Results 1 to 5 of 5

Math Help - Using disk method to find a value of x

  1. #1
    Newbie
    Joined
    Jul 2010
    Posts
    6

    Using disk method to find a value of x

    Hey guys this problem has me stumped.

    The region bounded by y = \sqrt{x}, y = 0, and y = 4 is revolved about the x-axis.

    (a) Find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

    (b) Find the values of x in the interval [0,4] that divide the solid into three parts of equal volume.

    I suspect the easiest way would be to use the disk method; however, I'm not sure how to get an x value that would result in 2 or 3 parts of equal volume. I can guess and check for two, but there has to be an easier way.

    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,621
    Thanks
    426
    Quote Originally Posted by Poptimus View Post
    Hey guys this problem has me stumped.

    The region bounded by y = \sqrt{x}, y = 0, and y = 4 is revolved about the x-axis.

    (a) Find the value of x in the interval [0,4] that divides the solid into two parts of equal volume.

    (b) Find the values of x in the interval [0,4] that divide the solid into three parts of equal volume.

    I suspect the easiest way would be to use the disk method; however, I'm not sure how to get an x value that would result in 2 or 3 parts of equal volume. I can guess and check for two, but there has to be an easier way.

    Thanks
    first, find the overall volume, V, using the disk method from x = 0 to x = 4

    then set up the equation ...

    \displaystyle \pi \int_0^a (\sqrt{x})^2 \, dx = \frac{V}{2}

    solve for a


    set up the equations ...

    \displaystyle \pi \int_0^b (\sqrt{x})^2 \, dx = \frac{V}{3}

    \displaystyle \pi \int_0^c (\sqrt{x})^2 \, dx = \frac{2V}{3}

    solve for b and c
    Last edited by skeeter; July 26th 2010 at 04:04 PM. Reason: had area on the brain instead of volume ... fixed
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    Did you make a typo? Did you mean bound by y=\sqrt{x}, ~ y=0, ~ x=4 ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2010
    Posts
    354
    The inside of skeeter's integrals was not correct, but the important part was how he set up the endpoints. Just make sure you use the correct integrals though.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jul 2010
    Posts
    6
    Yeah that was a typo, it should have been  x = 4 . I'll see what I come up with, thanks guys.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Disk method.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 3rd 2010, 02:51 PM
  2. Replies: 5
    Last Post: January 22nd 2010, 05:50 AM
  3. [SOLVED] The Disk Method, Hwk
    Posted in the Calculus Forum
    Replies: 2
    Last Post: June 24th 2009, 06:36 PM
  4. Volume (disk method)
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 25th 2008, 07:08 PM
  5. Replies: 2
    Last Post: August 17th 2008, 12:02 PM

Search Tags


/mathhelpforum @mathhelpforum