# urgent help please - sphere

• Jan 15th 2007, 11:07 PM
Jenny20
Show why (x,y,z) = (a*sin(phi)* cos(theta) , a*sin(phi)* sin(theta), sqrt(3)*a*sin(phi) ) lies on the sphere of radius 2a centered at origin.
• Jan 16th 2007, 04:40 AM
topsquark
Quote:

Originally Posted by Jenny20
Show why (x,y,z) = (a*sin(phi)* cos(theta) , a*sin(phi)* sin(theta), sqrt(3)*a*sin(phi) ) lies on the sphere of radius 2a centered at origin.

I'm not getting a result that says it's on this sphere:

Find the distance between this point and the origin:

$\displaystyle d = \sqrt{(asin(\phi)cos(\theta) - 0)^2 + (asin(\phi)sin(\theta) - 0)^2 + (\sqrt{3}asin(\phi) - 0)^2}$

$\displaystyle d = \sqrt{a^2sin^2(\phi)cos^2(\theta) + a^2sin^2(\phi)sin^2(\theta) + 3a^2sin^2(\phi)}$

$\displaystyle d = \sqrt{a^2sin^2(\phi)[cos^2(\theta) + sin^2(\theta)] + 3a^2sin^2(\phi)}$

$\displaystyle d = \sqrt{a^2sin^2(\phi) + 3a^2sin^2(\phi)}$

$\displaystyle d = \sqrt{4a^2sin^2(\phi)}$

$\displaystyle d = |2asin(\phi)|$

For this point to be on the sphere centered at the origin with a radius of 2a this distance would have to be 2a.

-Dan
• Jan 16th 2007, 05:01 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
$\displaystyle d = |2asin(\phi)|$

For this point to be on the sphere centered at the origin with a radius of 2a this distance would have to be 2a.

I think this shows that,
$\displaystyle -1\leq d\leq 1$.
Those all the points in in the sphere though not necessarility on it.
• Jan 16th 2007, 08:16 AM
Jenny20
Thank you very much !