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PROBLEM 4: The moment of inertia about the z-axis of the solid T of equation

T = {T2 + 42 + 22 < R, 220}

with density

6(r, y, 2) = 2 + y^2

is given by the triple integral âˆ«âˆ«âˆ«(2 + y^2) dxdydz. Simplify first the integral over T by replacing the value of 6(r, y, 2), then compute the integral I by performing the spherical change of coordinates Ï = âˆš(x^2 + y^2 + z^2), Î¸ = arccos(z/Ï), Ï† = arctan(y/x) and applying the formula âˆ«âˆ«âˆ«(2 + y^2) Ï^2 sinÎ¸ dÏdÎ¸dÏ†, where you can assume that G, the transformed region of T, is given by

G = {0 < Ï < R, 0 < Î¸ < 2, 0 < Ï† < 2Ï€}.

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08:45

Express the moment of inertia $I_{z}$ of the solid hemisphere$x^{2}+y^{2}+z^{2} \leq 1, z \geq 0,$ as an iterated integral in (a) cylindrical and (b) spherical coordinates. Then (c) find $I_{z}$ .

13:20

00:53

In Problems, find the polar moment of inertia$$I_{0}=\iint_{R} r^{2} \rho(r, \theta) d A=I_{x}+I_{y}$$of the lamina that has the given shape and density.$r=a ; \rho(r, \theta)=k$ (constant) [Hint: Use Problem 17 and the fact that $I_{x}=I_{y}$.]

01:14

(a) Recall that if L is a plane lamina, then the moment of inertia of L about an axis is given byI = âˆ«(r^2) dm where r is the distance to the axis and m is the density. Suppose that L is a rectangular plate with mass M, uniform density, and sides of length a and b, and that the axis is orthogonal to and passes through one of the corners of L as shown in the diagram below. Determine the moment of inertia in terms of M and the axis. [5 marks]

(b) Let C be the curve y = 12x from the point (0,0) to the point (2,4). Evaluate the line integral âˆ«(z^2) dy + u dx. [4 marks]

Let C be the ellipse x^2/9 + y^2/4 = 1 traversed anticlockwise. Use Green’s theorem to evaluate the line integral âˆ«(dy + u dx).

07:52

(6 marks) A triangular lamina has vertices (0, 0), (0, 6), (6, 0), and uniform density. Calculate the center of mass and moment of inertia about the x-axis.2. Corrected_text: (12 marks) Use a double integral to find the volume of the region bounded by the elliptic paraboloid z = 4 – x^2 – y^2 and the plane z = 0.4 (Hint: Use the symmetry in the octants and trigonometric substitutions to complete the integration.)3. Corrected_text: (10 marks) Use a triple integral in cylindrical coordinates to find the volume of the solid inside the sphere x^2 + y^2 + z^2 = 6 and above the paraboloid z = x^2 + y^2.

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