Describe the surface in cylindrical coordinates?

Colts
Messages
76
Reaction score
0

Homework Statement


The surface is x^2/y*z=10. Put this into cylidrical coordinates. in the form r=f(theta,z)


Homework Equations


No clue


The Attempt at a Solution


No clue
 
Physics news on Phys.org
in your book, you will probably find equations for x, for y in terms of theta and z. don't have my book here with me.
 
I don't understand what r=f(theta,z) means and how to write my answer in that form
 
apparently, this is the conversion from cartesian coordinates (x,y,z) to cilindrical (rho, phi, z):

x = \rho \cos \varphi
y = \rho \sin \varphi
\rho = \sqrt{x^{2}+y^{2}} and

\varphi = \begin{cases} 0 &amp; \mbox{if } x = 0 \mbox{ and } y = 0\\ \arcsin(\frac{y}{\rho}) &amp; \mbox{if } x \geq 0 \\ -\arcsin(\frac{y}{\rho}) + \pi &amp; \mbox{if } x &lt; 0\\ \end{cases} <br />
 
Colts said:
I don't understand what r=f(theta,z) means and how to write my answer in that form

Substitute ##x=r\cos\theta,\, y=r\sin\theta## in the equation and solve it for ##r##.
 
f(theta, z) means that you should equate radius to a function of theta and z. in this case theta is the angle \phi [/phi] in the equations above. this angle, simply put, is the same as the angle in polar coordinates. the only difference between polar coordinates and cilindrical, is that with cilindrical, you have height (z) as well
 
damabo said:
in this case theta is the angle \phi [/phi] in the equations above
<br /> ##\theta## is given by \theta, not \phi.
 
Got it. Thanks guys
 
Back
Top