Preview Activity 13.12.1.
In this activity, we will look at how we can apply the ideas about circulation along overlapping curves from the beginning of Subsection 13.8.2 to curves in space.
(a)
For this part, consider the curves in Figure 13.12.1, where the yellow curve is \(Y\text{,}\) the blue curve is \(B\text{,}\) and the magenta curve is \(M\text{.}\)
You should also go back and refamiliarize yourself with our notation for combining paths (as used in line integrals) from Convention 13.2.13. In our convention, \(Y+M\) would be closed loop, but \(Y-M\) would not make sense because the segment \(-M\) does not begin where \(Y\) begins.
(i)
Using the three segments in Figure 13.12.1, write out at least four different closed curves in terms of \(B\text{,}\) \(Y\text{,}\) and \(M\text{.}\) (Remember to consider orientation!)
(ii)
(iii)
Write a couple of sentences explaining how the circulation around \(C_1+C_2\) would compare to the circulation around \(C_1\) and the circulation around \(C_2\text{.}\) Write an equation in terms of \(\int_{C_1} \vF \cdot d\vr \text{,}\) \(\int_{C_2} \vF \cdot d\vr \text{,}\) and \(\int_{C_1+C_2} \vF \cdot d\vr \text{.}\)
(iv)
Explain how your arguments or equations from any of the parts above would or would not change if you considered the curves depicted in Figure 13.12.2.
(b)
Let \(C\) be the simple closed curve consisting of the yellow and magenta curves in Figure 13.12.1. You can see \(C\) plotted in red in Figure 13.12.3. The drop-down allows you to select three different surfaces. You can visually verify that each of the three surfaces contains \(C\text{.}\) Notice that the scale on the \(z\)-axis changes as you select different surfaces.
The simple closed curve consisting of the yellow and magenta curves in Figure 13.12.1 can be parameterized by \(\langle \cos(t), \sin(t), \cos(2t)\rangle \) with \(0\leq t\leq 2 \pi\text{.}\) Let \(C_3 = Y+M\text{.}\) Use the given parameterization of \(C_3\) to show that \(C_3\) is on each of the following surfaces:
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\(\displaystyle x^2-y^2=z\)
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\(\displaystyle z=x^4-y^4\)
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\(\displaystyle z=1-2y^2\)
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\(\displaystyle z=-\cos(\pi \sqrt{x^2+y^2})(x^2-y^2)\)
