Results 1 to 7 of 7

Math Help - Two curves, centroid, integration

  1. #1
    Member
    Joined
    Sep 2011
    Posts
    75

    Two curves, centroid, integration

    26-4-15
    Find centroid of region bounded by following curves:
    y = 4x^2
    y = 2x^3

    I found two y intercepts: x = 2 and x = -2
    obvious x intercept is zero

    Attempt:
    A = (int) 2x^3 - 2x^2
    boundary 0 and 2
    1/2x^4 - 2/3x^3 =
    1/2(2^4) - 2/3(2^3) = 8 - 2.666 = 5 1/3

    X centroid:
    1/2 * 1/(5.333) (int) [(2x^3)^2 - (4x^2)^2]
    boundaries 2 and -2
    = .09375 (int) 4/7x^7 - 16/5x^5
    plugging in 2
    = 4/7(2^7) - 16/5(2^5) = -29.26

    completely wrong... so where'd I go wrong? Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853

    Re: Two curves, centroid, integration

    Quote Originally Posted by togo View Post
    26-4-15
    Find centroid of region bounded by following curves:
    y = 4x^2
    y = 2x^3

    I found two y intercepts: x = 2 and x = -2
    obvious x intercept is zero
    Those really aren't relevant. What is relevant is where the two curves intersect- where y= 4x^2= 2x^3. Obviously, one root is x= 0. If x is not 0 we can divide by 2x^2 to get x= 2.

    Attempt:
    A = (int) 2x^3 - 2x^2
    boundary 0 and 2
    Yes, the boundary is 0 and 2. But where did you get 2x^3- 2x^2? Your two functions are y= 4x^2 and 2x^3. Further, for all x between 0 and 2, 4x^2> 2x^3 so you want to integrate 4x^2- 2x^3, not 2x^3- 2x^2.

    1/2x^4 - 2/3x^3 =
    Be careful about your notation. Many people would interpret "1/2x^4" as 1/(2x^4) not (1/2)x^4.
    In any case, the integral of 4x^2- 2x^3 is (4/3)x^3- (1/2)x^4 which, evaluated at 2 and 4 is 32/3- 8= 32/3- 24/3= 8/3.

    1/2(2^4) - 2/3(2^3) = 8 - 2.666 = 5 1/3

    X centroid:
    1/2 * 1/(5.333) (int) [(2x^3)^2 - (4x^2)^2]
    boundaries 2 and -2
    What? The integral should be just x times the area integrand: x(2x^3- 4x^2). I have no idea where those squares came from

    [tex]= .09375 (int) 4/7x^7 - 16/5x^5
    plugging in 2
    = 4/7(2^7) - 16/5(2^5) = -29.26

    completely wrong... so where'd I go wrong? Thanks[/QUOTE]
    Pretty much everything is wrong! You need to review "centroid".
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2011
    Posts
    75

    Re: Two curves, centroid, integration

    thanks I will be reworking the question this morning. I often copy numbers and signs wrong and forget to check for simple mistakes like that. I really appreciate getting an answer on this post, have made many posts in the past and they were ignored by the community.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2011
    Posts
    75

    Re: Two curves, centroid, integration

    I get -2.666 for area

    A = (int) 2x^3 - 4x^2
    = 1/2(2^4) - 4/3(2^3)
    = 8 - 10.666
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853

    Re: Two curves, centroid, integration

    Are you even reading my posts? I already told you that, for x between 0 and 2, 4x^2> 2x^3. The area is the integral of 4x^2- 2x^3, NOT "2x^3- 4x^2". In fact, I have already told you that the area is 8/3= 32/3- 8.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Sep 2011
    Posts
    75

    Re: Two curves, centroid, integration

    Yes thanks for that. Continuing on

    finding centroid of X
    1/(8/3) (int) x[2(x^3) - 4(x^2)]
    = 2(x^4) - 4(x^3)
    = 2/5(x^5) - 1/2(x^4)
    = 2/5(2^5) - 1/2(2^4) = 6.4-8

    using boundaries of 2 and -2, being plugged into X, the 2 and -2 with exponents on them (5 and 4 respectively) turn into the same thing so you end up with (6.4 - 8) - (6.4 - 8) which is zero and incorrect.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,413
    Thanks
    1853

    Re: Two curves, centroid, integration

    You are still integrating 2x^3- 4x^2 when I have told you twice that 4x^2> 2x^3. You need to use 4x^2- 2x^3.

    The boundaries are NOT "-2 and 2". The two curves intersect at x= 0 and at x= 2.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: September 15th 2012, 01:18 PM
  2. Integration- Area under curves
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 13th 2009, 02:16 PM
  3. Are between two curves (integration)
    Posted in the Calculus Forum
    Replies: 3
    Last Post: September 17th 2008, 04:11 PM
  4. Triple Integration. Coordinates of Centroid.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 3rd 2008, 07:49 AM
  5. Replies: 4
    Last Post: August 30th 2006, 10:02 PM

Search Tags


/mathhelpforum @mathhelpforum