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 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 \( \pi \)!

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