Recent content by NuclearMeerkat

Rz = (0,0,1) and Rθ = (-asinθ,acosθ,0) use -Rz because it the bottom face giving dΣ = (-acosθ,asinθ,0) therefore the integral is ∫dθ∫dz (3x,-y,-2(z+y))⋅(-acosθ,asinθ,0)

∇×A in cylindrical coordinates is (3acosθ,-asinθ,-2(z+asinθ)) and then I integrate w.r.t. θ and z?

Hi. I'm Martin and I'm studying Nuclear Engineering at University in the UK, the degree is excellent and all I hoped, but the maths runs me into the ground sometimes. I'm hoping to have a career in the fusion research area.

Homework Statement Homework Equations Stoke's Theorem: The Attempt at a Solution ∇×A = (3x,-y,-2(z+y)) I have parametric equation for wall and bottom: Wall: x(θ,z) = acosθ ; y(θ,z) = asinθ ; z(θ,z) = z [0≤θ≤2π],[0≤z≤h] Bottom: x(θ,r) = rcosθ ; y(θ,r) = rsinθ ; z(θ,r) = 0 [0≤θ≤2π],[0≤r≤a]...