If you have a symmetric object like a cylinder in a uniform electric field (either parallel or perpendicular to axis), you can easily see flux would be zero due to symmetry.
Suppose now you put a non symmetric object like a hemisphere in a unform electric field (field is paralle to axis), flux should still be zero.? whatever line of force enter the hemishere has to somehow leave the hemisphere so net should be zero. The actual answer for it is not zero it is π R squared E.
Best Answer
The main difference between the two cases you bring up is that the flux is zero through a closed surface like a cylinder since what comes in must come out in a uniform field, adding up to no flux overall. The answer is different for a surface that is not closed such as a sheet of paper or a hemisphere, since what goes through the surface never reemerges through "the other side," adding up to nonzero flux.
Let the radius of the hemisphere be $R$, and assume that it sits upon the top-half of the $x,y$-plane in $3D$ space. Suppose that the uniform field $\vec E$ points upwards in the $z$-direction. Then, the flux through the hemisphere is exactly the same as the flux through the "opening" of the hemisphere, that is the disk of radius $R$ sitting in the $x,y$-plane, since what comes in through that disk must go through the hemisphere. Hence the flux through the hemisphere $\phi_H$ is the same as the flux through the disk $\phi_D$ of area $A$, which is $$ \phi_D = \vec E\cdot \vec A = E\cdot(\pi R^2). $$ In general, to determine the flux $\phi$ through a surface $S$ with a nonuniform field, we employ a so-called vector surface integral: $$ \phi = \iint_S \vec E\cdot d\vec S. $$ It's something to keep in mind.
Just something else to add to the already complete discussion, note that the flux through the closed surface that is the disk plus the hemisphere is zero. This is always true for closed surfaces in uniform fields. In addition, what comes in through one surface exits through the other. In the case of the closed surface that is the disk plus the hemisphere, the flux $\phi$ is $$ 0 = \phi = \phi_D + \phi_H $$ so $\phi_D = -\phi_H$, which makes sense since the flux comes "in" through the disk, and leaves, or "enters negatively" (it's a ridiculous way of putting it, I'm aware) through the hemisphere. This is because $\phi_H$ is actually negative, indicating that there is a net flux out of the surface through the hemisphere. So the positive flux $\phi_D$ through the disk is the same as the negative of the negative flux $\phi_H$ through the hemisphere. In other words,
what comes in must come out.