Proove this: x^2 + y^2 + z^2 = p^2 (p is actually "rho")?

Question

This is for spherical coordinates, so:
x=psin(a)cos(b)
y=p(a)sin(b)
z=pcos(a)

(a and b are subsituted in for theta and phi, for obvious typing reasons.)

I feel i've wasted enough paper trying to figure this out myself.

haha, i copied it down wrong in class, no wonder i couldn't solve it!!!! lol, thanks.

Answer 1

A little correction.... y = psin(a)sin(b)

x^2 + y^2 + z^2 = p^2
(psin(a)cos(b))^2 + (psin(a)sin(b))^2 + (pcos(a))^2 = p^2
p^2 (sin^2(a)cos^2(b)+ sin^2(a)sin^2(b) +cos^2(a)) = p^2
sin^2(a)cos^2(b)+sin^2(a) sin^2(b) + cos^2(a) = 1
factor out sin^2(a) from first 2 terms:
sin^2(a) (cos^2(b)+sin^2(b)) + cos^2(a) = 1
sin^2(a) (1) + cos^2(a) = 1
sin^2(a) + cos^2(a) = 1,
which is true by the Pythagorean Identity.

Answer 2

rho (p) is the length of the vector from the centre to the surface of the sphere.
when the vector is pointing at an angle "A" above the x-y plane and "B" from the x axis (the first quadrant going towards the positive extension of hte y axis), then:

The length of the projection of the vector onto the x-y plane is p*cosA
Therefore, the projection onto the x axis is
(p*cosA)*cosB
If the angles are both 0, then the vector is along the x axis, therefore the length of the projection on the x axis is p*1*1.

The projection on the y-axis is (p*cosA)*sinB

The projection on the z axis is p*sinA

To do x^2+y^2+z^2, we'll proceed in this way:

{x^2+y^2}+z^2

{[(p*cosA)^2 *(cosB)^2 ] + [(p*cosA)^2 *(sinB)^2 ]} + z^2
{ (p*cosA)^2 *[(cosB)^2 + (sinB)^2)]} + z^2

Using cos^2 + sin^2 = 1, we get
{ (p*cosA)^2 * 1 } + z^2

p^2 * (cosA)^2 + z^2

filling in with z = p*sinA

p^2 * (cosA)^2 + (p*sinA)^2
p^2*(cosA)^2 + p^2*(sinA)^2
p^2 *[ (cosA)^2 + (sinA)^2]
p^2* [1]
p^2

QFD

Answer 3

This is between spherical & rectangular coordinates,
The derivation can be found at

http://mathworld.wolfram.com/SphericalCoordinates.html