Infinite long cylinder for Gauss' law

zezima1
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Sometimes you want to find the electric field of a loong cylinder carrying for instance a uniform volume charge. Then for easy calculations you approximate the cylinder with an infinitely long one and then put a gaussian surface over the entire thing. Symmetry then dictates that the field point in the same direction as the area elements of the cylinder and then you can find E. However, the ends of the cylinder are always accounted for in a way that I don't like. Because my solutions manual simply says that E is perpendicular to the area elements here. Why is that? Surely I can see that for an infinite cylinder at each point will have a field pointing radially out, but as soon as you consider what's happening at the end you are more or less actually assuming that the cylinder is finite.

Can someone explain this ambiguity and why the procedure works even though the steps are pointless (in my opinion)?
 
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As you said, it really only works in the approximation that the cylinder looks infinite at a particular field point, i.e. the distance from the end of the cylinder to the field point is much greater than the perpendicular distance. Still, I invite you to calculate the field on a symmetric axis through the cylinder, and compare it to the infinite version. If you're really up for some math, you can try the off axis calculation of the cylinder.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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