Divergence Theorem - Confused :s (2 problems)

[halfoflessthan5]

Question

Evaluate both sides of the divergence theorem for

V =(x)i +(y)j

over a circle of radius R

Correct answer

The answer should be 2(pi)(R^2)

My Answer

the divergence theorem is

**integral** (V . n ) d(sigma) = **double intergral** DivV d(tau)

in 2D. Where (sigma) is the curve bounding the area (tau.) n is the unit normal vector to the curve.

I did the RHS of the divergence theorem fine and got 2(pi)(R^2)

*******
For the LHS:

First I need to work out n, the unit normal vector to the circle.

The equation of the circle is y=(1 - (x^2)^1/2)

To get the unit normal you use the nabla (gradient) function to get a vector that points away from the line and then normalize it.

First problem: In the solution they took grad(x^2 + y^2.) You end up with n (normalised) as (x/R)i + (y/R)j. . But i don't understand why you take grad(x^2 + y^2) when the line has the function f(x)=(1 - (x^2)^1/2)

Pretending i understood that bit and moving on you get V.n = (x^2/R) + (y^2R) = R^2/R =R

So I need to work out *integral* R d(sigma)

i presume d(sigma) (the elemental boundary length) is d(theta.) Then you do the line integral from 0 to 2(pi) . R is a constant so you can take it out then you just get 2(pi) for the integral

=> *integral* (V.n) d(sigma) = 2(pi)R

which is wrong (missing an R)

******

Sorry for the length of that. Some help on those two problems would be awesome.

 

[Mystic998]

Well, the big problem with the first part you're having trouble with is that f(x) = (1 - x^2)^(1/2) only represents the top part of the circle.

The second problem is that d(sigma) isn't d(theta).

 

[halfoflessthan5]

The second problem is that d(sigma) isn't d(theta).

Yes! its R d(theta.) The length of the arc subtended. Which would provide the extra R. Excellent, thankyou

Well, the big problem with the first part you're having trouble with is that f(x) = (1 - x^2)^(1/2) only represents the top part of the circle.

Okay point taken. I am not sure how else to write the function though. f(x) = +/-(1 - x^2)^(1/2) would give the top and the bottom but i can't do anything with that.

(just noticed something, that should be a an R not a 1. Doesnt affect anything)

Since in the solution you do grad(x^2 + y^2.) you must be taking grad f(x,y) where

f(x,y)= x^2 + y^2

but that's the equation of a 3D surface (i don't know what surface, its not a sphere.)

How does it represent my circle of radius R? (there's no R in it for a start)

 

[Mystic998]

Technically, what you should do is find a tangent vector for your circle, then get a vector perpendicular to that to find the normal. But, as far as I remember, taking the gradient is essentially a cheat.

Basically, what you're doing is saying that every level set of f(x,y) = x^2 + y^2 represents a circle of some radius. So what you do is find the outward normal of the surface (which is (fx, fy, -1), then project it into the plane (level set) where z = r^2, and that gives you the outward normal for the circle in that plane.

But don't quote me on that. I never got a really rigorous treatment of that kind of stuff because the few times it was covered in my classes, I was swamped with other work. I just used it if it worked.

 

[halfoflessthan5]

okay got it. So i can think of f(x,y) as like a contour map with a load of concentric circles each representing a value of f(x,y.)

So you workout grad f(x,y) and normalize it then set (x^2 + y^2) to R^2... and it works!

thankyou for all the help mystic. Youre a legend