Answer to Question #176722 in Calculus for John

Question #176722

A solid of constant density is bounded below by the plane z = 0 , on the sides by the ellptical cylinder x^2 + 4y^2 = 4 , and above by the plane z = 2 - x .

Calculate

a) Volume

b) Center of mass

c) Moment of inertial about x-axis , y-axis and z-axis


1
Expert's answer
2021-04-29T17:37:41-0400

Sorry but it is the same question Q176709 which was done by me an is still under checking answer. I repeat here the same solution.

Solution

Let Q is the given solid. For ellipse in xy plane y = ±√(1-x2/4)

Therefore all calculations in this task are reduced to calculation of triple integrals

"\\iiint_Qf(x,y,z)dxdydz = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}f(x,y,z)dz)dy)dx"

"\\iiint_Qf(x,y,z)dxdydz = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}f(x,y,z)dz)dy)dx""\\iiint_Qf(x,y,z)dxdydz = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}f(x,y,z)dz)dy)dx"

Solid mass for unit density or volume of the solid is


"m = V = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)dx = 2\\int_{-2}^2\\sqrt{4-x^2}dx = 4\\pi"

"m = V = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)dx = 2\\int_{-2}^2\\sqrt{4-x^2}dx = 4\\pi""m = V = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)dx = 2\\int_{-2}^2\\sqrt{4-x^2}dx = 4\\pi""m = V = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)dx = 2\\int_{-2}^2\\sqrt{4-x^2}dx = 4\\pi"

Its moments about the xy-plane the xz-plane and the yz-plane are

"M_{xy} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}zdz)dy)dx = \\frac{1}{2}\\int_{-2}^2(4-4x+x^2)\\sqrt{4-x^2}dx =5\\pi"

"M_{xy} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}zdz)dy)dx = \\frac{1}{2}\\int_{-2}^2(4-4x+x^2)\\sqrt{4-x^2}dx =5\\pi""M_{xy} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}zdz)dy)dx = \\frac{1}{2}\\int_{-2}^2(4-4x+x^2)\\sqrt{4-x^2}dx =5\\pi""M_{xy} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}zdz)dy)dx = \\frac{1}{2}\\int_{-2}^2(4-4x+x^2)\\sqrt{4-x^2}dx =5\\pi"

"M_{xz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)ydy)dx = 0"

"M_{xz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)ydy)dx = 0""M_{xz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)ydy)dx = 0"

"M_{yz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)xdx = \\int_{-2}^2(2-x)\\sqrt{4-x^2}xdx =-2\\pi"

"M_{yz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)xdx = \\int_{-2}^2(2-x)\\sqrt{4-x^2}xdx =-2\\pi""M_{yz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)xdx = \\int_{-2}^2(2-x)\\sqrt{4-x^2}xdx =-2\\pi""M_{yz} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}dz)dy)xdx = \\int_{-2}^2(2-x)\\sqrt{4-x^2}xdx =-2\\pi"

The center of mass of the object is the point (xc,yc,zc)

xc = Myz/m = -0.5, yc = Mxz/m = 0, zc = Mxy/m = 1.25

The moments of inertia

"I_{x} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(y^2+z^2)dz)dy)dx = \\frac{1}{6}\\int_{-2}^2(20+11x^2)\\sqrt{4-x^2}dx =11\\pi"

"I_{x} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(y^2+z^2)dz)dy)dx = \\frac{1}{6}\\int_{-2}^2(20+11x^2)\\sqrt{4-x^2}dx =11\\pi""I_{x} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(y^2+z^2)dz)dy)dx = \\frac{1}{6}\\int_{-2}^2(20+11x^2)\\sqrt{4-x^2}dx =11\\pi""I_{x} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(y^2+z^2)dz)dy)dx = \\frac{1}{6}\\int_{-2}^2(20+11x^2)\\sqrt{4-x^2}dx =11\\pi"

"I_{y} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+z^2)dz)dy)dx = \\frac{4}{3}\\int_{-2}^2(2+3x^2)\\sqrt{4-x^2}dx =40\\pi\/3"

"I_{y} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+z^2)dz)dy)dx = \\frac{4}{3}\\int_{-2}^2(2+3x^2)\\sqrt{4-x^2}dx =40\\pi\/3""I_{y} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+z^2)dz)dy)dx = \\frac{4}{3}\\int_{-2}^2(2+3x^2)\\sqrt{4-x^2}dx =40\\pi\/3""I_{y} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+z^2)dz)dy)dx = \\frac{4}{3}\\int_{-2}^2(2+3x^2)\\sqrt{4-x^2}dx =40\\pi\/3"

"I_{z} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+y^2)dz)dy)dx = \\frac{1}{12}\\int_{-2}^2(8+22x^2)(2-x)\\sqrt{4-x^2}dx =5\\pi"

"I_{z} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+y^2)dz)dy)dx = \\frac{1}{12}\\int_{-2}^2(8+22x^2)(2-x)\\sqrt{4-x^2}dx =5\\pi""I_{z} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+y^2)dz)dy)dx = \\frac{1}{12}\\int_{-2}^2(8+22x^2)(2-x)\\sqrt{4-x^2}dx =5\\pi""I_{z} = \\intop_{-2}^2 ( \\int_{-\\sqrt{1-x^2\/4}} ^{\\sqrt{1-x^2\/4} }(\\int_0^{2-x}(x^2+y^2)dz)dy)dx = \\frac{1}{12}\\int_{-2}^2(8+22x^2)(2-x)\\sqrt{4-x^2}dx =5\\pi"

Answer

a) Volume V = 4π

b) Center of mass (xc,yc,zc) = (-0.5, 0, 1.25)

c) Moments of inertial about x-axis , y-axis and z-axis Ix = 11π, Iy = 40π/3, Iz = 5π.


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