Surface integral (divergence theorem ) Evaluate: $\int \int \bar{F}\cdot d\bar S $,
$ \bar{F}=\left(2-x^{2}yz+y^{3},xy^{2}z+ye^{z},y^{2}+z-e^{z}\right)$
$\gamma : y^{2}+z^{2}=x^{2} $ between the planes $x=1$ and $x=2$ , The normal pointing away from the $x$-axis. Thanks in advance.
Edit : I did try to solve this for hours , I have problem with the concept of orientation. I will post my attempt when am done , It is still hard for me to write using Latex but I will spend some time writing. I dont have the solution to this problem, and I dont want just the solution I want a way to master all this kind of questions , preparing for my exam. meanwhile I started a bounty to whomever can make me understand how to work these questions out . ( I have no background on physics ) thanks again
 A: Solving the problem
The Divergence Theorem states
$$\oint_{\partial V}\vec{F}\cdot\text{d}\vec{S} = \int_{V}\vec{\nabla}\cdot\vec{F}\text{d}V$$
In our case, $S$ isn't a closed surface, so we'll have to subtract the flux through the surfaces formed on the planes $x=1$ and $x=2$, which we'll call $S_1$ and $S_2$.
$$\int_S \vec{F}\cdot\text{d}\vec{S} = \int_{V}\vec{\nabla}\cdot\vec{F}\text{d}V - \int_{S_1}\vec{F}\cdot\text{d}\vec{S}_1 - \int_{S_2} \vec{F}\cdot\text{d}\vec{S}_2$$
We calculate the divergence of $\vec{F}$:
\begin{align*}
\vec{\nabla}\cdot\vec{F} &= \frac{\partial}{\partial x}(2-x^2 yz + y^3) + \frac{\partial}{\partial y}(xy^2z  ye^z) + \frac{\partial}{\partial z}(y^2 + z +e^z) \\
&=-2xyz +2xyz+e^z+1-e^z\\
&=1
\end{align*}
The triple integral of the constant function $1$ just gives you the volume of the area enclosed by the surface - here, since our surface isn't closed, we first consider the case that it is and then subtract the surface integral through the two surfaces that close our solid, but aren't a part of the surface that we want to consider.
For any value of $x$, $y^2 + z^2 = x^2$ is a circle of radius $x$ about the $x$-axis at that point. Since $x$ ranges from $1$ to $2$, your surface is the surface of the body of rotation formed by rotating the graph of $f(x) = x$ about the x-axis.
We can use the formula for calculating the volume of solids of rotation to compute our surface integral: for the volume of the solid of rotation formed by rotating the graph of $f(x)$ about the $x$-axis on the interval $[a,b]$, we have
$$V =\pi\int_{a}^{b}f^2(x)\text{d}x$$
Therefore, we have:
\begin{align*}
\oint_{\partial S}\vec{F}\cdot\text{d}\vec{S} = V &=\pi\int_1^2 x^2\text{d}x \\
&= \frac{\pi}{3}(8-1) \\
&= \frac{7}{3}\pi
\end{align*}
Now, we consider the surface integral on $S_1$ and $S_2$, starting with $S_1$.
On $S_1$, $x = 1$ and the unit outward normal vector $\hat{n}$ is just $-\hat{x}$. On the surface, $\vec{F}$ is given by
$$\vec{F} = (2 - yz + y^3, y^2 z + e^z,y^2 - z - e^z)$$
Due to the fact that we are integrating over a circle of radius $1$, we transform to polar coordinates with $(y,z)\rightarrow(r\cos\varphi, r\sin\varphi)$. Our surface element is $\text{d}\vec{S} = -r\hat{x}\text{d}r\text{d}\varphi$.
Since the inner product with a unit vector just gives us that component of a vector field, we have:
\begin{align*}
\int_{S_1}\vec{F}\text{d}\vec{S} &= -\int_0^1\text{d}r\int_0^{2\pi}\text{d}\varphi r(2-r^2\cos\varphi\sin\varphi + r^3\cos^3\varphi)\\
&=-\int_0^1\text{d}r\int_0^{2\pi}\text{d}\varphi r\left[2-\frac{r^2}{2}\sin\left(\frac{\varphi}{2}\right) + r^3\cos\varphi\right]\\
&=-4\pi\int_0^1 r\text{d}r\\
&=-2\pi
\end{align*}
In the second line, the integrals of the trigonometric functions evaluates to $0$ because they are both $2\pi$-periodic.
In the same manner, we treat $S_2$. Here the unit outward normal vector $\hat{n}$ is $\hat{x}$, not $-\hat{x}$, and $x=2$ everywhere.
$$\vec{F} = (2 - 4yz + y^3, 2y^2 z + e^z,y^2 - z - e^z)$$
Once again, we switch to polar coordinates and only consider the $x$-component. We now have:
\begin{align*}
\int_{S_2}\vec{F}\text{d}\vec{S} &=
\int_0^2\text{d}r\int_0^{2\pi}\text{d}\varphi r\left[2-2r^2\sin\left(\frac{\varphi}{2}\right) + r^3\cos\varphi\right]\\
&=4\pi\int_0^2 r\text{d}r\\
&=8\pi
\end{align*}
Putting these results together, we obtain
\begin{align*}
\int_S \vec{F}\cdot\text{d}\vec{S} &= \oint_{\partial V}\vec{F}\cdot\text{d}\vec{S} - \int_{S_1}\vec{F}\cdot\text{d}\vec{S} - \int_{S_2}\vec{F}\cdot\text{d}\vec{S}_2\\
&=\frac{7}{3}\pi - 8\pi + 2\pi\\
&=-\frac{11}{3}\pi
\end{align*}
General remarks on this type of problem
It is hard to speak about a general case, becase a lot of things can happen. You might encounter a surface integral which is easier to solve by evaluating it directly. Sometimes, it might be easier to use the divergence theorem, even if a surface isn't closed, like in our example, if the expression for the divergence is sufficiently nice. Given how easy it is to calculate the divergence for most vector fields that you would encounter on a test, I'd always try calculating it and see if that makes your life easier.
The best general tip that I can give you to always try to sketch your region that you are integrating over. Almost any kind of symmetry that this region has can be exploited to make your life easier - coordinate transforms are often helpful here. Especially on tests, I've found that cleverly choosing coordinates or exploiting symmetries will often make the integration that you have to perform easy.
Beyond that, the only thing you can really do is practice lots of these kind of problems, so that you've seen lots of different cases and approaches by the time you have your exam. Sadly, there isn't one approach that always works best for this type of problem.
A: The divergence of the vector field is $1$. Therefore, by the divergence Theorem
$$
\iint_{\partial D}\overline{F}\cdot\mathrm{d}\overline{S}
=\iiint_D\underbrace{\nabla\cdot\overline{F}}_1\,\mathrm{d}x
$$
The integral on the right is just the volume of $D$, the frustrum of the cone, $y^2+z^2\le x^2$ between $x=1$ and $x=2$. Using $V=\frac\pi3hr^2$:
$$
\frac\pi3\left(2\cdot2^2-1\cdot1^2\right)=\frac{7\pi}3
$$
The integral on the left is the integral in the question plus the integral of $\overline{F}\cdot\mathrm{d}\overline{S}$ over the circular end caps of the frustrum of the cone:
$$
\begin{align}
&\overbrace{\iint_{y^2+z^2\le4}\left(2-4yz+y^3\right)\mathrm{d}y\,\mathrm{d}z}^{x=2\text{ and }\mathrm{d}\overline{S}\,=\,(1,0,0)\,\mathrm{d}y\,\mathrm{d}z}\ \ \overbrace{-\iint_{y^2+z^2\le1}\left(2-yz+y^3\right)\mathrm{d}y\,\mathrm{d}z}^{x=1\text{ and }\mathrm{d}\overline{S}\,=\,(-1,0,0)\,\mathrm{d}y\,\mathrm{d}z}\\[9pt]
&=6\pi
\end{align}
$$
By symmetry, the $yz$ and $y^3$ terms integrate to $0$ over the disks of radius $1$ and $2$. Therefore, using $A=\pi r^2$, the integrals above are $2$ times the area of the circle of radius $2$ ($8\pi$) minus $2$ times the area of the circle of radius $1$ ($2\pi$), that is $6\pi$.
Therefore, the integral in the question is
$$
\frac{7\pi}3-6\pi=\bbox[5px,border:2px solid #C0A000]{-\frac{11\pi}3}
$$
