A doubt

If you got Int(dA) and A=xy, which are variables, how would that integration become in terms of x and y?

wat


do you mean dA=xy?

\int dA = A

No matter what "A" is.  lrn2 calculus

>>2
No, I mean A=xy, The integral in term of xy, how would it be?
>>3
You're an idiot.

In this case, Int(dA)=Int(Int(dx))dy), but how is it in a general case?

your question makes no sense

>>5
Why not?

This is the actual exercise:

\int ((K*Cos[x])/(r^2))dA
Where
\r=h/Cos[x]
\A=((R^2)*y)/2
\R=h*Tan[x]
0<y<2*Pi
0<x<a
and K,h are constants.

I have to say I agree with >>5.  What the hell is your notation?

Are you trying to integrate a function of x,y in 2D over some area of the xy plane?  What the hell is \A=((R^2)*y)/2 supposed to mean?  Are you integrating over a curved 2 surface, where the area element depends on the coordinates x,y?

>>8
x and y are angles.
It's just a simple electrostatics exercise, damn.
Anyway, already solved it.

>>9
So you're trying to calculate the surface integral of a vector field in 3D?  I see now.  You could have made that more clear.

>>10
The question remains, how the hell is dA in terms of x and y.
In this case dA=R*dR*dy, A=((R^2)*y)/2 (a circle of A area)
In the case where A=x*y, dA=dx*dy (a square of A area)
But isn't it supposed to be a sum of the partial derivatives?

>>11
Not really... It depends on the coordinates/metric you're using.

The full differential geometry answer is that dA = sqrt(det(metric))

polar coordinates: metric = ds^2 + s^2 dphi^2
dA = s^2 ds dphi

in higher dimensions...
spherical coordinates: metric = dr^2 + r^2 dtheta^2 + (rsin(theta))^2 dphi^2
dV = r^2 sin(theta) dr dtheta dphi

If you're doing electrostatics, you're probably concerned with an integral over a 2D surface in 3D Euclidean space; that is a far more complicated issue.

If you're brute forcing Gauss' law, you'll probably integrate over a spherical shell of radius R in spherical coordinates... except you're not because of

>> \A=((R^2)*y)/2
polar coordinates? surface of a cylinder at fixed z? ???

>> \R=h*Tan[x]
if x = elevation angle, then you're almost certainly integrating over a surface of constant z = h

Continuing my reverse engineering, it seems you have a charged particle at the origin, and you're integrating over the surface z=h in a region bounded by a circle of radius htan(a)

(cylindrical coordinates)
integral(K/r^2 rhat * dA) =
integral(|K/r^2| (rhat * zhat) |dA|) =
integral(|K/r^2| h/r |s ds dphi|) =
K h 2pi integral(s/sqrt(h^2 + s^2)^3 ds)

If you insist on doing this in spherical coordinates, I can't tell you the answer off the top of my head, because dA will be a horrible function of theta.

tl;dr
Be clear about what you're asking about.
Integrating over a 2D surface in a 2D world is very different from integrating over a 2D surface in a 3D world.

>>12
Alright, made a quick mspaint of the problem.
http://g.imagehost.org/0549/untitled_4.jpg
It's a charged flat disc and a punctual charge at h height perpendicular to the disc's plane from the center.

I misinterpreted your question, but the integral I gave gives the right answer for the (sort of) wrong reasons.

integral((K/r^2) cos(phi) dA) =
integral((K/r^2) (h/r) R dR dtheta) =
K h 2pi integral(R/sqrt(h^2 + R^2)^3 dR)

In going from standard polar coordinates (= cylindrical coordinates at z=h) -> sort of spherical coordinates, the theta variable doesn't change.  Thus, all you're really doing is changing R -> phi using R = h tan(phi).  Since this is a change of variables in 1 dimension, it's just a u (trig) substitution.

dR = h dphi / cos^2(phi)

K h 2pi integral(h tan(phi)/sqrt(h^2 + [h tan(phi)]^2)^3 h dphi / cos^2(phi)) =
K 2pi integral(tan(phi)/sqrt(1 + tan^2(phi))^3 dphi / cos^2(phi)) =
K 2pi integral(tan(phi)/sqrt(1/cos^2(phi))^3 dphi / cos^2(phi)) =
K 2pi integral(tan(phi)/[cos^2(phi)/cos^3(phi)] dphi) =
K 2pi integral(sin(phi) dphi)...

If you're changing both theta and phi, it gets a lot more complicated: http://en.wikipedia.org/wiki/Jacobian_determinant