Question

Greetings, let's say we have a ring, and we apply a field vector F, is it true that the surface intagral of this product becomes a double integral of the lower directional components? for an example check below at my "response"

Answer 1

\[\int\limits_{surface}^{}Fda = \int\limits_{}^{}\int\limits_{}^{}gdφdl\] where f = Cxx^hat cyy^hat

Answer 2

(let's suppose the ring is moving linear within this vector field) so dl is the length on x axis and df is the rotation vector

Answer 3

\[ ds = dx \times dy \text{ - in rectangular coordinate system} \\
ds = r dr d\theta \text{ - in rectangular coordinate system} \\\]
It's natural that it has double integral from ... unless you know F(s) directly as well as S

Answer 4

As you are implying F is a vector \( \vec F\)
So integration must be of
\[ \int_s \vec F \times \hat n ds \text { or } \int_s \vec F * \hat n ds \]

Answer 5

r dr d\theta <--- this was supposed to be in polar coordinate system

Answer 6

so this would become \[\int\limits_{0}^{a}\int\limits_{0}^{2π} (Crcosθ\hat{j} - Crsinθ\hat{j}) \hat{n} rdθdr\]

?

Answer 7

to clarify; you want the surface integral of a vector field acting on the ring, or through it (like flux)?
is this an E&M problem?

Answer 8

(Crcosθjˆ−Crsinθjˆ) (dot or cross??) nˆr

Answer 9

TuringTest, actually the second :P
yes it is the mathematic portion of the problem i stated on the E&M part, though i need good maths for this. Well since I am interested in infinitesimal flux, I suppose only side-flux is counted, since the main surface area is removed during infinitesimal changes of the magnetic flux

experimentX: dot

Answer 10

**actually the first

Answer 11

The basic sketch of the problem



in r,θ axes i suppose this would become: