Answer for Stokes’ Example from Lecture 22.

F= Function[{x,y,z},
{2y,3z,z}
];
f=Function[{x,y}, (1-(x^2))(1-(y^2))];

Line integral around the outside ∮F·dr.

(∫_(-1)^1)F[1,t,0].{0,1,0}\[DifferentialD]t+(∫_(-1)^1)F[-t,1,0].{-1,0,0}\[DifferentialD]t+(∫_(-1)^1)F[-1,-t,0].{0,-1,0}\[DifferentialD]t+(∫_(-1)^1)F[t,-1,0].{1,0,0}\[DifferentialD]t

Using Stokes’.

(Subscript[∇, {x,y,z}])\[Cross]F[x,y,z]

Curl[F[x,y,z],{x,y,z}]

{-3,0,-2}.{-D[f[u,v],u],-D[f[u,v],v],1}

(∫_(-1)^1)(∫_(-1)^1)(-2-6 u (1-(v^2)))\[DifferentialD]u\[DifferentialD]v