Results 1 to 7 of 7

Math Help - Integral not sure how to do

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    3

    Integral not sure how to do

    my teacher gave this to me on a test that i failed i would like to know how to do this kind of problem in case it appears in the future

    a region bounded by y=(-x^2)+4, y=4, and x=2 is revolved around the y-axis. use cylindrical shells to find the volume of the solid generated.

    hope someone can help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Apr 2008
    Posts
    9
    I don't know what the cylindrical shells means, but I would just do

    Pi ( Integral from 0 to 4 of (4 -(4-y)) dy)

    I'll explain how I got that if you want.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    3
    please
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Mathstud28's Avatar
    Joined
    Mar 2008
    From
    Pennsylvania
    Posts
    3,641
    Quote Originally Posted by unique_name1 View Post
    my teacher gave this to me on a test that i failed i would like to know how to do this kind of problem in case it appears in the future

    a region bounded by y=(-x^2)+4, y=4, and x=2 is revolved around the y-axis. use cylindrical shells to find the volume of the solid generated.

    hope someone can help.
    V=2\pi\int_0^{4}(2-x)\cdot{(-x^2+4)}dx..that is using the shell method
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    3
    ok thank you
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    Posts
    9
    Okay, so a graph of the three functions gives an enclosed region that kind of looks like a right triangle, with the right angle in the top right, and the hypotenuse being a curve, in the first quadrant.

    If you are to revolve this around the y-axis you will get a disc shape because of the space between the area and the axis.

    If you remember how to find the area of the disc, it is the outer radius squared, minus the inner radius squared, times Pi.

    Since you are revolving around the y-axis, your change is going to be along the y axis; hence dy. To integrate with respect to y, you need to change your equation to x=

    This gives you x= +- sqrt[4-y] (Sorry I do not know how to use math tags)

    You only need the positive portion of this graph since we only need to worry about y from 0 to 4 as bounded by the fuctions.

    When you imagine the disc that will be generated. x=2 becomes your outer radius, and x = sqrt[4-y] your inner radius.

    You integrate starting at y = 0 because this is where the graphs of the two previous functions intersect. You end at 4 because your third function bounds the area here.

    This all gives you the integral I gave you before.

    Tell me if anything doesn't make sense.

    EDIT:

    nevermind.. didn't know there is a shell method
    Follow Math Help Forum on Facebook and Google+

  7. #7
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    See here

    By the method of shells. the volume is given by:

    V = 2 \pi \int_0^2 x[4 - (4 - x^2)]~dx
    Attached Thumbnails Attached Thumbnails Integral not sure how to do-region.jpeg  
    Last edited by Jhevon; April 8th 2008 at 05:56 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 31st 2010, 07:38 AM
  2. Replies: 1
    Last Post: June 2nd 2010, 02:25 AM
  3. Replies: 0
    Last Post: May 9th 2010, 01:52 PM
  4. Replies: 0
    Last Post: September 10th 2008, 07:53 PM
  5. Replies: 6
    Last Post: May 18th 2008, 06:37 AM

Search Tags


/mathhelpforum @mathhelpforum