Wednesday
March 22, 2017

Post a New Question

Posted by on Thursday, May 31, 2012 at 6:55pm.

Find the center of gravity enclosed by y^2=4x,x=4,y=0 if its density is given by รค(x.y)=ky

  • calculus - , Thursday, May 31, 2012 at 7:32pm

    If we are dealing with real numbers x must be >/=0 becaue y^2 may not be negative
    find vertical cg
    y = 2 sqrt |x|
    integrate from (0,0) to (4,4)

    y of cg
    = int dy y (4-x) (ky) / int dy(4-x) (ky)

    numerator (the moment)
    4 k y^2 dy - k dy y^2 (y^2/4)
    4 k y^3/3 - k y^5/20
    at y = 4
    85.33 k - 51.2 k = 34.1 k
    denominator (the mass)
    dy(4-x) (ky) = 4k y dy - k y (y^2/4)dy
    = 2 k y^2 - k y^4/16
    at y = 4
    32 k - 16 k = 16 k
    so
    34.1 k / 16 k = 2.13
    You can do the stripes the other way for the Xcg

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question