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

Qf(x,y,z)dxdydz=22(1x2/41x2/4(02xf(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

Qf(x,y,z)dxdydz=22(1x2/41x2/4(02xf(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)dxQf(x,y,z)dxdydz=22(1x2/41x2/4(02xf(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=22(1x2/41x2/4(02xdz)dy)dx=2224x2dx=4πm = 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=22(1x2/41x2/4(02xdz)dy)dx=2224x2dx=4πm = 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\pim=V=22(1x2/41x2/4(02xdz)dy)dx=2224x2dx=4πm = 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\pim=V=22(1x2/41x2/4(02xdz)dy)dx=2224x2dx=4πm = 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

Mxy=22(1x2/41x2/4(02xzdz)dy)dx=1222(44x+x2)4x2dx=5πM_{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

Mxy=22(1x2/41x2/4(02xzdz)dy)dx=1222(44x+x2)4x2dx=5πM_{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\piMxy=22(1x2/41x2/4(02xzdz)dy)dx=1222(44x+x2)4x2dx=5πM_{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\piMxy=22(1x2/41x2/4(02xzdz)dy)dx=1222(44x+x2)4x2dx=5πM_{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

Mxz=22(1x2/41x2/4(02xdz)ydy)dx=0M_{xz} = \intop_{-2}^2 ( \int_{-\sqrt{1-x^2/4}} ^{\sqrt{1-x^2/4} }(\int_0^{2-x}dz)ydy)dx = 0

Mxz=22(1x2/41x2/4(02xdz)ydy)dx=0M_{xz} = \intop_{-2}^2 ( \int_{-\sqrt{1-x^2/4}} ^{\sqrt{1-x^2/4} }(\int_0^{2-x}dz)ydy)dx = 0Mxz=22(1x2/41x2/4(02xdz)ydy)dx=0M_{xz} = \intop_{-2}^2 ( \int_{-\sqrt{1-x^2/4}} ^{\sqrt{1-x^2/4} }(\int_0^{2-x}dz)ydy)dx = 0

Myz=22(1x2/41x2/4(02xdz)dy)xdx=22(2x)4x2xdx=2πM_{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

Myz=22(1x2/41x2/4(02xdz)dy)xdx=22(2x)4x2xdx=2πM_{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\piMyz=22(1x2/41x2/4(02xdz)dy)xdx=22(2x)4x2xdx=2πM_{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\piMyz=22(1x2/41x2/4(02xdz)dy)xdx=22(2x)4x2xdx=2πM_{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

Ix=22(1x2/41x2/4(02x(y2+z2)dz)dy)dx=1622(20+11x2)4x2dx=11πI_{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

Ix=22(1x2/41x2/4(02x(y2+z2)dz)dy)dx=1622(20+11x2)4x2dx=11πI_{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\piIx=22(1x2/41x2/4(02x(y2+z2)dz)dy)dx=1622(20+11x2)4x2dx=11πI_{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\piIx=22(1x2/41x2/4(02x(y2+z2)dz)dy)dx=1622(20+11x2)4x2dx=11πI_{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

Iy=22(1x2/41x2/4(02x(x2+z2)dz)dy)dx=4322(2+3x2)4x2dx=40π/3I_{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

Iy=22(1x2/41x2/4(02x(x2+z2)dz)dy)dx=4322(2+3x2)4x2dx=40π/3I_{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/3Iy=22(1x2/41x2/4(02x(x2+z2)dz)dy)dx=4322(2+3x2)4x2dx=40π/3I_{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/3Iy=22(1x2/41x2/4(02x(x2+z2)dz)dy)dx=4322(2+3x2)4x2dx=40π/3I_{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

Iz=22(1x2/41x2/4(02x(x2+y2)dz)dy)dx=11222(8+22x2)(2x)4x2dx=5πI_{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

Iz=22(1x2/41x2/4(02x(x2+y2)dz)dy)dx=11222(8+22x2)(2x)4x2dx=5πI_{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\piIz=22(1x2/41x2/4(02x(x2+y2)dz)dy)dx=11222(8+22x2)(2x)4x2dx=5πI_{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\piIz=22(1x2/41x2/4(02x(x2+y2)dz)dy)dx=11222(8+22x2)(2x)4x2dx=5πI_{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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