Cylinders

By changing our co-ordinate system using parametric surfaces we can represent other types of shapes more easily. Recall that if we want to represent a circle in cartesian co-ordinates we would have to use an implicit function, such as:

x^{2} + y^{2} = 1

$x^2 + y^2 = 1$

Such functions are very inconvenient to work with, because if we were to represent the function in terms of $x$ or $y$ it would have square roots in it, which can have either a positive or negative value.

y = \sqrt{1 + x^{2}}

$y = \sqrt{1 + x^2}$

Instead, it is much easier to represent circular geometry using a polar co-ordinates in terms of $r$ (the radius) and $\theta$ (the angle between 1 and $2\pi$ ). Then we can can use a the following change of co-ordinates to express our circle:

$x = r\cos(\theta)$

$y = r\cos(\theta)$

Or the other way around:

$r = \sqrt{x^2 + y^2}$

$theta = \tan^-1(y / x)$

So, converting from cartesian co-ordinates to polar co-ordinates, our circle is:

(r, θ) = (1, θ)

$(r, \theta) = (1, \theta)$

Representing our co-ordinate system in this way has important advantages. Since there are no square roots, it is a lot easier to take the integral of a circle and find its area:

\int_{0}^{2 π} \int_{0}^{1} r d r d θ

$\int_0^{2\pi} \int_0^1 r \, dr \, d\theta$

\int_{0}^{2 π} \frac{1}{2} r^{2} d θ

$\int_0^{2\pi} \frac{1}{2}r^2 \, d\theta$

π r^{2}

$\pi r^2$

This should look familiar!

We can now extend this to three dimensions to define a cylinder using cylindrical co-ordinates. Cylindrical co-ordinates are effective a cartesian extension of polar co-ordinates, so there is a radius, $\theta$ and a $z$ co-ordinate.

s (r, θ, z)

$s(r, \theta, z)$

The translation from cylindrical to cartesian co-ordinates is as follows:

$x = r\cos(\theta)$

$y = r\cos(\theta)$

$z = z$

And from cartesian to cylindrical we have:

$r = \sqrt{x^2 + y^2}$

$\theta = \tan^{-1}(\frac{y}{x})$

$z = z$

For instance, a cylinder with height 1, and radius 2 is defined as:

s (r, θ, z) = (2, θ, 1)

$s(r, \theta, z) = (2, \theta, 1)$

If we want to find the volume of a cylinder, we must use a triple integral as the input space has three parameters.

\int_{0}^{1} \int_{0}^{2 π} \int_{0}^{2} r \times d r d θ d z

$\int_0^1 \int_0^{2\pi} \int_0^2 r \times \, dr \, d\theta \, dz$

Note also that we need to apply a density correction which is the determinant of the Jacobian matrix for each component. In the case of a cylinder, that is $r$ , as we have:

r \cos^{2} (θ) + r \sin^{2} (θ) = r

$r\cos^2(\theta) + r\sin^2(\theta) = r$

Finding the surface area of a cylinder requires us to find a parameterization in terms of two parameters. Because this cylinder has a constant radius, we can use a parameterization that keeps $r$ constant.

s (u, v) = (2, 2 π v, u)

$s(u, v) = (2, 2\pi v, u)$

From there we can use the same approach to find the length of the normal vector at every point.

So from those points, we can find the surface area.