## Changing from rectangular coordinate to sperical

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?

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age
 Recognitions: Homework Help Science Advisor How did you conclude that sin^2(phi)*cos^2(theta) + sin^2(phi)*sin^2(theta) equals sin^2(phi) + sin^2(phi)??

 Quote by tnutty 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?
Look at the third to the last line again:

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

hope that helps.

Recognitions:
Gold Member
Homework Help

## Changing from rectangular coordinate to sperical

 Quote by tnutty But the book has it as : cos^2(phi) = sin^2(phi) ?? what gives?
Once you correct your mistake that Dick has pointed out, you might notice that your book's answer:

$$\cos^2(\phi) = \sin^2(\phi)$$

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

 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?

Recognitions:
Gold Member
Homework Help
 Quote by tnutty Yes I see what I did wrong. It is cos^2(phi) = sin^2(phi) = cos(phi) = sin(phi)
You mean $|\cos(\phi)| = |\sin(\phi)|$
 >> . 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?
Nope. Guess again. Think about $|\tan(\phi)| = 1$.

 I mean pi/4.

Recognitions:
Gold Member
Homework Help
 Quote by tnutty I mean pi/4.
Remember $\tan(\phi)$ can be ± 1. $\pi/4$ will get you the top half of the cone. What value of $\phi$ will get you the bottom half?

 -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

Recognitions:
Gold Member
Homework Help
 Quote by 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
So if phi is between 0 and pi, what value gives the bottom half of the cone? Visualize the surface.

 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.
 Recognitions: Gold Member Homework Help 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 $\phi$ that gives the bottom half.
 Thread Tools

 Similar Threads for: Changing from rectangular coordinate to sperical Thread Forum Replies Calculus & Beyond Homework 6 Classical Physics 3 Precalculus Mathematics Homework 2 Introductory Physics Homework 7