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 dont 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.