Let $\mathbf{v}=\left\langle x^{3} \cos z, 5-3 x^{2} y \cos z-3 y z^{2} \sin x, z^{3} \sin x\right\rangle$ be the velocity field of a fluid. Compute the flux of $\mathbf{v}$ across the surface $x^{2}+y+z^{2}=1$ where $y>0$ and the surface is oriented away from the origin. Hint: Use the Divergence Theorem.

Step 1 ：Let \(\mathbf{v}=\left\langle x^{3} \cos z, 5-3 x^{2} y \cos z-3 y z^{2} \sin x, z^{3} \sin x\right\rangle\) be the velocity field of a fluid. We are asked to compute the flux of \(\mathbf{v}\) across the surface \(x^{2}+y+z^{2}=1\) where \(y>0\) and the surface is oriented away from the origin.

Step 2 ：We can use the Divergence Theorem to solve this problem. The Divergence Theorem states that the flux of a vector field across a closed surface is equal to the triple integral of the divergence of the field over the volume enclosed by the surface.

Step 3 ：In this case, we need to compute the divergence of the vector field \(\mathbf{v}\), and then integrate it over the volume enclosed by the surface \(x^{2}+y+z^{2}=1\) where \(y>0\).

Step 4 ：The divergence of a vector field \(\mathbf{v}=\left\langle P, Q, R\right\rangle\) is given by \(\nabla \cdot \mathbf{v} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\).

Step 5 ：So, we need to compute the partial derivatives of the components of \(\mathbf{v}\) with respect to \(x\), \(y\), and \(z\), and then sum them up to get the divergence of \(\mathbf{v}\).

Step 6 ：Let's compute the divergence of \(\mathbf{v}\):

Step 7 ：\(P = x^{3}\cos(z)\)

Step 8 ：\(Q = -3x^{2}y\cos(z) - 3yz^{2}\sin(x) + 5\)

Step 9 ：\(R = z^{3}\sin(x)\)

Step 10 ：The divergence of the vector field \(\mathbf{v}\) is zero.

Step 11 ：This means that the flux of \(\mathbf{v}\) across the surface \(x^{2}+y+z^{2}=1\) where \(y>0\) and the surface is oriented away from the origin is also zero, according to the Divergence Theorem.

Step 12 ：Final Answer: The flux of \(\mathbf{v}\) across the surface is \(\boxed{0}\).