Let and let S be the cylinder Verify the Divergence Theorem by evaluating as a surface integral and as a triple integral.

Question

Let  and let S be the cylinder  Verify the Divergence Theorem by evaluating  as a surface integral and as a triple integral.

Quizplus · Accepted Answer

By the Divergence Theorem, we have:
$\iint_S\textbf{F}\cdot d\textbf{S}=\iiint_E(\nabla\cdot\textbf{F})dV$
where $\textbf{F}$ is the vector field, $S$ is the boundary surface of the region $E$, and $d\textbf{S}$ and $dV$ are the surface element and volume element, respectively.

In our case, $\textbf{F}=\langle x,y,z\rangle$, and $E$ is the cylinder $11eaa595\_61e3\_558a\_a696\_6f67c19e91ea\_{TB4584\_11}\_{TB4584\_11}$, which can be described as $0\leq\theta<2\pi$, $0\leq z\leq 4$, and $1\leq\rho\leq 2$. Here, $\theta$ is the angle in the $xy$-plane, $z$ is the height, and $\rho=\sqrt{x^2+y^2}$ is the distance from the $z$-axis.

To use the Divergence Theorem, we need to find $\nabla\cdot\textbf{F}$:
$$\nabla\cdot\textbf{F}=\frac{\partial}{\partial x}x+\frac{\partial}{\partial y}y+\frac{\partial}{\partial z}z=3.$$
Therefore, we have:
$$\iiint_E(\nabla\cdot\textbf{F})dV=\int_0^{2\pi}\int_1^2\int_0^4 3\rho dzd\rho d\theta=18\pi.$$

Next, we need to evaluate the surface integral $\iint_S\textbf{F}\cdot d\textbf{S}$, where $S$ is the curved part of the cylinder. To do this, we need to parameterize $S$:
$$\textbf{r}(\theta,z)=\langle\rho\cos\theta,\rho\sin\theta,z\rangle,$$
where $1\leq\rho\leq 2$, $0\leq z\leq 4$, and $0\leq\theta<2\pi$.

The unit normal vector $d\textbf{S}$ to the surface $S$ is given by:
$$d\textbf{S}=\frac{\partial \textbf{r}}{\partial\rho}\times\frac{\partial \textbf{r}}{\partial z}d\rho dz=\langle -\rho\cos\theta,-\rho\sin\theta,1\rangle d\rho dz.$$

Hence, we have:
$$\iint_S\textbf{F}\cdot d\textbf{S}=\int_0^{2\pi}\int_1^2\langle\rho\cos\theta,\rho\sin\theta,z\rangle\cdot\langle -\rho\cos\theta,-\rho\sin\theta,1\rangle\rho d\rho dz d\theta.$$

After simplification, we get:
$$\iint_S\textbf{F}\cdot d\textbf{S}=\int_0^{2\pi}\int_1^2 -\rho^3 d\rho dz d\theta=-\frac{15}{2}\pi.$$
Therefore, by the Divergence Theorem, we have:
$$\int_S\textbf{F}\cdot d\textbf{S}=\iiint_E(\nabla\cdot\textbf{F})dV=18\pi,$$
which is consistent with the surface integral above.

Let and Let S Be the Cylinder Verify