# Why is Green's theorem a special case of Stokes theorem?

$$\oint_C \mathbf{A} \cdot d\mathbf{r} = \iint_S (\nabla x \mathbf{A})\cdot \mathbf{n}$$ dS

Let $$A \leq P,Q,0>$$

Then $$\nabla x \mathbf{A}= <0,0,(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} )>$$.

So $$\iint_S (\nabla x \mathbf{A})\cdot \mathbf{n}$$ dS $$=\iint_S <0,0,(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} )>\cdot \mathbf{n}$$ dS $$=\iint_S <0,0,(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} )>\cdot \mathbf{k}$$ dS$$=\iint_S (\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} )$$ dS

Now, here people replace dS with dA and then say its equal to green's theorem but why would be able to do that? Isn't dS =$$|r_u x r_v|$$ dA

or this second screenshot. Why do we take unit normal vector. We don't do that when going from dS to dA as seen in this screenshot.

in this third screenshot, Wikipedia does it but doesn't explain the step I struggle with.

• Surely not answering your question in a way that's immediately satisfying to you, but, still, is a sort of "larger" answer: the general pattern of (generalized) Stokes' theorem is that (in tersest notation) $\int_{\partial D} \omega = \int_D d\omega$, where $\partial D$ is the boundary of a domain $D$, and $d\omega$ is the exterior derivative of the differential form $\omega$. Explanation of the latter concept is both an explanation of why the concept is correct, and of how to prove such a thing. In any case, both classical Green's and other similarly things can be understood as examples. :) Apr 12, 2022 at 2:10
• On a surface inside the $xy$ plane you can use $\mathbf{r}=\langle x,y,0 \rangle$ and compute that $|\mathbf{r}_x \times \mathbf{r}_y|=1$ so $dS=dA$ (this is the scalar $dS$ not the vector $dS$). (Similarly you can use $\mathbf{r}=\langle x,y,z_0 \rangle$ for a surface in the plane $z=z_0$). Let me know whether this completely answers your question, if it does then I will post it as an answer.
– Ian
Apr 12, 2022 at 2:23
• oh ok ty guys. I think I understand a bit better. I just wanted to make sure I wasn't going crazy. I will accept the answer of anyone who makes an answer because I think I get it thanks to you guys. Apr 12, 2022 at 2:27
• What Ian said; the form including $|\mathbf{r}_x \times \mathbf{r}_y|$ is useful for a general parameterization of a surface, but for just the flat plane, you don't always need a parameterization. You might use one like polar coordinates anyway, but integrals can be converted with a change of coordinates with a Jacobian factor, which for two variables is essentially $|\mathbf{r}_x \times \mathbf{r}_y|$. Apr 12, 2022 at 2:30
• Personally I prefer to think of parametrizing the flat surface by $\langle x,y,0 \rangle$ and then potentially change variables in that double integral (for example if the flat surface is a disk). But it is an interesting little exercise to show how the Jacobian determinant that you study in the "double integral" unit is identical to the factor $|\mathbf{r}_u \times \mathbf{r}_v|$ that you get from a parametrization even when you bake the change of variables into the parametrization (e.g. $\mathbf{r}(r,\theta)=\langle r\cos(\theta),r\sin(\theta),0 \rangle$).
– Ian
Apr 12, 2022 at 3:47

The correct formulation is the following. First it is true that when you parametrize a surface with $$r(u,v)$$, we have $$d\mathbf{S} = |r_u \times r_v| dA$$. Then $$\mathbf{n} = \frac{r_u \times r_v}{|r_u \times r_v|}$$ so $$\oint_C F\cdot dr = \iint_S (\nabla \times F) \cdot \mathbf{n} \,d\mathbf{S} = \iint_S (\nabla \times F) \cdot \frac{r_u \times r_v}{|r_u \times r_v|}|r_u \times r_v| dA = \iint_S (\nabla \times F) \cdot (r_u \times r_v) \,dA.$$ Now when your surface is on a coordinate plane, usually the $$xy$$-plane, then $$r(u,v) = \langle u,v,0\rangle$$ where $$z = 0$$ because the surface is on the $$xy$$-plane. Now $$r_u \times r_v = \langle 1,0, 0\rangle \times \langle 0, 1, 0\rangle = \langle 0,0,1\rangle = \vec k$$. What this is saying is the normal vector to a surface entirely in the $$xy$$-plane is the vector pointing up i.e. $$\vec k$$.
Now if $$F = \langle P, Q, R\rangle$$ then $$\nabla \times F = \langle R_y - Q_z, P_z - R_x, Q_x - P_y\rangle$$ so $$\iint_S (\nabla \times F)\cdot (r_u \times r_v)\,dA = \iint_S Q_x - P_y\, dA$$ which is exactly Green's Theorem.