Changing from rectangular coordinate to sperical

tnutty

1. The problem statement, all variables and given/known data

change from rectangle to spherical coordinate :

z^2 = x^2 + y^2


I know that :

z = pcos(phi)

x = psin(phi)cos(theta)

y = psin(phi)sin(theta)

there fore

z^2 = x^2 + y^2 in spherical coordinate is

p^2cos(phi)^2 = (psin(phi)cos(theta))^2 + (psin(phi)sin(theta))^2

=

cos^2(phi) = sin^2(phi)*cos^2(theta) + sin^2(phi)*sin^2(theta)

=

cos^2(phi) = sin^2(phi) + sin^2(phi)

=

cos^2(phi) = 2*sin^2(phi)


But the book has it as : cos^2(phi) = sin^2(phi) ?? what gives?

Dick

How did you conclude that sin^2(phi)*cos^2(theta) + sin^2(phi)*sin^2(theta) equals sin^2(phi) + sin^2(phi)??

Legion81

Look at the third to the last line again:

cos^2(phi) = sin^2(phi)[cos^2(theta) + sin^2(theta)]

hope that helps.

LCKurtz

Once you correct your mistake that Dick has pointed out, you might notice that your book's answer:

[tex] \cos^2(\phi) = \sin^2(\phi)[/tex]

while correct, isn't completely simplified. Can you see that the final answer should be of the form [itex]\phi = C[/itex] for the appropriate value of the constant C, and that the surface is actually one of the parameter surfaces for spherical coordinates?

tnutty

Yes I see what I did wrong.

It is cos^2(phi) = sin^2(phi)

=

cos(phi) = sin(phi)

>>
. Can you see that the final answer should be of the form phi = C.
Yes I guess pi/2 would be correct.

So the equation z^2 = x^2 + y^2 is phi = pi/2 in spherical coord?

LCKurtz

You mean [itex]|\cos(\phi)| = |\sin(\phi)|[/itex]
Nope. Guess again. Think about [itex]|\tan(\phi)| = 1[/itex].

tnutty

I mean pi/4.

LCKurtz

Remember [itex]\tan(\phi)[/itex] can be ± 1. [itex]\pi/4[/itex] will get you the top half of the cone. What value of [itex]\phi[/itex] will get you the bottom half?

tnutty

-pi/4 ?

would this be ok :


plus/minus : n*pi/4n : where n is a integer. OR no because phi is limited to 0 <= phi <= pi

LCKurtz

So if phi is between 0 and pi, what value gives the bottom half of the cone? Visualize the surface.

tnutty

Is there a bottom half? I don't think it will be within bounds because
only pi/4 would solve | tan(phi) | = 1 where 0<= phi <= pi.

| tan( -pi/4) |= 1 but thats no within bounds and

|tan( 5pi/4 ) |= 1 but thats also not within bounds, and those 2 points are the
next points which satisfies the equation.

LCKurtz

Yes, there is a bottom half. In your original xyz equation z can be negative. Draw a sketch of the cone, top and bottom. Just look at it. There obviously is a [itex]\phi[/itex] that gives the bottom half.