Answer to Question #176709 in Calculus for John

Question

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

Answer 1

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

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

Solid mass for unit density or volume of the solid is

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

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

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

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

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

The center of mass of the object is the point (x_c,y_c,z_c)

x_c = M_yz/m = -0.5, y_c = M_xz/m = 0, z_c = M_xy/m = 1.25

The moments of inertia

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

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

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

Answer

a) Volume V = 4π

b) Center of mass (x_c,y_c,z_c) = (-0.5, 0, 1.25)

c) Moments of inertial about x-axis , y-axis and z-axis I_x = 11π, I_y = 40π/3, I_z = 5π.