W04 - Examples

Arc length

Arc length of chain, via position

Arc length of chain, via position

A hanging chain describes a catenary shape. (‘Catenary’ is to hyperbolic trig as ‘sinusoid’ is to normal trig.) The graph of the hyperbolic cosine is a catenary:

$$y=f(x)=\cosh x$$

Let us compute the arc length of this catenary on the portion from $x=0$ to $x=a$.

Solution

(1) Arc-length function:

$$s(a)={\int }_0^a\sqrt{1+(f^{\prime }{)}^2}dx$$

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(2) Compute $f^{\prime }(x)$:

$$\frac{d}{dx}\cosh (x)=\sinh (x)$$

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(3) Plug into arclength formula:

$$s(a)={\int }_0^a\sqrt{1+{\sinh }^2(x)}dx$$

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(4) Hyperbolic trig identity:

$$\begin{array}{c}{\cosh }^{2}x-{\sinh }^{2}x=1\\ \\ 1+{\sinh }^{2}x={\cosh }^{2}x\end{array}$$

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(5) Simplify integrand and integrate:

$$\begin{array}{c}{\int }_0^a\sqrt{1+{\sinh }^2(x)}dx\gg \gg {\int }_0^a\sqrt{{\cosh }^{2}x}dx\\ \\ \gg \gg {\int }_0^a\cosh xdx\gg \gg \sinh a\end{array}$$

Coincidence?

The arc length of a catenary curve matches the area under the catenary curve!

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Surface areas of revolution

Surface area of a sphere

Surface area of a sphere

Using the fact that a sphere is given by revolving a semicircle, verify the formula $S=4\pi r^2$ for the surface area of a sphere.

Solution

(1) Describe sphere as surface of revolution:

Upper semicircle:

$$x^2+y^2=r^2,y\ge 0$$

As function of $x$:

$$y=f(x)=\sqrt{r^2-x^2},x\in [-r,r]$$

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(2) Surface area formula:

$$S={\int }_a^{b}2\pi f(x)\sqrt{1+(f^{\prime }{)}^2}dx$$

Our bounds are $x=-r$ and $x=+r$. Function is $f(x)=\sqrt{r^2-x^2}$:

$$S\gg \gg {\int }_{-r}^{+r}2\pi \sqrt{r^2-x^2}\sqrt{1+(f^{\prime }{)}^2}dx$$

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(3) Work out integrand:

Power rule and chain rule:

$$\begin{array}{c}{f}^{\prime }(x)\gg \gg \frac{1}{2}(r^2-x^2{)}^{-1/2}(-2x)\\ \\ \gg \gg -x(r^2-x^2{)}^{-1/2}\end{array}$$

Therefore:

$$\begin{array}{c}(f^{\prime }{)}^2\gg \gg \frac{x^2}{r^2-x^2}\\ \\ 1+(f^{\prime }{)}^2\gg \gg \frac{r^2}{r^2-x^2}\end{array}$$

Integrand:

$$\sqrt{r^2-x^2}\sqrt{1+(f^{\prime }{)}^2}\gg \gg \sqrt{r^2-x^2}\sqrt{\frac{r^2}{r^2-x^2}}\gg \gg r$$

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(4) Compute integral:

$$\begin{array}{c}S\gg \gg {\int }_{-r}^{+r}2\pi rdx\\ \\ \gg \gg 2\pi rx{|}_{-r}^{+r}\gg \gg 2\pi rr-2\pi r(-r)\gg \gg 4\pi r^2\end{array}$$

This is the expected surface area formula: $S=4\pi r^2$.

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