Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
    
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
1  
$z\geq 0$ 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
add comment

1 Answer

up vote 4 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.