# 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:

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

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

$x=r\mathrm{cos}(\theta )$

$y=r\mathrm{cos}(\theta )$

Or the other way around:

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

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

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

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

$${\int}_{0}^{2\pi}{\int}_{0}^{1}r\phantom{\rule{thinmathspace}{0ex}}dr\phantom{\rule{thinmathspace}{0ex}}d\theta $$$${\int}_{0}^{2\pi}\frac{1}{2}{r}^{2}\phantom{\rule{thinmathspace}{0ex}}d\theta $$$$\pi {r}^{2}$$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.

$$s(r,\theta ,z)$$The translation from cylindrical to cartesian co-ordinates is as follows:

$x=r\mathrm{cos}(\theta )$

$y=r\mathrm{cos}(\theta )$

$z=z$

And from cartesian to cylindrical we have:

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

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

$z=z$

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

$$s(r,\theta ,z)=(2,\theta ,1)$$If we want to find the volume of a cylinder, we must use a triple integral as the input space has three parameters.

$${\int}_{0}^{1}{\int}_{0}^{2\pi}{\int}_{0}^{2}r\times \phantom{\rule{thinmathspace}{0ex}}dr\phantom{\rule{thinmathspace}{0ex}}d\theta \phantom{\rule{thinmathspace}{0ex}}dz$$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:

$$r{\mathrm{cos}}^{2}(\theta )+r{\mathrm{sin}}^{2}(\theta )=r$$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.

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