Question

In Exercises \(5-16,\) use the Divergence Theorem to evaluate \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S\) and find the outward flux of \(F\) through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results. $$ \begin{aligned} &\mathbf{F}(x, y, z)=\left(x y^{2}+\cos z\right) \mathbf{i}+\left(x^{2} y+\sin z\right) \mathbf{j}+e^{z} \mathbf{k}\\\ &S: z=\frac{1}{2} \sqrt{x^{2}+y^{2}}, z=8 \end{aligned} $$

Short answer

To find the outward flux, calculate the given triple integral after converting everything to cylindrical coordinates.

Step-by-step solution

Identify the vector field and the solid region

We first identify the vector field \(\mathbf{F}(x, y, z) = \left(x y^{2}+\cos z\right) \mathbf{i}+\left(x^{2} y+\sin z\right) \mathbf{j}+e^{z} \mathbf{k}\) and the solid region \(D\) bounded by \(S_1: z=\frac{1}{2} \sqrt{x^{2}+y^{2}}\) and \(S_2: z=8\).

Write the divergence of the vector field

The divergence of \(\mathbf{F}\) is calculated using the formula \(\nabla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\). This gives us \(\nabla \cdot \mathbf{F} = 2xy + x^{2} + e^{z}\).

Convert to cylindrical coordinates

We change to cylindrical coordinates to make the integral simpler. In cylindrical coordinates, \(x = r\cos\theta, y = r\sin\theta,\) and \(z = z\). Our divergence becomes \(\nabla \cdot \mathbf{F} = 2r^{2}\cos\theta\sin\theta + r^{2}\cos^{2}\theta + e^{z}\). The lower surface in cylindrical coordinates becomes \(z = \frac{1}{2}r\) and upper surface remains \(z = 8\). The region \(D\) in cylindrical coordinates is then \(0 \leq r \leq 16\), \(0 \leq \theta \leq 2\pi\), and \(\frac{1}{2}r \leq z \leq 8\).

Calculate the triple integral

With all the components in place, we can now evaluate the triple integral using the divergence theorem. The divergence theorem states that \(\iint_{S} \mathbf{F} \cdot d\mathbf{S} = \iiint_{D} \nabla \cdot \mathbf{F} \,dV\) which, in this case, yields the triple integral \(\iiint_{D} 2r^{2}\cos\theta\sin\theta + r^{2}\cos^{2}\theta + e^{z} \, dzdrd\theta\). Evaluating this integral should yield the outward flux through the solid surface.