# Parameterizing the upper hemisphere of a sphere with an upward pointing normal

Can someone explain how to do this?

area we're dealing with:

$x^2 + y^2 + z^2 = a^2, z \geq 0$

I'm aware that the answer is:

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

$y = a \sin(\phi) \sin(\theta)$

$z = a \cos(\phi)$

I'm just not quite sure how to get there.

Thanks!

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You're aware of spherical coordinates? Anyway, here's an idea: imagine a segment of length $a$ above the x-y plane, and note the angles it makes with the z-axis, and the angle made by its "shadow" in the x-y plane and the x-axis. – J. M. Apr 27 '11 at 19:46
I was able to find the answer, I just want to know why that is the answer, if that makes sense. – Sir Winford Apr 27 '11 at 19:47
$z\geq 0$ obviously leads to the condition $\phi\leq\pi/2$. Together with the usual restriction on the parameters of spherical coordinates, you get $\theta \in [0,2\pi)$ and $\phi \in [0,\pi/2]$. – Fabian Apr 27 '11 at 19:49

Intuitively, $\theta$ represents longitude on the (hemi-)sphere, and $\phi$ represents co-latitude. If you let $a$ vary, then it would describe altitude.

To see that these coordinates actually describe the hemisphere is straightfoward: all you have to do is check that \begin{align*} x & = a \sin \phi \cos \theta \\ y &= a \sin \phi \sin \theta \\ z &= a \cos \phi \end{align*} actually satisfy $x^2 + y^2 + z^2 = a^2$.

But how would one actually derive these coordinates? Essentially, it works the same as in 2-dimensional polar coordinates: draw the right triangle and use "soh cah toa."

So which right triangle do we draw? Let $O$ denote the origin, let $P = (x,y,z)$ denote the point on the (hemi-)sphere in question, and let $Q = (x,y,0)$ denote the projection of $P$ directly down onto the $xy$-plane. Look at the triangle $\Delta OPQ$. Mathworld's page on spherical coordinates has an illustration of this.

The height of this triangle (side $PQ$) has length $z$, the base (side $OQ$) has length $r = \sqrt{x^2 + y^2}$, and the hypotenuse (side $OP$) is the radius of the sphere $a$. Then by "soh cah toa," \begin{align*} r &= a \sin \phi \\ z &= a \cos\phi. \end{align*} Now we examine the point $Q$, and use 2-dimensional polar coordinates. Since the length of $OQ$ is $r = a\sin\phi$, we have \begin{align*} x &= (a \sin \phi)\cos \theta \\ y &= (a \sin\phi)\sin\theta, \end{align*} which is what we wanted.

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In the actual setup of conventional spherical coordinates, $\phi$ is actually co-latitude, the complement of the latitude. Some care is needed in comparing results in the literature, since some references use $\theta$ for the co-latitude and $\phi$ for the longitude. – J. M. Apr 27 '11 at 23:48
True. Fixed. (character limit) – Jesse Madnick Apr 28 '11 at 3:07