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.
Best Answer
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}