#23 Please help Find the center of mass of the region with the given mass densit
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#23 Please help
Find the center of mass of the region with the given mass density delta. Region bounded by y = 4 - x, x = 0, y = 0; delta(x, y) = x Region bounded by y^2 = x + 4 and x = 0; delta(x, y) = |y| Region |x| + |y| lessthanorequalto 1; delta(x, y) = (x + 1)(y + 1) Semicircle x^2 + y^2 lessthanorequalto R^2, y greaterthanorequalto 0; delta(x, y) = y Find the z-coordinate of the center of mass of the first octant of the unit sphere with mass density delta(x, y, z) = y (Figure 17). FIGURE 17Explanation / Answer
23) We have given y=4-x,y=0,x=0 with density(x,y)=x
Let f(x)=4-x,g(x)=0
intersection point of y=4-x=0 implies x=4
Area of the region is
A=integration of (x=0 to 4) (4-x)dx =[4x-x^2/2] from x=0 to 4 =[16-16/2]=8
A=8
now the moments are
Mx=integration of (x=0 to 4) [density(x,y)1/2*(4-x)^2]dx and My=integration of (x=0 to 4) [density(x,y)*x*(4-x)]dx
Mx=integration of (x=0 to 4) [density(x,y)1/2*(4-x)^2]dx
=integration of (x=0 to 4) [x/2*(4-x)^2]dx since density(x,y)=x
=integration of (x=0 to 4) [x/2*(16-8x+x^2)]dx
=integration of (x=0 to 4) [(8x-4x^2+(x^3)/2)]dx
=[4x-(4/3)x^3+(x^4)/8] from x=0 to 4
=[16-256/3+32]=[48-256+96]/3 =-112/3
Mx=-112/3
My=integration of (x=0 to 4) [density(x,y)*x*(4-x)]dx
=integration of (x=0 to 4) [x*x*(4-x)]dx since density(x,y)=x
=integration of (x=0 to 4) [4x^2-x^3)]dx
=[(4/3)x^3-x^4/4] from x=0 to 4
=[256/3 -64]=[256-192]/3=64/3
My=64/3
center of mass is (x bar,y bar)=(My/M,Mx/M)
My/M =1/A* integration of (x=0 to 4)[x*(4-x)]dx =1/8*integration of (x=0 to 4)[(4x-x^2)]dx
=1/8*[2x^2-x^3/3] from x=0 to 4
=1/8*[32-64/3]=1/8*[96-64]/3 =32/24
My/M =4/3
Mx/M =1/A* integration of (x=0 to 4)[1/2*(4-x)^2]dx =1/8* integration of (x=0 to 4)[1/2*(4-x)^2]dx
=1/8*integration of (x=0 to 4)[1/2*(16-8x+x^2)]dx
=1/8*(1/2)*[16x-4x^2+x^3/3] from x=0 to 4
=1/16*[64-64+64/3]
Mx/M=4/3
center of mass of the region is (x bar,y bar)=(4/3,4/3)
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