# spherical coordinates

• Sep 10th 2009, 05:03 PM
holly123
spherical coordinates
does anyone know how to plot (6, 11pi/12, 0) ? i tried transforming it to rectangular coordinates and got (0,0,1/2) is this right?

also, for cylindrical coordinates, how do you plot (3, pi/4, 2)? i'm not sure if i did that one right.

greatly appreciated :)
• Sep 10th 2009, 06:23 PM
seld
cylindrical coordinates are : $\displaystyle (r,\theta, z)$ right?

spherical coordinates are: $\displaystyle (\theta, \phi, \rho)$ right?

so in your second question the r = 3, theta = pi/4 and z = 2

r and theta are on the xy plane. You're looking at a point that has a radius 3 and an angle of 45 degrees measured from the x axis. So that's roughly $\displaystyle (\sqrt{1.5},\sqrt{1.5})$ on the x-y plane. Then your Z is your height.

So you just have a point 2 spaces above that.

About your first question if you could be a little clearer, it seems like it should also be cylindrical but you named the thread as "spherical" coordinates.
• Sep 10th 2009, 06:30 PM
Calculus26
(6, 11pi/12, 0)

in rect coord x= psin(phi)cos(theta) =0
y= psin(phi)sin(theta) = 0
z= pcos(phi) = 6cos(0) = 6

For (3, pi/4, 2) See attachment
• Sep 10th 2009, 06:31 PM
seld
whoa, what'd you use to generate that image?
• Sep 10th 2009, 06:35 PM
Calculus26
spherical are listed ( p,theta,phi)

disappoiningly I just used paintbrush
• Sep 10th 2009, 06:40 PM
holly123
whoa thanks! so in your picture is the point the part in towards the bottom right?
• Sep 10th 2009, 06:43 PM
Calculus26
See attachment
• Sep 10th 2009, 06:45 PM
holly123
yay thank you thats where i had it i think i'm finally getting the hang of this!