(Problem)
Evaluate the surface integral, where $\vec{F}(x,y,z)=(x^2-y^2+y)\vec{i}+2xy\vec{j}+(y^2-z^2)\vec{k}$ and $S$ is the surface of the cylinder $x^2+y^2=1,0 \leqq z \leqq 1$
I solved this question,but I'm worried if the answer is correct.
I would like to check my answer.
(My answer)
Let $V$ be the region enclosed by the surface $S$,and let $\vec{n}$ be the outward pointing unit normal.Then,the following formula holds.
$$\displaystyle\int\int_{S} \vec{F} \cdot \vec{n}\,dS=\int\int\int_{V} div\vec{F}dV$$
When calculating $\operatorname{div}\vec{F}$, it becomes as follows:
\begin{align} \operatorname{div}\vec{F} &=\displaystyle\frac{\delta}{\delta x}(x^2-y^2+y)+\frac{\delta}{\delta y}(2xy)+\frac{\delta}{\delta z}(y^2-z^2)\\ &=2x+2x-2z\\ &=4x-2z \end{align}
Let $s(r,\theta,z)=(r\cos\theta,r\sin\theta,z)$ be the parameterization of $S$.
The determinant of the Jacobian matrix is calculated as follows:
$$\left| \begin{matrix} \frac{dx}{dr} & \frac{dy}{dr} & \frac{dz}{dr} \\ \frac{dx}{d\theta} & \frac{dy}{d\theta} & \frac{dz}{d\theta} \\ \frac{dx}{dz} & \frac{dy}{dz} & \frac{dz}{dz} \\ \end{matrix} \right| = \left| \begin{matrix} \cos\theta & \sin\theta & 0 \\ -r\sin\theta & r\cos\theta & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right|=r$$
Therefore, $dV=dxdydz=rdrd\theta dz.$
\begin{align} \displaystyle\int\int\int_{V} \operatorname{div}\vec{F}dV &=\displaystyle\int\int\int_{V} (4x-2z)dxdydz\\ &=\displaystyle\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{1}(4r\cos\theta-2z)rdrd\theta dz\\ &=\displaystyle\int_{0}^{1}\int_{0}^{2\pi}\left[\frac{4}{3}r^{3}\cos\theta-r^{2}z\right]^{1}_{0}d\theta dz\\ &=\displaystyle\int_{0}^{1}\int_{0}^{2\pi}\left(\frac{4}{3}\cos\theta -z\right)d\theta dz\\ &=\displaystyle\int_{0}^{1}\left[\frac{4}{3}\sin\theta -z\theta \right]_{0}^{2\pi} dz\\ &=\displaystyle\int_{0}^{1}\left(-2\pi z \right) dz\\ &=\displaystyle\left[-\pi z^{2} \right]_{0}^{1}=-\pi \end{align}
I think the answer is $-\pi$.Is this correct?