Results 1 to 4 of 4

Math Help - Double Integral Using Polar Coordinates to Find Mass of Disk

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    5

    Double Integral Using Polar Coordinates to Find Mass of Disk

    Problem: A disk of radius 5 cm has density 15 gm/cm62 at its center, density 0 at its edge. Assume its density is a linear function of the distance from the center.

    Set up the integral to find the total mass of the disk.

    My Attempt So Far:

    I understand how to set up the integral itself in polar coordinates just fine.

    The integral I set up is as follows:

    $\int_{0}^{2\pi}$\int_{0}^{5} f(r, \theta) rdrd\theta

    The part I'm actually having trouble with is determining what the function I'm integrating is!

    In the problem, it says that the density of this disk is 15gm/cm^2 at the center, and 0gm/cm^2 at the edge. It says that the function is a linear function of the distance from the center.

    This is what I drew based on that, with z sort of representing the density.

    Double Integral Using Polar Coordinates to Find Mass of Disk-densitydiagramthing.png

    This graph is in the xy plane (I forgot to label it). I'm not really sure where to go from here, or if this is the right way to look at it. How do I get an equation representing the change in density of the disk?

    Edit:

    I'm still playing with this and I've set up the following idea:

    I know that when r = 0, density = 15, and when r = 5, density= 0, which gives me two points (0,15) and (5,0). If I make these into a linear function, I get -3r. Is that anywhere close to right?
    Last edited by abadscene; October 16th 2010 at 01:58 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    No, I'd say you need to try again with your density function. The density-intercept of -3r is not the required 15. I think you have the right slope, though. If the density is f(r) = -3r + b, find b. Plug that into your integral.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Sep 2010
    Posts
    5
    Thank you so much! I can't believe I didn't see that before...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    You're welcome. Have a good one!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Double integral of polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 22nd 2010, 10:16 AM
  2. double integral, mass using polar co-ordinates
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 24th 2009, 08:09 AM
  3. Double integral in polar coordinates 2
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2008, 07:07 PM
  4. Double integral in polar coordinates 3
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2008, 06:17 PM
  5. Double integral in polar coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 24th 2008, 05:33 PM

Search Tags


/mathhelpforum @mathhelpforum