Heidi T. answered 09/25/19
Experienced tutor/teacher/scientist
To solve this one you will need to convert the Cartesian coordinates (x,y,a) to cylindrical (r,θ,z).
x = r cosθ
y = r sinθ
z = z
In this case, r = 1 because x2 + y2 = 1 and this is the equation of a circle of radius 1.
Parameterize the curve in terms of r and θ: r(θ) = (cos θ, sin θ, 0) and dr = (-sinθ, cosθ, 0) dθ. 0 ≤ θ ≤ 2π. Next convert B to cylindrical coordinates with r = 1: B = (z, -3x, 2z) = (z, -3 cosθ, 2z)
On the curve, z = 0, so B reduces to B = (0, -3 cosθ, 0)
The line integral around the curve becomes: ∫02π (B(θ)) • dr(θ), integrating with respect to θ
The paraboloid is extraneous only in the sense that the surface attached to the curve can be any surface...so it doesn't matter if it is that specific paraboloid or some other surface.
We can choose to use the surface that is the disk enclosed by the curve at z = 0. Since this surface is in the x-y plane, the normal to the surface is in the z-direction, so the normal is n = ((0, 0, 1)
If you actually use the paraboloid, the normal will be pointed upward and away from the z-axis (draw the picture and this becomes clear).
In either case, the unit disc is x2 + y2 ≤ 1, z ≥ 0 To find the normal, write r(x,y) = (x, y, 1 - x2 - y2) (the value given for z in the equation) and find the tangent lines in the x and y directions (derivative of r wrt x and derivative of r wrt y) The cross product of the tangent lines is perpendicular to the surface by definition of cross product.
Now convert n and curl B back to cylindrical coordinates using the definitions above. This time you can't assume r = 1 and z = 0 because r varies with z. Limits of integration are 0 to 2π for dθ and 0 to 1 for dr, dS = dxdy = rdrdθ
This should get you started.