- #1
halfoflessthan5
- 16
- 0
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.
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.