Posted by **calculs3** 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 -
**Damon**, 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

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