# Surface integral to Lateral integral?

by KingBigness
Tags: integral, lateral, surface
 P: 94 1. The problem statement, all variables and given/known data I have some working out my lecture gave me to a problem and I don't think I understand part of it. Hoping you could help me. It's using Gauss' Law to find the capacitance of a cylindrical capacitor of length L but this information shouldn't matter for my question. $\lambda=\frac{∂q}{∂l}$ $\oint E \bullet \Delta A = \frac{\sum q}{\epsilon}$ He then jumps to. $E \int \Delta A = \frac{\lambda L}{\epsilon}$ First question, how does he go from the surface integral to the normal integral? Has it got anything to do with removing the dot product? He then changes $E \int \Delta A$ to E(2πrL) I understand this is taking the integral of dA which becomes A, which is the area of a circle, hence 2πr, but how does he then bring the L into play...would this not be the Volume, not the area? Thanks for any help.
 HW Helper P: 3,309 could it be that E is constant in the integral and always makes the same angle with the surface normal? This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals
P: 94
 Quote by lanedance could it be that E is constant in the integral and always makes the same angle with the surface normal? This is generally the best way to make use of Guass's theorem - exploit the symmetry of the problem to simplify any integrals
So when an angle doesn't change a surface integral can be turned into a lateral integral?
Sorry I haven't done much study on surface integrals so I don't know a lot about them. I'll look into them.

HW Helper
P: 3,309

## Surface integral to Lateral integral?

so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A
P: 94
 Quote by lanedance so if the relative angle, and field strength both don't change the scalar result of the vector dot product is just EdA, everywhere over the surface. As E is constant, the integral is just E.A
Perfect sense, thanks so much for that

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