Solving Density Integral for Solid Cylinder

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Homework Statement



The density of material at any point in a solid cylinder, y^2 + z^2 ≤ a^2, 0 ≤ x ≤ b is proportional to the distance of the point from the x-axis. Find the density. Use k as the proportionality constant.


Homework Equations





The Attempt at a Solution



Can someone guide me here? is x here a?
 
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No x isn't a. That's the parametric representation of a cylinder whose axis coincides with the x-axis. Now from what is given you should be able to tell that the equation describes a circle on the y-z plane. What then is the equation of any point within that circle? That, and using pythagoras theorem should give you the answer.
 
how can I get the equation of any point within the circle?
 
the point (x,y) you mean here are you referring that to a point in the circle or.. if I visualize this correctly the cylinder is not standing on it's base.. right?
 
Well it depends on what you mean by "base". The base of the cylinder is the y-z plane. So from the conventional orientation of x-y-z plane it's lying on its curved side.
 
Like this:

http://img22.imageshack.us/img22/5278/terip.jpg
 
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ok.. so now I need to find the equation within the circle... I am really confused how this can be achievedf
 
If the coordinates of a point are (x, y, z), what is the distance of this point from the x-axis?
 
which part of the x-axis? the origin?
 
is the answer just k(y^2+z^2)?
 
You forgot the square root. The density is k√(y² + z²). You might also want to mention that the density is 0 for √(y² + z²) > a.
 
hmm.. yea I fogot the sqrt,, how can I express this in terms of r?
 
how can I express this in terms of r?

Are you going to use polar/cylindrical coordinates? Use the polar coordinates (r,theta) for the yz plane instead of the usual xy plane.
 
polar coordinates... how can I do that?
 
-EquinoX- said:
polar coordinates... how can I do that?

?

:confused:
 
I need to represent the density above in terms of r... not y and z
 
What is r? What does r mean?
 
I believe it's the radius
 
Well, you already know the radius is given by [tex]r=\sqrt{y^2 + z^2}[/tex]. So going from this to an answer in terms of r just requires to replace that latter expression with r.
 
I am asked to set up the integral to find the mass of such a cylinder with radius 6 and length 11.

f(x, r, θ) here is just the density? in terms of x, r and theta?
 
well I am confused on how to represent the density in terms of f(x, r, θ)
 
one was in terms of x,y and one was in terms of r... but not f(x,r,theta).. I am mostly confused with the theta here
 
is it just something like this if the cylinder has a radius of 6 ad length 11:

[tex]\int_0^{11}\int_0^{2\pi} \int_0^6 kr^2 dr d\theta dz[/tex]