# Spherical equation to Rectangular

1. Jan 29, 2016

### RyanTAsher

1. The problem statement, all variables and given/known data

An equation is given in spherical coordinates

$\rho sin(\phi) = 8cos(\theta)$

Express the equation in rectangular coordinates

a) $(x-4)^2 + y^2 = 16$
b) $x^2 + y^2 + z^2 = 16$
c) $x^2 + (y-4)^2 = 16$

2. Relevant equations

$x = \rho sin \phi cos \theta, y = \rho sin \phi sin \theta, z = \rho cos \phi$

$\rho = \sqrt {x^2 + y ^2 + z^2}, tan \theta = \frac y x, cos \phi = \frac {z} {\sqrt {x^2 + y^2 + z^2} }$

3. The attempt at a solution

Everytime I try to do this problem, I cannot get any of the clean answers is has in the multiple choice.

In the relevent equations above, I plug in the $\rho$ for the $\rho$, solve for $\theta$ and $\phi$ and plug those in as well, and I get this giant mess of an equation that has no means of simplifying down further.

I've tried using trigonometric identities and such to simplify, but I don't think I understand exactly what it wants me to do.

Last edited by a moderator: Jan 29, 2016
2. Jan 29, 2016

### ehild

Do the opposite, substitute the equation $\rho sin(\phi) = 8cos(\theta)$ into the expression for x and y and z and then try which equation from a, b, c is valid.

3. Jan 30, 2016

### RyanTAsher

Okay, I get what you mean, but I'm not seeing how to plug the equation into the x, y, and z expressions, there doesn't seem any common substitution grounds between the whole equation and one of the x,y, or z's.

4. Jan 30, 2016

### ehild

You wrote that $x = \rho sin \phi cos \theta, y = \rho sin \phi sin \theta, z = \rho cos \phi$. The problem states that $\rho sin(\phi) = 8cos(\theta)$. You can substitute $8 \cos(θ)$ for $ρ\sin(\phi)$, so $x=8 \cos^2(θ)$ and $y=8 \cos(θ)\sin(θ)$. Is $(x-4)^2+y^2=16$ true?

5. Jan 30, 2016

### RyanTAsher

v

After doing the substitution I determined that equation wasn't true, I ended up getting $4sin(2\theta) - 64cos(\theta)^2 + 64cos(\theta)^4 = 16$ and I'm not seeing much of anyway to reduce that any further.

My question in response to this, is from getting $x = 8cos^2(\theta)$ and $y = 8cos(\theta)sin(\theta)$ how would I deduce the formula in x, y, z form, without having to look at those multiple choice answers?

6. Jan 30, 2016

### ehild

You did something wrong.
Well, let's go back to the direct way.
$\rho = \sqrt {x^2 + y ^2 + z^2}, tan \theta = \frac {y} {x}, cos \phi = \frac {z} {\sqrt {x^2 + y^2 + z^2} }$. Write cos(θ) in terms of x, y.

7. Jan 30, 2016

### RyanTAsher

$cos(\theta) = \frac {xsin(\theta)} {y}$ ?

8. Jan 30, 2016

### ehild

Last edited: Jan 30, 2016
9. Jan 30, 2016

### RyanTAsher

Oh okay... Do you mean $cos(\theta) = \frac {x} {\sqrt {x^2 + y^2}}$

10. Jan 30, 2016

### ehild

Correct. Now, substitute the expressions for ρ,sinΦ and cosθ into the equation ρsin(Φ)=8cos(θ).

11. Jan 30, 2016

### RyanTAsher

Okay, so when I did this I found an expression for $sin(\phi)$ as well before substituting, because I didn't know how else to do it.

$psin(\phi) = 8cos(\theta)$
$\sqrt{x^2 + y^2 + z^2} \frac {\sqrt{x^2 + y^2}} {\sqrt{x^2 + y^2 + z^2}} = 8 \frac {x} {\sqrt{x^2 + y^2}}$
$\frac {x^2 + y^2} {x} = 8$

not really sure what else I can do here besides divide the x through the left hand side.

12. Jan 30, 2016

### ehild

Multiply the equation by x, and arrange it so as the right hand side is zero. Complete square for x.

13. Jan 30, 2016

### RyanTAsher

Okay, I've got it!

After completing the square I've arrived at $(x-4)^2 + y^2 = 16$

Thank you a lot for your time, I was really frustrated with this problem, and wasn't following it. I appreciate it.

14. Jan 30, 2016

### ehild

Well done!

The other method would have been a bit easier. Substituting $x = 8\cos^2(\theta)$ and $y=8 \cos(θ)\sin(θ)$ into the expression $(x-4)^2 +y^2$ : $(8\cos^2(\theta)-4)^2+(8 \cos(θ)\sin(θ))^2 = 64\cos^4(\theta)-64\cos^2(\theta)+16 + 64 \cos^2(\theta)\sin^2(\theta) = 64 \cos^2(\theta)(\cos^2(\theta)+\sin^2(\theta)) -64\cos^2(\theta) +16 = 64 \cos^2(\theta) -64\cos^2(\theta) +16 =16$ , that is, a) proved to be true.

15. Jan 31, 2016

### WWGD

If it is a matter of choosing, you can throw away the middle choice $x^2+y^2+z^2=16$ , which would have equation $\rho=4$. But this is the easy part, ehild took care of the hard 99%.