Spherical coordinates equation

[yitriana]

Homework Statement

Identify surface whose equation in spherical coordinates is given
p = sin(theta)*sin(fi)

The Attempt at a Solution

I know that y = r*sin(theta)*sin(fi). and thus, y = rp.

This yields y = (x² + y²)^0.5*(x² + y²+z²)^0.5

However, this is rather ugly. The answer is supposed to be "a sphere with radius 0.5, center (0. 0.5, 0)" however, I don't see how that results from this expression.

 

[Office_Shredder]

What is p supposed to be?

 

[yitriana]

p is ro. p = (x^2 + y^2 + z^2)^0.5

 

[Fenn]

I'm guessing that the expression should read something like

<br /> r = \sin\theta\sin\phi<br />

where \theta and \phi are your polar angles. First, look in the x-y plane, where \theta=90^\circ. This simplifies to

<br /> r = \sin\phi.<br />

Now, apply r to the equations linking polar and Cartesian coordinates, for \theta=90^\circ, as

<br /> x = r\cos\phi<br />

<br /> y = r\sin\phi.<br />

You'll see they come out to be

<br /> x = \sin\phi\cos\phi<br />

<br /> y = \sin^2\phi.<br />

Next, recall the double-angle formulas that \sin(2x) = 2\sin x\cos x and \cos(2x) = 1 - 2 \sin^2 x. I'll leave it as an exercise to substitute these back into the expressions for x and y. What you should notice is that your values should now look like

<br /> x = A \sin(2\phi)<br />

and

<br /> y = A \cos(2\phi) + B<br />

where A and B are numbers. You should recognize this as the parametric representation of a circle, at the coordinate (0,B).

Now, this doesn't exactly answer your question, but it should hopefully get you to visualize how the surface should be a sphere. Particularly, if you repeat this exercise in the y-z plane, where \phi=90^\circ, you should find find another case where the result is a circle, offset from the origin by some distance B.

I think that, once you've identified what B is, you can look into the coordinate transformation of

<br /> x&#039; = x, \quad y&#039; = y + B, \quad z&#039; = z<br />

and then determine r&#039; = \sqrt{x&#039;^2 + y&#039;^2 + z&#039;^2} which should be a constant. If this is the case, it shows that this surface is a sphere centered at (0,B,0).