Find the mass and center of mass of the lamina that occupies the region D and ha

Question

Find the mass and center of mass of the lamina that occupies the region D and has the given density function rho. D = {(x, y) | 1 lessthanorequlato x lessthanorequlato 3, 1 lessthanorequlato y lessthanorequlato 4}; rho (x, y) = ky^2 D = {(x, y) | 0 lessthanorequlato a, 0 lessthanorequlato y lessthanorequlato b} rho (x, y) = 1 + x^2 + y^2 D is the triangular region with vertices (0, 0), (2, 1), (0, 3); rho(x, y) = x + y D is the triangular region enclosed by the lines x = 0, y = x, and 2x + y = 6; rho(x, y) = x^2 D is bounded by y = 1 - x^2 and y = 0; p(x, y) = ky D is bounded by y = x^2 and y = x + 2; p(x, y) = kx D = {(x, y) |0 lessthanorequlato y lessthanorequlato sin(pi x/L), 0 lessthanorequlato x lessthanorequlato L}; rho 9x, y) = y D is bounded by the parabolas y = x^2 and x = y^2; rho (x, y) = Squareroot x A lamina occupies the part of the disk x^2 + y^2 l in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the x-axis.

Explanation / Answer

9) 0<=y<=sin(x/L) ,0<=x<=L , (x,y) =y

mass m=[0 to L][0 to sin(x/L)] y dy dx

mass m=[0 to L][0 to sin(x/L)] (1/2)y2 dx

mass m=[0 to L](1/2)[(sin(x/L))2-02] dx

mass m=[0 to L](1/2)(sin(x/L))2 dx

mass m=[0 to L](1/4)(1-cos(2x/L)) dx

mass m=[0 to L](1/4)(x-(L/2)sin(2x/L))

mass m=(1/4)(L-(L/2)sin(2L/L)) -(1/4)(0-(L/2)sin(2*0/L))

mass m=(1/4)(L-0) -(1/4)(0-0)

mass m=(1/4)L

moment about y axis ,My=[0 to L][0 to sin(x/L)] xy dy dx

My=[0 to L][0 to sin(x/L)] x(1/2)y2 dx

My=[0 to L]x(1/2)[(sin(x/L))2-02]  dx

My=[0 to L]x(1/2)(sin(x/L))2 dx

My=[0 to L]x(1/4)(1-cos(2x/L)) dx

My=[0 to L](1/4)(x- xcos(2x/L)) dx

My=[0 to L](1/4)((1/2)x2- [x(L/2)sin(2x/L) -(L/2)sin(2x/L) dx])

My=[0 to L](1/4)((1/2)x2- [(L/2)xsin(2x/L) +(L/2)2cos(2x/L) ])

My=[0 to L](1/4)((1/2)x2- (L/2)xsin(2x/L) -(L/2)2cos(2x/L))

My=(1/4)((1/2)L2- (L/2)Lsin(2L/L) -(L/2)2cos(2L/L)) -(1/4)((1/2)02- (L/2)0sin(0/L) -(L/2)2cos(0/L))

My=(1/4)((1/2)L2- 0 -(L/2)2) -(1/4)(0- 0 -(L/2)2)

My=(1/8)L2

moment about x axis ,Mx=[0 to L][0 to sin(x/L)] yy dy dx

moment about x axis ,Mx=[0 to L][0 to sin(x/L)] y2 dy dx

moment about x axis ,Mx=[0 to L][0 to sin(x/L)] (1/3)y3 dx

moment about x axis ,Mx=[0 to L](1/3)[sin3(x/L) -03]dx

moment about x axis ,Mx=[0 to L](1/3)[sin2(x/L)](sin(x/L))dx

moment about x axis ,Mx=[0 to L](1/3)[1-cos2(x/L)](sin(x/L))dx

moment about x axis ,Mx=[0 to L](1/3)[sin(x/L) -cos2(x/L)sin(x/L)]dx

moment about x axis ,Mx=[0 to L](1/3)[-(L/)cos(x/L) +(L/3)cos3(x/L)]

moment about x axis ,Mx=(1/3)[-(L/)cos(L/L) +(L/3)cos3(L/L)] -(1/3)[-(L/)cos(0) +(L/3)cos3(0)]

moment about x axis ,Mx=(1/3)[(L/) -(L/3)] -(1/3)[-(L/) +(L/3)]

moment about x axis ,Mx=(4L/9)

center of mass (x,y)=(My/m ,Mx/m)

center of mass (x,y)=(((1/8)L2)/((1/4)L) ,(4L/9)/((1/4)L))

center of mass (x,y)=((1/2)L ,16/9)

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