Find the mass and center of mass of the solid E with the given density function

Question

Find the mass and center of mass of the solid E with the given density function rho. E is the tetrahedron bounded by the planes x = 0, y = 0, z = 0, x + y + z = 2; p (x, y, z) = by m = (x bar, y bar, z bar) = The average value of a function f(x, y, z) over a solid region E is defined be f_ave = 1/V(E) tripleintegral_E f(x, y, z) dV where V(E) is the volume of E. For instance, id rho is a density function, the rho_ave is the average density of E. Find the average value of the function f(x, y, z) = 3x^2 z + 3y^2 z over the region enclosed by the paraboloid z = 9 - x^2 - y^2 and the plane z = 0.

Explanation / Answer

9)

x = 0, y = 0, z = 0, x + y + z = 2

=>0<=x<=2, 0<=y<=2-x, 0<=z<=2-x-y

given density (x, y, z) = 8y

mass m=[0 to 2][0 to 2-x][0 to 2-x-y]8y dz dy dx

mass m=[0 to 2][0 to 2-x][0 to 2-x-y]8yz dy dx

mass m=[0 to 2][0 to 2-x]8y(2-x-y-0) dy dx

mass m=[0 to 2][0 to 2-x]8(2y-xy-y2) dy dx

mass m=[0 to 2][0 to 2-x]8(y2-(1/2)xy2-(1/3)y3) dx

mass m=[0 to 2]8((2-x)2-(1/2)x(2-x)2-(1/3)(2-x)3)-0 dx

mass m=[0 to 2]8((1/6)(2-x)3) dx

mass m=[0 to 2](4/3)(-1/4)(2-x)4

mass m=[0 to 2](-1/3)(2-x)4

mass m=(-1/3)(2-2)4 -(-1/3)(2-0)4

mass m=0+(1/3)(16)

mass m=16/3

Myz=[0 to 2][0 to 2-x][0 to 2-x-y]8yx dz dy dx

Myz=32/15

Mxz=[0 to 2][0 to 2-x][0 to 2-x-y]8y2 dz dy dx

Mxz=64/15

Mxy=[0 to 2][0 to 2-x][0 to 2-x-y]8yz dz dy dx

Mxy=32/15

centre of mass(x,y,z)=(Myz/m,Mxz/m,Mxy/m)

centre of mass(x,y,z)=((32/15)/(16/3),(64/15)/(16/3),(32/15)/(16/3))

centre of mass(x,y,z)=((32/15)*(3/16),(64/15)*(3/16),(32/15)*(3/16))

centre of mass(x,y,z)=(2/5,4/5,2/5)

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