Contour Integration
This calculus technique bridges the gap between calculus 1 and multivariable calculus along with many other fields. In this section, we will discuss vector fields, rectifiable curves, contours, contour integrals and connections across many disciplines that go beyond the scope of calculus. This section is FUNDAMENTAL to much of modern mathematics and almost all of modern physics and engineering.
1. Vectors Fields
A vector field is both a way of making sense of higher dimensional object that is hard for humans to perceive the whole of with mathematics and a way to bring our intuition to bear on an object that would perhaps, otherwise, seem quite foreign to us. The way we most often try to constructive think about vector fields is they behave like smooth, continuous, space-filling fluids. If I run my hand across the surface of a pond, the water ripples smoothly over the surface, but forces are also transferred downwards and outwards in every direction through that fluid at each point I touched it. Each of those forces is itself a 3-vector that has a dependence on where it is located in time and space away from my touch. For this course and for simplicity’s sake, we will suppress the time dimension forcing our vector fields to be, at most, a function of 3 spatial dimensions.
The thing that makes vector fields very tractable though is that we often traverse in a one dimension path (approximately!). As such vector fields capture concepts like work and energy very well.
2. Rectifiable Curves
A rectifiable curve is a curve that can be well-approximated by small line segments. Contour integration is often referred to as path or line integration and the reason we might call it line integration is because we principally use rectifiable curves. Mathematically, we can easily construct curves that are not rectifiable- many fractals, even ones that have application to the world in which we live like brownian motion are not rectifiable.
Let us make our idea clearer: we can partition our curve, C, into segments and join the endpoints of those segments with line segments. The produces a new curve, $S_1$, whose arclength differs from that of our actual curve by some number, call it $\varepsilon_1$. This satisfies the inequality $|\text{Length}(C) – \text{Length}(S_1)| \leq \varepsilon_1$. We can repeat this process with each segment joining them all together to form a new curve $S_2$ that differs by a smaller number $\varepsilon_2$ and satisfies the inequality $|\text{Length}(C) – \text{Length}(S_2)| \leq \varepsilon_2 < \varepsilon_1$. We call the second collection of points joined by line segments a refinement of the first partition.
This sort of process should appear somewhat familiar to the reader. It looks a lot like a kind of limit (and it is)! We notice in the diagrams given tot he right that the segmented partitions become closer to the curve with more subdivisions. It is possible that any given partition is longer or shorter the curve C, but the idea is that we should be able to keep finding $S_n$ such that our sequence $\varepsilon_n \rightarrow 0$ as $n \rightarrow \infty$.
3. Contours and Parameterization
4. Contour Integrals
Worked Problems for Contour Integration
1. Suppose there is a fence with height $h(x, y) = 4 – .04(x^2 – y^2)$ meters that encircles a circular garden of radius 10 meters. If our blue paint covers 100 $\frac{m^2}{L}$ then how much blue paint does it take to paint BOTH sides of the fence blue?
We parameterize the circle as $10sin(t) = y$ and $10cos(t) = x$. Now, we notice that
$\int_C h(x, y) ds =$
$\int_0^{2\pi}\left( 4 – .04(100cos^2(t) – 100sin^2(t))\right)\sqrt{100cos^2(t)+100sin^2(t)} dt =$
$\int_0^{2\pi}\left( 4 – 4(cos^2(t) – sin^2(t))\right)10dt =$
$10\int_0^{2\pi} 4 – 4cos(2t)dt =$
$40\int_0^{2\pi} 1 – cos(2t)dt =$
$40\left(\int_0^{2\pi}dt – \int_0^{2\pi}cos(2t)dt\right) =$
$40\int_0^{2\pi}dt – 40\cancel{\int_0^{2\pi}cos(2t)dt} =$
$80\pi$ But we have to paint two sides of the fence, so $2 \cdot 80\pi =160\pi$ $m^2$ And this implies that we need $1.6\pi \text{ L blue paint} \approx 5.0 \text{ L blue paint}$.
