# Sphere surface area

• Apr 9th 2010, 05:25 AM
floater
Sphere surface area
I was wondering if someone could help explain the following to me.

http://i39.tinypic.com/t9a2ax.jpg

$dw = sin(\theta) d\theta d\phi$

If only angles are given, how can you compute the surface area? And is it just trigonometry that is being used to computer the edge lengths?
• Apr 9th 2010, 07:31 AM
Hello floater
Quote:

Originally Posted by floater
I was wondering if someone could help explain the following to me.

http://i39.tinypic.com/t9a2ax.jpg

$dw = sin(\theta) d\theta d\phi$

If only angles are given, how can you compute the surface area? And is it just trigonometry that is being used to computer the edge lengths?

The sphere is assumed to have a radius of 1 unit. $\theta$ and $\phi$ are the 'latitude' and 'longitude' of any point on its surface.

$\theta$ is the angle measured upwards from the horizontal ( $x$- $z$) plane through the centre of the sphere - the point's 'latitude', as it were.

$\phi$ is the 'longitude' of the point. This is the angle, measured anticlockwise around the 'equator', between two planes: the vertical plane containing the
great circle* through the point and the $y$- $z$ plane.

A neighbouring point $(\theta+d\theta, \phi+d\phi)$ is chosen, defining an area of the sphere's surface which, for small increases in $\theta$ and $\phi$, is approximately rectangular.

Simple trigonometry will show that the radius of the 'circle of latitude' is $\sin\theta$. So an angle of $d\phi$ at its centre is subtended by an arc of length $\sin\theta d\phi$ on its circumference.

The angle $d\theta$ is subtended at the centre of the great circle by an arc of length $d\theta$, since this circle has unit radius.

Hence the dimensions of the approximate rectangle are $\sin\theta d\phi \times d\theta$. Its area is therefore given by
$d\omega = \sin\theta d\theta d\phi$