M = ∫[0,4]∫[x/2+4,√x+4]ρ dy dx
= ρ∫[0,4] (√x+4)-(x/2+4) dx
= ρ∫[0,4] √x - x/2 dx
= ρ (2/3)x^(3/2) - (1/4)x^2 [0,4]
= ρ (2/3)(8)-(1/4)(16)
= 4/3 ρ
Mx = ∫[0,4]∫[x/2+4,√x+4]ρy dy dx = 20/3 ρ
My = ∫[0,4]∫[x/2+4,√x+4]ρx dy dx = 32/15 ρ
so, the center of mass is at (Mx/M,My/M) = (5,8/5)
Hmmm. I'd expect it to be above the line y=4. Better double-check my math.
Read good article here:
http://www.ltcconline.net/greenl/courses/202/multipleIntegration/MassMoments.htm
Find Mx, My, and (x, y) for the laminas of uniform density ρ bounded by the graphs of the equations.
y=sqrt(x)+4
y=(1/2)x+4
The I keep getting 20/3 for Mx but webassign acts like it is wrong.
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