Flux through the curved surface of the cylinder in the first octant. I am given a vector field $\vec{F}=z\hat{i}+x\hat{j}+y\hat{k}$.
The flux of the vector field passing through the curved surface of the cylinder $x^2+y^2=a^2$ in the first octant and below $Z=h$ plane is calculated to be :- $\frac{1}{2}ah(a+h)$ by me, but the answer discussed by the fellow mates is coming somewhat to be $h^2$.
I need your opinion on this.
If I am wrong, what possible mistake I would be making. 
 A: I'm getting the same result as you.
Parameterize the surface $S$ using cylindrical coordinates as $(\phi, z) \mapsto (a\cos\phi, a\sin\phi, z)$ on $\left[0, \frac{\pi}2\right] \times [0,h]$.
The normal vector is clearly $\vec{n}(\phi,z) = (a\cos\phi, a\sin\phi,0)$ so
\begin{align}
\int_S \vec{F}\cdot d\vec{A} &= \int_{\left[0, \frac{\pi}2\right] \times [0,h]} \vec{F}(a\cos\phi, a\sin\phi, z)\cdot\vec{n}(\phi,z) \;dz\,d\phi\\
&= \int_{\phi = 0}^{\frac\pi2} \int_{z=0}^h (z, a\cos\phi, a\sin\phi)\cdot (a\cos\phi, a\sin\phi,0)\;dz\,d\phi\\
&\int_{\phi = 0}^{\frac\pi2} \int_{z=0}^h (az \cos\phi + a^2\cos\phi\sin\phi) \;dz\,d\phi\\
&= \frac12 ah^2+\frac12 a^2h\\
&= \frac12 ah(a+h)
\end{align}
A: You can also use the Stokes theorem to get the same result. Notice that for the vector field $G(x,y,z) = \left(xz, xy-\frac{z^2}2, 0\right)$ we have $\vec{\nabla} \times \vec{G} = \vec{F}$ so we can turn the surface integral over $S$ into a line integral over the boundary $\partial S$:
$$\int_S \vec{F}\cdot d\vec{A} = \int_S (\vec{\nabla} \times \vec{G})\cdot d\vec{A} 
= \int_{\partial S} \vec{G}\cdot d\vec{\ell}$$
The boundary consists of two circular arcs at the bottom and top and two line segments on the sides. The line integral over the two line segments is zero because we can parameterize them by $z \mapsto (0,a,z)$ and $z \mapsto (a,0,h-z)$ with derivatives $(0,0,\pm1)$, and the $z$-component of $G$ is zero.
The only contributions are from the circular arcs. For the bottom one use the parameterization
$$\vec{r}_1(t) = (a\cos t, b\sin t, 0), \quad t \in \left[0, \frac\pi2\right]$$
and for the top one use 
$$\vec{r}_2(t) = (a\sin t, b\cos t, 0), \quad t \in \left[0, \frac\pi2\right]$$
Then the integral is
$$ \int_{\partial S} \vec{G}\cdot d\vec{\ell} = \int_0^{\frac\pi2} G(\vec{r}_1(t))\cdot \vec{r}_1'(t)\,dt + \int_0^{\frac\pi2} G(\vec{r}_2(t))\cdot \vec{r}_2'(t)\,dt = \frac12ah(a+h)$$
