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Arc Length

Definition

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.

Technique

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 $$s = \int_0^\pi\, d\theta = \pi. \, _\square$$

Application and Extensions

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 shell 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{align} 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{align}

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

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 $$\pi$$!

Note by Arron Kau
3 years, 10 months ago

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