# Thread: converting from spherical coordinates

1. ## converting from spherical coordinates

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

Can anyone help me at all?

Thanks again.

2. 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?

3. 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.

4. 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.

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

6. 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.

7. 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 ($\displaystyle \theta$, $\displaystyle \phi$, $\displaystyle \rho$) coordinates is
$\displaystyle x= \rho sin(\phi)cos(\theta)$
$\displaystyle y= \rho sin(\phi)sin(\theta)$
$\displaystyle z= \rho cos(\phi)$.

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

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

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

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

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

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### convert p(2,3,-1) into spherical co-ordinates

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