Gabriel's Horn

This simple shape can be cut up into vertical disks of radius $y = \frac{1}{x}$ and thickness $dx$ whose volume is given by:

$dV = \pi \bigg( \frac{1}{x} \bigg)^2 dx =\frac{\pi}{x^2}dx$

Integrating over the required range:

$V = \int_{1}^{\infty} \frac{\pi}{x^2}dx = \pi \bigg[ - \frac{1}{x} \bigg]_{1}^{\infty} = - \pi \bigg( \frac{1}{\infty} - \frac{1}{1} \bigg) = \pi$

So the volume is $V = \pi$

BONUS

We can once again use the vertical disks but this time focus on their differential surface area that makes up the surface of Gabriel's horn, that is:

$dA = 2 \pi \frac{1}{x} dx$

Integrating gives:

$A = 2 \pi \int_{1}^{\infty} \frac{1}{x} dx = 2 \pi \bigg[ \ln x \bigg]_{1}^{\infty} =2 \pi \bigg( \ln(\infty) - \ln(1) \bigg) = \infty$

So the surface area of Gabriel's horn is in fact infinite!