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The sphere is simply the set of points equidistant from a given point. The same exact approach can work in any dimension...

An $n$-dimensional sphere of radius $r$ centered at point $a$ is the set of points in ${\bf R}^{n+1}$ that lie at the distance of $r$ from point a:

$$\{x \in {\bf R}^{n+1}: ||x-a|| = r \}.$$

Exercise. What is a 0-sphere?

In topology we talk about the $n$-dimensional sphere denoted by ${\bf S}^n$. One can still think of it as the boundary of an $(n+1)$-dimensional ball:

$${\bf S}^n = \partial {\bf B}^{n+1}.$$

As a cell complex

The $n$-sphere can be build as a cell complex from just two cells: a $0$-cell and a $n$-cell.

N-sphere as cell complex.jpg

Exercise. Find other representations of the $n$-sphere as a cell complex?

Exercise. What about a cubical complex?

See how the sphere can be represented as a parametric surface.



$$I = \int_D ( 1 - x^2 - y^2 )^{\frac{1}{2}} dA,$$


$$D = \{ ( x, y ): x^2 + y^2 \leq 1 \}.$$


$$z = ( 1 - x^2 - y^2 )^{\frac{1}{2}}$$

$\rightarrow x^2 + y^2 + z^2 = 1$, sphere.

Use iterated integration. Then

$$I = \int_{-1}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} ( 1 - x^2 - y^2 )^{\frac{1}{2}} dy dx.$$

To evaluate

$$\int ( 1 - x^2 - y^2 )^{\frac{1}{2}} dy$$

we need trigonometric substitution. This means we have to carry out separate trigonometric substitution for each $x$. Instead we do trigonometric" substitution for $( x, y )$ which turns into the "change of variables" and the new variable are the polar coordinates:

$$( x, y ) \rightarrow ( r, \theta ) $$

$$x = r \cos \theta \longleftrightarrow r = ( x^2 - y^2 )^{\frac{1}{2}}$$

$$y = r \sin \theta \longleftrightarrow \theta = \arctan \frac{y}{x}.$$