \( \mathbf{F}(x, y, z)=\left\langle 4 y^{2} z, \frac{7}{48} y^{3}, \frac{7}{16} x^{2} z\right\rangle \) over the surface of \( D \). Consider cylindrical coordinates. Write the exact answer. Do not round.
Solution
Step 1 ：\(\nabla \cdot \mathbf{F} = 2y^{2} + \frac{21}{16}x^{2} \)
Step 2 ：\( \Phi(\mathbf{F}) = \int_{0}^{2\pi} \int_{2}^{4} \int_{0}^{r} \left(2y^{2} + \frac{21}{16}x^{2}\right)rdzd\theta drdy = \int_{0}^{2\pi} \int_{2}^{4} \int_{0}^{r} \left(2r^{2} + \frac{21}{16}r^{2}\right)rdzdr \)
Step 3 ：\(\Phi(\mathbf{F}) = \int_{0}^{2\pi} \int_{2}^{4} \left(\frac{35}{8}r^{4}\right) dr d\theta = \frac{140}{3}(\pi - 0) = \frac{140\pi}{3}\)
