Gabriel's Horn, or sometimes known as Torricelli's Trumpet, is the name given to a surface of revolution that has some puzzling properties.
The surface resembles a horn, open at both ends, hence the name and has the strange property that we can fill it with paint, but that amount
of paint is not enough to paint the inside of the surface...

The Problem

Let's construct Gabriel's Horn and then you can try to prove that it has a finite volume (and to calculate it) and that
it has an infinite surface area.

Let's start with ??f(x) = \frac{1}{x}?? between ??x = 1?? and ??x = \infty??.

Now imagine that this is rotated through 360° around the ??x?? axis. It would make a shape that resembles a horn, open at both ends, as shown in the figure below (it is the surface area of this shape that we wish to find later):

Finally, consider the area between the function ??f(x)?? and the ??x?? axis, as shown in the figure below, if this were rotated it would produce
a 'filled' horn, which has the volume that we are interested in.

Can you find expressions for the volume and surface area of Gabriel's Horn and evaluate the volume and show that the surface area is infinite?

Before we begin, we should just note that:
$$ f(x) = \frac{1}{x} $$

and that the differential of ??f(x)??, which we will denote ??f'(x)??:
$$ f'(x) = \frac{d}{dx} \frac{1}{x} $$
$$ f'(x) = \frac{d}{dx} x^{-1} $$
$$ f'(x) = (-1) \times x^{-2} $$
$$ f'(x) = \frac{-1}{x^{2}} $$

Let's start by finding the volume. Recall, that the formula for the volume (??V??) of revolution is:
$$V = \pi \int_{1}^{\infty}f^{2}(x) dx$$

We can substitute in ??f(x)?? and evaluate the integral to find the volume:
$$V = \pi \int_{1}^{\infty} \left (\frac{1}{x} \right )^2 dx$$
$$V = \pi \int_{1}^{\infty} \frac{1}{x^{2}} dx$$
$$V = \pi \int_{1}^{\infty} x^{-2} dx$$
$$V = \pi \left ( (-1) \times x^{-1} \right ) \Big|_{1}^{\infty}$$
$$V = \pi \left ( \frac{-1}{x} \right ) \Big|_{1}^{\infty}$$
$$V = \pi \left [(\frac{-1}{\infty}) - (\frac{-1}{1}) \right ]$$
$$V = \pi[(0) - (-1)]$$
$$V = \pi[1]$$
$$V = \pi$$

So that means that it takes ??\pi?? units of paint to fill the horn.

Next we can find the surface area. Recall, that the formula for the surface area (??S??) of revolution is:
$$S = 2\pi \int_{1}^{\infty}f(x) \sqrt{1 + [f'(x)]^2} dx$$

We can substitute in for ??f(x)?? and ??f'(x)?? and evaluate the integral to find the surface area:
$$S = 2\pi \int_{1}^{\infty} \left (\frac{1}{x} \right ) \sqrt{1 + \left (\frac{-1}{x^{2}}\right )^{2} } dx$$
$$S = 2\pi \int_{1}^{\infty} \left (\frac{1}{x} \right ) \sqrt{1 + \left (\frac{1}{x^{4}}\right ) } dx$$

Now, let's divert a little and note that for the integral ??I??:
$$I = \int_{1}^{\infty} \left (\frac{1}{x} \right ) dx$$
$$I = \left ( ln(x) \right ) \Big|_{1}^{\infty}$$

Furthermore, we can note that, because the second term is always positive and greater than 1:
$$\frac{1}{x} \sqrt{1 + \frac{1}{x^4}} \geq \frac{1}{x}$$

Therefore we can say that:
$$S = 2\pi \int_{1}^{\infty} \left (\frac{1}{x} \right ) \sqrt{1 + \left (\frac{1}{x^{4}}\right ) } dx$$
$$S \geq 2\pi \int_{1}^{\infty} \left (\frac{1}{x} \right ) dx$$
$$\therefore S \geq \infty$$

And that yields the paradoxical answer that we were looking for, we can fill the horn with ??\pi?? units of paint, but
we could not paint the surface area of the horn, unless we had an infinite amount of paint.