The arc length, \( s \) along the graph of a parametric function \( (x,y) = \left(x(t),y(t)\right) \) from \( t=a \) to \( t=b \) is given by:

$s = \int_a^b \sqrt{\left( \frac{dx}{dt} \right)^2+\left( \frac{dy}{dt} \right )^2 }\, dt$

We can see why by examining a linear approximation of the arclength:

$\Delta s^2 = \Delta x^2 + \Delta y^2$ by the Pythagorean Theorem, which means that for very small $\Delta s$, $ds \approx \Delta s = \sqrt{\Delta x^2 + \Delta y^2}$.

Also, notice that in the the case where $x(t)=t$, we have $(x,y) = (t,y(t))$, and the arc length becomes:

$s = \int_a^b \sqrt{1+\left( \frac{dy}{dt} \right )^2 }\, dt$

which we can use for non-parametrized functions.

## Show that the arc length of a semi-circle of radius 1 is $\pi$.

Let $(x,y)=(\cos \theta, \sin \theta)$, which describes the unit-circle. Then the arc length around the the semi-circle is given by:

$s = \int_0^\pi \sqrt{\left(\sin \theta\right)^2 + \left(- \cos \theta\right)^2}\, d\theta$

But $sin^2 \theta + \cos^2 \theta = 1$ by the Pythagorean Identity, so $s = \int_0^\pi\, d\theta = \pi. \, _\square$

## Gabriel's Horn

Consider the surface area and volume of of the solid formed by rotating the region bounded by the $x$-axis, $y=1$, and $y = \frac{1}{x}$, around the $x$-axis. This solid is called Gabriel's Horn.

## Volume

The volume of the solid of revolution can be found using the disk method:

$V = \pi \int_0^a \frac{dx}{x^2} = \pi \left( 1 - \frac{1}{a} \right)$

Now consider what happens as we allow $a$ to approach infinity:

$\displaystyle \lim_{a \to \infty} \pi \left( 1 - \frac{1}{a} \right) = \pi$

## Surface Area

The surface area of a solid of revolution is given by the formula

$S = 2\pi \int_a^b r(x)\sqrt{1+\left( \frac{dy}{dx} \right )^2 } \, dx$

In this case, since $\frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{1}{x^2}$ that gives us:

$\begin{aligned} S &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+\left( -\frac{1}{x^2} \right )^2 } \, dx \\ &= 2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx \end{aligned}$

This integral is hard to evaluate, but since in our interval $\sqrt{1+ \frac{1}{x^4}} \geq 1$ and $\frac{1}{x} > 0$:

$2\pi \int_1^a \frac{1}{x} \sqrt{1+ \frac{1}{x^4} } \, dx \geq 2\pi \int_1^a \frac{1}{x} \, dx$

It follows that:

$\begin{aligned} S &\geq 2\pi \int_1^a \frac{1}{x} \, dx \\ S &\geq 2\pi \ln a \end{aligned}$

But $\displaystyle \lim_{a \to \infty} 2\pi \ln a = \infty$, which means that Gabriel's Horn has infinite surface area, but a volume of only $\pi$!

No vote yet

1 vote

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in`\(`

...`\)`

or`\[`

...`\]`

to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

There are no comments in this discussion.