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Section 3.5 Fundamental Theorems

Subsection 3.5.1 General Stokes’ Theorem

The one-variable Fundamental Theorem of Calculus is the case \(p=1\text{.}\) Let \(f\) be a differentiable function on an interval containing \([a,b]\text{,}\) and let \(c(t)=a+t(b-a)\) for \(0 \le t \le 1\text{.}\) Then \(f\) is a \(0\)-form and
\begin{equation*} \partial c = c(1)-c(0)=b-a \end{equation*}
is a \(0\)-chain. By Theorem 3.5.1,
\begin{equation*} \int_c df = \int_{\partial c} f. \end{equation*}
The left-hand side is the usual integral of the derivative:
\begin{equation*} \int_c df = \int_0^1 \frac{d}{dt}(f(c(t)))\,dt = \int_a^b f'(x)\,dx. \end{equation*}
The right-hand side is integration of a \(0\)-form over a \(0\)-chain, which by definition is just evaluation with signs:
\begin{equation*} \int_{\partial c} f = \int_{c(1)-c(0)} f = f(c(1))-f(c(0)) = f(b)-f(a). \end{equation*}
Therefore general Stokes’ theorem gives
\begin{equation*} \int_a^b f'(x)\,dx = f(b)-f(a), \end{equation*}
which is exactly the Fundamental Theorem of Calculus in \(\R^1\text{.}\)

Subsection 3.5.2 Classical Special Cases

The familiar theorems from vector calculus are obtained by choosing the degree of the form and the dimension of the chain appropriately.

Example 3.5.3.

Use Green’s theorem to compute the circulation of \(\mathbf{F}(x,y)=\langle -y,x \rangle\) around the unit circle \(C\text{,}\) oriented counterclockwise.
Solution.
Let \(D\) be the unit disk, so \(\partial D=C\text{.}\) The corresponding \(1\)-form is
\begin{equation*} \omega = -y\,dx + x\,dy. \end{equation*}
Its exterior derivative is
\begin{equation*} d\omega = dx \wedge dy - (-dy \wedge dx) = 2\,dx \wedge dy. \end{equation*}
Therefore Green’s theorem gives
\begin{equation*} \oint_C (-y\,dx+x\,dy) = \iint_D 2\,dA = 2\operatorname{area}(D) = 2\pi. \end{equation*}
In vector field notation,
\begin{equation*} \oint_C \mathbf{F}\cdot d\mathbf{r} = 2\pi. \end{equation*}

Example 3.5.5.

Let \(C\) be the unit circle in the plane \(z=0\text{,}\) oriented counterclockwise as viewed from above. Use Stokes’ theorem to compute the circulation of \(\mathbf{F}(x,y,z)=\langle -y,x,0 \rangle\) around \(C\text{.}\)
Solution.
Let \(S\) be the unit disk in the plane \(z=0\) with upward orientation, so \(\partial S=C\text{.}\) The corresponding \(1\)-form is
\begin{equation*} \omega = -y\,dx + x\,dy. \end{equation*}
As above,
\begin{equation*} d\omega = 2\,dx \wedge dy. \end{equation*}
So Stokes’ theorem gives
\begin{equation*} \oint_C (-y\,dx+x\,dy) = \int_S 2\,dx \wedge dy = 2\iint_S dA = 2\pi. \end{equation*}
Since \(\nabla \times \mathbf{F}=\langle 0,0,2 \rangle\text{,}\) the same computation in vector field notation is
\begin{equation*} \oint_C \mathbf{F}\cdot d\mathbf{r} = \iint_S (\nabla \times \mathbf{F})\cdot \vn\,dS = 2\pi. \end{equation*}

Example 3.5.7.

Use the divergence theorem to compute the outward flux of the vector field \(\mathbf{F}(x,y,z)=\langle x,y,z \rangle\) across the sphere of radius \(a\text{.}\)
Solution.
Let \(B_a\) be the solid ball of radius \(a\text{,}\) so \(\partial B_a\) is the sphere. The corresponding \(2\)-form is
\begin{equation*} \eta = x\,dy \wedge dz + y\,dz \wedge dx + z\,dx \wedge dy. \end{equation*}
Its exterior derivative is
\begin{equation*} d\eta = (1+1+1)\,dx \wedge dy \wedge dz = 3\,dV. \end{equation*}
Therefore the divergence theorem gives
\begin{equation*} \iint_{\partial B_a}\eta = \iiint_{B_a} 3\,dV = 3\operatorname{vol}(B_a) = 3\cdot \frac{4}{3}\pi a^3 = 4\pi a^3. \end{equation*}
Since \(\nabla \cdot \mathbf{F}=3\text{,}\) this is
\begin{equation*} \iint_{\partial B_a}\mathbf{F}\cdot \vn\,dS = 4\pi a^3. \end{equation*}
Green’s theorem, the classical Stokes’ theorem, and the divergence theorem are therefore not three unrelated results. They are three instances of the single identity \(\int_c d\omega=\int_{\partial c}\omega\text{.}\)
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