Converting Cartersian coordinate term to spherical

• Jun 10th 2011, 08:12 PM
laguna92651
Converting Cartersian coordinate term to spherical
I am working through a problem where the vector has a Cartesian term and want to convert it to spherical::
(Sorry, I couldn't figure out how to use the equation editor)

sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) (x component of vector)

I know:
r=sqrt(x^2+y^2+z^2)
x=rsin\theta cos\phi
y=rsin\theta sin\phi
z=rcos\theta

I also know the answer is:
(rsin\phi )/r = sin\phi

I can't figure out how rsin\phi = sqrt(x^2=y^2) in the answer.
• Jun 10th 2011, 08:28 PM
Also sprach Zarathustra
Quote:

Originally Posted by laguna92651
I am working through a problem where the vector has a Cartesian term and want to convert it to spherical::
(Sorry, I couldn't figure out how to use the equation editor)

sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) (x component of vector)

I know:
r=sqrt(x^2+y^2+z^2)
x=rsin\theta cos\phi
y=rsin\theta sin\phi
z=rcos\theta

I also know the answer is:
(rsin\phi )/r = sin\phi

I can't figure out how rsin\phi = sqrt(x^2=y^2) in the answer.

How this is possible that the first red is NOT an equation and the second red is an equation?
• Jun 10th 2011, 08:55 PM
laguna92651
The second red, just goes one step further to simplify the answer.
• Jun 10th 2011, 09:02 PM
Also sprach Zarathustra
Quote:

Originally Posted by laguna92651
The second red, just goes one step further to simplify the answer.

Let me put my question in other words, where is the "=" sign in the first red?
• Jun 10th 2011, 09:10 PM
laguna92651
I'll ask the question is way, how did the left convert to the right side? Cartesian to spherical

sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) = [r(sin(phi) ]/r
• Jun 11th 2011, 03:49 AM
HallsofIvy
Quote:

Originally Posted by laguna92651
I'll ask the question is way, how did the left convert to the right side? Cartesian to spherical

sqrt(x^2+y^2)/sqrt(X^2+y^2+z^2) = [r(sin(phi) ]/r

sqrt(x^2+ y^2+ z^2)= r. that's where the denominator on the right comes from.

x= r cos(theta) sin(phi), y= r sin(theta) sin(phi) so x^2= r^2 cos^2(\theta) sin^2(phi) and y^2= r^2 sin^2(\theta) sin^2(\phi).

x^2+ y^2= r^2 cos^2(theta) sin^2(phi)+ r^2 sin^2(theta) cos^2(phi)= r^2 sin^2(phi)(cos^2(theta)+ sin^2(theta))
= r^2 sin^2(\phi). That's the numerator.
• Jun 11th 2011, 10:15 AM
laguna92651
Thanks for the help.

Did you mean for the second term in the x^2+y^2 equation to be sin^2(phi) instead of cos^2(phi)?
Second term: r^2 sin^2(theta) cos^2(phi)

You used; x= r cos(theta) sin(phi)
My electromagnetics book uses; x= r cos(phi) sin(theta)

So I get sin(theta) as an answer, the correct answer is sin(phi). Am I missing something?

Theta is defined as the angle between the z-axis the position vector.
Phi is measured from the x-axis to the plane of the vector.