I have a vector field $\vec{H} = (8z,0,-4x^3)$
Naturally, $\nabla \times \vec{H} = (0,8+12x^2,0)$
Stokes theorem says: $$ \int_s{\nabla \times \vec{H}} \cdot \vec{dS} = \oint_l{\vec{H} \cdot dl} $$
When evaluating across the closed rectangular path defined by $$P(2,3,4) \space, Q(4,3,4) \space, R(4,3,1) \space,S(2,3,1)$$
$$\int_s{\nabla \times \vec{H}} \cdot \vec{dS} = \int_1^4\int_2^4{8+12x^2 \space dx \space dz} = 720 \\$$
$$and$$
$$\oint_l{\vec{H} \cdot dl} = \int_P^Q{\vec{H} \cdot (dx \space \hat{ax})} \space + \int_Q^R{\vec{H} \cdot (-dz \space \hat{az})} \space + \int_R^S{\vec{H} \cdot (-dx \space \hat{ax})} \space + \int_S^P{\vec{H} \cdot (dz \space \hat{az})} \space$$
$$= \int_2^4{8 *4 \space dx}\space+ \int_4^1{4*4^3 \space dz} \space + \int_4^2{-8*1 \space dx} \space + \int_1^4{-4*2^3 \space dz} \space$$
$$= 64 \space-\space 768 \space+\space 16 \space-96\space$$
$$=-784$$
I am almost certain about the signs of the path integrals but if I change the signs of two terms in the last expression to $$64+768-16-96$$ I get 720, which is supposed to be the right answer.