calculus
posted by calculs3 .
Find the center of gravity enclosed by y^2=4x,x=4,y=0 if its density is given by ä(x.y)=ky

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 (4x) (ky) / int dy(4x) (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(4x) (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
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