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Week 6A: Area under a function

Consider the Cartesian $x-y$ plane. The area $A$ of some surface ($\Gamma$) may be computed as integral of infinitely small surface elements $\mathrm{d}A = \mathrm{d}x\mathrm{d}y$,

$$A = \iint_{\Gamma}\mathrm{d}x\mathrm{dy}$$

Lets say we want to compute the surface area of the blue-shaded area in the figure. This area is bound by the function $y = f(x) > 0$ and $y = 0$ for $x \in \langle a, b\rangle$. We can write

$$A = \iint_{\Gamma}\mathrm{dy}\mathrm{d}x = \int_{a}^{b}\int_0^{f(x)}\mathrm{d}y\mathrm{d}x$$ $$= \int_{a}^{b}f(x)\mathrm{d}x.$$

This results corresponds to what you already knew from your single-variable Calculus course.

Thank you multiple integration! for degenerating into single-variable calculus in this limit.

Week 6B: Work done by a force field

Consider a particle moving in a circular trajectory with radius $R$, angular frequency $\omega$, and time parameter $t$,

$$\vec{r}(t) = R\cos(\omega t)\vec{e}_x + R\sin(\omega t)\vec{e}_y.$$

It is affected by a force field ($\vec{F}(x,y)$) given by,

$$\vec{F}(x,y) = -y\vec{e}_x.$$

We can compute the work ($W$) done on this particle during a single rotation as the line integral of infinitely small work sections $\mathrm{d}W = \vec{F}\cdot\mathrm{d}\vec{r}$. Because we know we want to integrate force over a distance, it would be nice if we could integrate over the displaced (from 0 to $2\pi R$). As such we look for the paramterization in terms of its intrinsic parameter (i.e. path length). For this simple paramterization, we can rewrite the coordinate $\vec{r}$ as a function of time into a function of the traveled distance ($s$), such that $\| \frac{\mathrm{d}\vec{r}}{\mathrm{d}s}\| = 1$. Fortunately there is a simple invertible relation $s(t)$,

$$\begin{align} s(t) &= \int_0^{t} \|\frac{\mathrm{d}\vec{r}(\tau)}{\mathrm{d}\tau}\| \mathrm{d}\tau \\ &= \left[ \sqrt{R^2\omega^2\sin^2 (\omega \tau) + R^2\omega^2\cos^2 (\omega \tau)}\right]_0^t\\ &= \left[R\omega \sqrt{\sin^2 (\omega \tau) +\cos^2 (\omega \tau)}\right]_0^t\\ &= R\omega [1]_0^t = R\omega t. \end{align}$$

and its inverse,

$$t(s) = \frac{s}{R\omega}.$$

Such that

$$\vec{r}(s) = R\cos(\frac{s}{R})\vec{e}_x + R\sin(\frac{s}{R})\vec{e}_y.$$

So for the work we may write,

$$W = \int \vec{F}\cdot \mathrm{d}\vec{r} = \int_0^{2\pi R} \vec{F}\cdot \vec{t}\mathrm{d}s.$$

With $\vec{t}$ the (unit) tangential vector and not to be confused with the time parameter ($t$),

$$\vec{t} = \frac{\mathrm{d}\vec{r}(s)}{\mathrm{d}s} = -\sin(\frac{s}{R})\vec{e}_x + \cos(\frac{s}{R})\vec{e}_y.$$

Such that the inner product

$$\vec{F}\cdot\vec{t} = y\sin(\frac{s}{R}),$$

We know from the parameterization $y(s) = R\sin(\frac{s}{R})$. Thus,

$$\vec{F}\cdot\vec{t} = R\sin^2(\frac{s}{R}).$$

Resulting in the integral,

$$\begin{align} W &= \int_0^{2\pi R} \vec{F}\cdot \vec{t}\mathrm{d}s = R\int_0^{2\pi R} \sin^2(\frac{s}{R}) \mathrm{d}s.\\ &= R\int_0^{2\pi R} \frac{1}{2}\left( 1-\cos(\frac{2s}{R}) \right)\mathrm{d}s.\\ &= \frac{R}{2} \left[ s - \frac{R}{2}\sin(\frac{2s}{R}) \right]_0^{2\pi R} \mathrm{d}s\\ &= \pi R^2. \end{align}$$

Thank you line integral, I guess. But isn’t there an easier way?

Continue to the examples of week 7…