
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 :)

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 xy 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.

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(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

whoa, what'd you use to generate that image?

spherical are listed ( p,theta,phi)
disappoiningly I just used paintbrush

whoa thanks! so in your picture is the point the part in towards the bottom right?

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yay thank you thats where i had it i think i'm finally getting the hang of this!