converting from spherical coordinates

**Nguyen** · October 10th 2010, 05:20 AM

Hey everyone, I really need some help on this one, I just can seem to find a connection anywhere between formulas for each coordinate system.

Convert $\phi = \frac{\pi}{6}$ from spherical coordinates in to both Cartesian coordinates and cylindrical coordinates.

Can anyone help me at all?

Thanks again.

**Nguyen** · October 10th 2010, 05:56 AM

Is it as simple as just plugging in phi=pi/6 into the relavant forumula relating spherical and cylindrical, and spherical and cartesian coordinates? And leaving as it is, with say rho and theta still in the equation?

**Pulock2009** · October 10th 2010, 06:59 AM

spherical coordinates refers to an angle inscribed on a circle of unit length. So the polar coordinates og your point will be (1,pi/6). Now we know that x^2+y^2=1 and tan^-1(x/y)=pi/6. solving both of them we get your point in cartesian system as (+-sqrt(3)/2,+-1/2). hope your are comfortable with my notations. i have no latex installed and having some problem with uploading bmp files as well.

**Pulock2009** · October 10th 2010, 07:12 AM

As for cylindrical coordinates is it the hyperbolic or elliptical system?? if that is the case then simply solve for the equation and tanh^-1(y/x)=pi/6. i suppose that would suffice.

**Nguyen** · October 10th 2010, 02:36 PM

Thanks Pulock2009, but we haven't done the hyperbolic system yet. Can anyone help me with the cylindrical coordinates?

**Nguyen** · October 10th 2010, 03:16 PM

Originally Posted by Pulock2009

spherical coordinates refers to an angle inscribed on a circle of unit length. So the polar coordinates og your point will be (1,pi/6). Now we know that x^2+y^2=1 and tan^-1(x/y)=pi/6. solving both of them we get your point in cartesian system as (+-sqrt(3)/2,+-1/2). hope your are comfortable with my notations. i have no latex installed and having some problem with uploading bmp files as well.

And isn't Cartesian coordinates supposed to be (x,y,z), I don't think we can find all 3 of these when we are only given phi.

**HallsofIvy** · October 10th 2010, 03:27 PM

The aren't asking for a specific point, but for the equation satisfied by all points that for which [tex]\phi= \frac{\pi}{6}.

The connection between Cartesian (x, y, z) coordinates and Spherical ( $\theta$ , $\phi$ , $\rho$ ) coordinates is
$x= \rho sin(\phi)cos(\theta)$
$y= \rho sin(\phi)sin(\theta)$
$z= \rho cos(\phi)$ .

With $\phi= \pi/6$ then $sin(\phi)= \frac{1}{2}$ and $cos(\phi)= \frac{\sqrt{3}}{2}$ so that
$x= \frac{\rho}{2}cos(\theta)$
$y= \frac{\rho}{2}sin(\theta)$
$z= \rho\frac{\sqrt{3}}2}$

Then $x^2+ y^2= \frac{\rho^2}{4}(cos^2(\theta)+ sin^2(\theta)= \frac{\rho^2}{4}$ and
$z^2= \frac{3\rho^2}{4}$ .

So $z^2= 3(x^2+ y^2)$ . That's the equation of the cone of all points satifying $\phi= \frac{\pi}{6}$ .

In cylindrical coordinates, $x^2+ y^2= r^2$ so that equation is $z^2= 3r^2$ in cylindrical coordinates.

**Pulock2009** · October 10th 2010, 05:37 PM

in your case z will always be zero because we are dealing with 2-D. polar coordinates generally refers to 2-D.

Thread: converting from spherical coordinates

LinkBack

Thread Tools

Display

converting from spherical coordinates

Similar Math Help Forum Discussions

Converting Cartersian coordinate term to spherical

Please help in transforming rectangle coordinates to spherical coordinates.

Converting a vector field from spherical to cylindrical coordinates

Converting from Rectangular to Spherical Coordinates

Integration first converting to spherical polars

Search Tags