# A problem about the area element in the Stokes' Theorem

Given a vector field $$F(x,y,z) = x^2 \hat i + 2x \hat{j} + z^2 \hat{k}$$ and a curve $$C: \text{the ellipse } 4x^2 + y^2 = 4 \text{ in the } xy- \text{plane}$$, I want to find $$\oint_{C} \vec F \cdot dr = \int\int_{S} \nabla \times \vec F \cdot \hat n d\sigma$$ via Stokes' Theorem. Here, $$S$$ is the surface on the $$xy$$-plane bounded by the curve $$C$$ above. I found $$\nabla \times \vec F = 2 \hat{k}$$ and I take $$\hat n = \hat k$$ as the surface is lying on the $$xy$$-plane. Thus, I end up with $$\oint_{C} \vec F \cdot dr = \int\int_{S} \nabla \times \vec F \cdot \hat n d\sigma =\int\int_{S} 2 d\sigma = 2(\text{area of the ellipse}) = 4 \pi.$$ However, for the surface area element, we should have $$d \sigma = \frac{\mid \nabla f|}{|\nabla f \cdot \hat k |}$$ where $$f$$ is the level surface that comes from the the surface $$4x^2 + y^2 = 4$$ in the $$xy$$-plane. So, if I let $$f(x,y,z)=4x^2 +y^2 -4$$ for example, I end up with $$\nabla f \cdot \hat k = 0$$ such that I cannot write $$d\sigma$$. What do I do wrong?

• So you're saying that the level set $f(x,y,z)=0$ contains the surface $S=\{4x^2+y^2\leq 1\}$? This would be true if you defined $f(x,y,z)=z$. With you definition of $f$ the level set $f(x,y,z)=0$ would be a an elliptic cylinder emanating from the $xy-$ plane. Jan 6 at 13:11

In this statement from Stokes' theorem, $$d\sigma$$ is the differential for area of the region of integration $$S$$. When using the formula for projection onto the $$x$$-$$y$$ plane

$$d\sigma = \frac{|\nabla f|}{|\nabla f \cdot \hat{k}|}\, dx\, dy,$$

function $$f$$ is the function whose level curves define that region $$S$$. The level curve from $$f(x,y,z) = 4x^2+y^2-4$$ gives the boundary ellipse $$C$$, and does not hold in the interior of $$S$$. $$S$$ is a piece of the $$x$$-$$y$$ plane, so the appropriate $$f$$ is $$f(x,y,z)=z$$. This gives $$d\sigma = dx\, dy$$. Not really a surprise: since $$S$$ is already in the $$x$$-$$y$$ plane, projecting it onto the $$x$$-$$y$$ plane is trivial.

A side note: When using this method projecting onto the $$x$$-$$y$$ plane and a formula $$f$$ whose level curve describes the surface of integration $$S$$, we also have

$$\hat{n} = \frac{\pm \nabla f}{|\nabla f|}$$

provided $$\nabla f \neq 0$$ everywhere on $$S$$, and choosing the sign so that $$\hat{n}$$ always has a positive $$z$$ component. So it can often be helpful to evaluate the theorem as

$$\oint_C \vec F \cdot d \vec r = \int_S \nabla \times F \cdot \frac{\pm \nabla f}{|\nabla f \cdot \hat{k}|}\, dx\, dy$$

and skip evaluating the canceled $$|\nabla f|$$ factor.