Spheres

Similar to cylinders, a sphere is an extension of polar co-ordinates, except this time it is a polar extension as opposed to a cartesian extension. This means that spherical co-ordinates have another angle $\phi$ .

s (r, θ, ϕ)

$s(r, \theta, \phi)$

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

$x = r\cos(\theta)\sin(\phi)$

$y = r\cos(\theta)\sin(\phi)$

$z = r\cos(\phi)$

And from cartesian to cylindrical we have:

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

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

$\phi = \tan^{-1}(\frac{y}{z})$

For instance, a sphere with 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 sphere, we must again use a triple integral as the input space has three parameters.

\int_{0}^{π} \int_{0}^{2 π} \int_{0}^{2} - r \sin (ϕ) d r d θ d ϕ

$\int_0^{\pi} \int_0^{2\pi} \int_0^2 -r\sin(\phi) \, dr \, d\theta \, d\phi$

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 $-r\sin(\theta)$ .

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

s (u, v) = (2 \cos (2 π u) \sin (π ϕ), 2 \sin (2 π u) \sin (π ϕ), 2 \cos (ϕ))

$s(u, v) = (2\cos(2\pi u)\sin(\pi \phi), 2\sin(2\pi u)\sin(\pi \phi), 2 \cos(\phi))$

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.