Electric flux through nonspherical surphace Here's an explanation from a book that I read:

Now, I have a few problems with this explanation as well as with the "cone method". I guess that all my problems boil down to the concept of infinitesimal.

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*The book says "a normal to $dA$ makes an angle $\phi$ with a radial line from $q$". Now, the problem is that I can take another radial line intersecting $dA$ which will make a different angle with the area.


*The strength of the electric field passing through the projected area ($dA \cos \phi$) is different from $\vec{E}$ because the projected area is closer to the charge, yet the book seems to ignore that. The same goes for the "cone method":

Books calculate proportion between the areas and the radii using the same radius for $dA \cos \phi$ and $dA$ although the distance of the projected area from the charge is different than from the original area to the charge.
So how can I solve these contradictions? Is there a strict mathematical proof that an electric flux through any area inside a cone (regardless of its orientation) is the same?
 A: Well the problem is your concept of infinitesimal. You see the picture with a macroscopic $dA$, it isn't a real situation, it's to make you understand.
You have to do $dA \rightarrow 0$, in this limit :
1-You can't take another radial line, if you do that the different angle $\Delta\phi \rightarrow 0$ in the limit $dA \rightarrow 0$  
2-The field $\vec{E}$ is different for a quantity $\Delta E \rightarrow 0$ in the limit $dA\rightarrow0$
You can do the calculation with macroscopic $dA$ but you will see that the difference will go to 0 in the infinitesimal limit. It's the same when you do surface integrals in a calculus course.
A: @Karozo essentially gave you the right answer - but just to help you convince yourself, I encourage you to write down the equations for the "difference" - the analysis that takes account of $\Delta \phi$ being finite. You will find that some additional terms appear in the answer - just as you were suspecting. However, you will see that as $\Delta \phi$ tends to zero, these additional terms will tend to zero faster than the answer - in other words, they become zero in the limit of infinitesimal angles. And that is precisely the point.
If you had difficulty writing down the geometrical equations let me know - but you will learn a lot more from trying this yourself. And you are already 75% of the way there because you drew a clear diagram.
