As we saw in

part b of Preview Activity 12.11.1 there is more than one surface that has

\(C\) as a boundary. We will calculate the flux integral (the right side of Stokes’ Theorem) for a different surface to help motivate why it will not matter which surface we use (as long as we have the correct orientation and boundary). For this part we will use the surface

\(z=x^2-y^2\text{,}\) which will be parameterized by

\(\vr(s,t)=\langle s\cos(t),s\sin(t), s^2 (\cos(t)^2-\sin(t)^2) \rangle \) with

\(0\leq s\leq 1\) and

\(0\leq t\leq 2\pi\text{.}\) Additionally, we will use the trig identity

\(\cos(2t)=\cos(t)^2-\sin(t)^2 \) to write the last component of our parameterization as

\(s^2 \cos(2t)\text{.}\)
Using our parameterization, we have the following for the partial derivative functions and the corresponding normal vector:

\begin{align*}
\vr_s(s,t) \amp =\langle \cos(t), \sin(t),2s\cos(2t)\rangle \\
\vr_r(s,t) \amp =\langle -s\sin(t), s\cos(t),-2s^2\sin(2t)\rangle \\
\vw=\vr_s \times \vr_t \amp =\langle -2s^2\sin(2t)\sin(t)-2s^2\cos(2t)\cos(t) , \\
\amp \quad \quad 2s^2\cos(t)\sin(2t)-2s^2\sin(t)\cos(2t), s \cos(t)^2 + s \sin(t)^2 \rangle \\
\amp = \langle -2s^2 \cos(t), 2s^2 \sin(t), s \rangle
\end{align*}

(There are a number of equivalent algebraic simpmlifications that can be done here by choosing different trigonometric identities.)

Now we are ready to set up the flux integral as the following iterated integral:

\begin{equation*}
\int_0^1 \int_0^{2\pi} \langle s\cos(t)-1,1-s\sin(t),0 \rangle \cdot \langle -2s^2 \cos(t), 2s^2 \sin(t), s \rangle \, dt\, ds
\end{equation*}

You probably noticed that this integral looks slightly more complicated than our work in the previous part, but if we are consistent, we should get the same result. Evaluating this dot product and computing each integral gives

\begin{align*}
\amp \int_0^1 \int_0^{2\pi} (-2s^2 \cos(t))(s\cos(t)-1) + (2s^2 \sin(t))(1-s\sin(t)) \, dt\, ds \\
=\amp \int_0^1 \int_0^{2\pi} -2s^3 \cos(t)^2 + 2s^2 \cos(t) + 2s^2 \sin(t) -2s^3\sin(t)^2 \, dt\, ds \\
=\amp \int_0^1 \int_0^{2\pi} -2s^3 + 2s^2 \cos(t) + 2s^2 \sin(t) \, dt\, ds \text{.}
\end{align*}

Because both sine and cosine will integrate to zero over the interval from \(0\) to \(2\pi\text{,}\) we only need to evaluate

\begin{equation*}
2 \pi \int_0^1 -2s^3 \, ds= -\pi\text{,}
\end{equation*}

which gives exactly the same result as our circulation integral *and* our flux integral with the other surface.