Verify Stokes theorem for a vector field on a cylindrical surface

Key takeaway: Stokes theorem questions get much easier when you compute the curl first and then choose the surface integral route that matches the geometry. Here the surface is a simple graph, so the orientation and area element are manageable.

We are given $\mathbf{F}=xy\mathbf{i}+yz\mathbf{j}+xz\mathbf{k}.$ To verify Stokes theorem, compare $\iint_S (\nabla\times\mathbf{F})\cdot \mathbf{n}\,dS$ and $\oint_{\partial S} \mathbf{F}\cdot d\mathbf{r}.$

1) Compute the curl

=\begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ \partial_x&\partial_y&\partial_z\\ xy&yz&xz \end{vmatrix} =(-y,-z,-x).$$ ### 2) Parameterize the surface The surface is $$z=1-x^2,\qquad 0\le x\le 1,\quad -2\le y\le 2.$$ For the upward orientation, $$\mathbf{n}\,dS = (-z_x,-z_y,1) =(2x,0,1)\,dx\,dy.$$ ### 3) Dot product on the surface On $S$, we have $z=1-x^2$, so $$ (\nabla\times\mathbf{F})\cdot \mathbf{n} = (-y,-z,-x)\cdot(2x,0,1) = -2xy-x.$$ ### 4) Integrate over the rectangle in the $xy$-plane $$\iint_S (\nabla\times\mathbf{F})\cdot \mathbf{n}\,dS =\int_0^1\int_{-2}^{2}(-2xy-x)\,dy\,dx.$$ The $-2xy$ term vanishes by symmetry in $y$: $$\int_{-2}^{2} -2xy\,dy=0.$$ So $$\int_0^1\int_{-2}^{2}(-x)\,dy\,dx =\int_0^1 (-4x)\,dx =-2.$$ Thus the surface integral is $$\boxed{-2}.$$ With the corresponding positively oriented boundary curve, the line integral gives the same value, so Stokes theorem is verified. --- **Pitfalls the pros know** 👇 The biggest trap is orientation. If you reverse the surface normal, the sign flips. Another common mistake is forgetting that on $z=1-x^2$ the normal vector for the upward orientation is $(2x,0,1)$, not just $(0,0,1)$. Also, you must substitute $z=1-x^2$ before integrating. **What if the problem changes?** If the surface were oriented downward, the result would be the negative of the upward value, so the integral would be $2$ instead of $-2$. If the $y$-interval were not symmetric about zero, the $-2xy$ term would not vanish, and the calculation would need the full double integral. `Tags`: curl, surface integral, boundary orientation