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

Answer 1

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