#$&* course Mth 277 December 3 around 12:45pm. Question: `q001. Find the center of mass of the part of the spherical solid with density delta = 2 described by x^2 + y^2 + z^2 <= 9, with x,y,z all >= 0.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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Given Solution: The region is described by 0 <= x <= 3, 0 <= y <= sqrt( 9 - x^2 ), 0 <= z <= sqrt( 9 - x^2 - y^2). The typical volume increment has dimensions `dx by `dy by `dz, hence volume `dV = `dx * `dy * `dz. The increment includes a sample point but it is irrelevant to the mass of the integral, which is `dm = density * `dV (mass = density * volume). Since density is always 2, the mass increment is `dm = 2 * `dV = 2 * `dx * `dy * `dz. If we form our Riemann sum, then take the limit, we find that the total mass is int(int(int( 2 dz, 0, sqrt(9 - x^2 - y^2)) dy, 0, sqrt(9 - x^2) ) dx, 0, 3) = 9 pi. Note: The portion of the sphere which corresponds to this region comprises 1/8 the volume of the total sphere. The volume of a sphere of radius 3 is 4 pi * 9 = 36 pi, so the volume of the present region is 9/2 pi. Multiplying this by the uniform density 2 gives us the total mass 9 pi. We now find the 'torque' of the region about the x axis, for which we will use the symbol tau_x: A typical volume increment, for which we now include a sample point (x_hat, y_hat, z_hat), lies at horizontal distance y_hat from the axis, so it contributes torque `dTau_x = y_hat * `dm = y_hat * 2 `dV = y_hat * 2 * `dx * `dy * `dz. If we form our Riemann sum and take the limit we get the integral tau_x = int(int(int( 2 y dz, 0, sqrt(9 - x^2 - y^2)) dy, 0, sqrt(9 - x^2) ) dx, 0, 3) = 81 pi / 8. The center of mass therefore occurs at y coordinate y_cMass = tau_x / mass = 81 pi / 8 / (9 pi) = 9/8. Compare to the central point y = 1.5 between y = 0 and y = 3. If the region was an 'enclosing cube' this would be the center of mass in the y direction. However the mass of the spherical region is concentrated closer to the x axis than would the mass of the cube, and as a result the center of mass is closer to the x axis. If we analyze the torques about the y and z axes, we get analogous results: tau_y = int(int(int( 2 x dz, 0, sqrt(9 - x^2 - y^2)) dy, 0, sqrt(9 - x^2) ) dx, 0, 3) = 81 pi / 8 tau_z = int(int(int( 2 x dz, 0, sqrt(9 - x^2 - y^2)) dy, 0, sqrt(9 - x^2) ) dx, 0, 3) = 81 pi / 8 each of which divided by the mass again yields the same result 9/8. We conclude that the center of mass is (9/8, 9/8, 9/8). Some of the integrals look like this: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, I used decimals instead of fractions though. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q002. Use double integration to find the center of mass when delta(x,y) = y over the region bounded by y = e^-x, x = 0, x = 3, y = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** [Int Int (y) dy, 0, e^(-x), dx, 0, 3] =(e^(6)-1)*(e^(-6))/4 =0.25 Torque about “x”=Int Int (y^2) dy, 0, e^(-x), dx, 0, 3 =0.11 Torque about “y”= Int Int (xy) dy, 0, e^(-x), dx, 0, 3 =0.12 Center of mass: Torque/total mass= (x): 0.11/0.25=0.44 (y): 0.12/0.25=0.48 (0.44, 0.48) #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The region is described by 0 <= x <= 3, 0 <= y <= e^(-x). The mass increment has area `dy by `dx and density delta(x_hat, y_hat) = y_hat is `dm = y_hat * `dy * `dx, leading in the usual manner to the integral int(int( y dy, 0, e^(-x))dx, 0, 3) = .249, approximately. The moment (or torque) of a mass increment about the x axis is y_hat `dm = y_hat * (y_hat * `dy * `dx) = y_hat^2 `dy `dx, leading to the integral tau_x = int(int( y ^2 dy, 0, e^(-x))dx, 0, 3) = .111. The moment (or torque) of a mass increment about the y axis is x_hat `dm = x_hat * (y_hat * `dy * `dx), leading to the integral tau_y = int(int( x y dy, 0, e^(-x))dx, 0, 3) = .122. The coordinates of the center are therefore x_cMass = tau_y / mass = .122 / .249 = .49, roughly, and y_cMass = tau_x / mass = .111 / .249 = .45, roughly. Standard-form representation of these integrals: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK, a few rounding differences. ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q003. Find I_x, the moment of inertia about the x-axis, of the lamina that covers the region bounded by the graph of y = 3 - x^2 and the x-axis, and with density delta(x,y) = x^2*y^2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** y=3-x^2 x=sqrt (3) I=mr^2 x-axis (r=y hat) ‘dm=(x hat^2)*(y hat^2) ‘dI=((x hat^2)*(y hat^2))*(y hat^2) [Int Int (x^2*y^4) dy, 0, (3-x^2), dx, -sqrt (3), sqrt (3)] =41472*sqrt(3)/5005 =14.4 [moment of inertia] #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The typical area increment has dimensions `dy by `dx and sample point (x_hat, y_hat). The mass of the region is `dm = density * area = x_hat^2 y^hat^2 `dx `dy, and its distance from the x axis is r = y_hat. Its moment of inertia is mass * r^2, giving us `dI = `dm * r^2 = (x_hat^2 y^hat^2 `dx `dy) * y_hat^2 = x_hat^2 y^hat^4 `dx `dy. This leads us to the integral int(int(x^2 y^4 dy, 0, 3 - x^2) dx, -sqrt(3), sqrt(3) ) = 14.35, approx.. This is reasonable. This region of the graph covers most of an 'enclosing rectangle' with dimensions 2 sqrt(3) * 3 = 10, very approximately. (1, 1) could be regarded as a 'typical point', at which density is 1 and distance from the x axis is 1, so the moment of inertia of a neighboring region would be is about 1 * area. Extending this to the entire region would result in moment of inertia 1 * 10 = 10, which puts our much more accurate answer of 14.35 in the right ballpark. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): PERFECT! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q004. Suppose the joint probability density function for the random variables X and Y is f(x,y) = {4e^(-2x)*e^(-y) if x,y >= 0, 0 otherwise.} Find the probability that X + Y <= 1. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** P[(X,Y) is in R]= Int Int (f (x,y) dA), over boundary R =(Int (Int (4e^(-2x)*e^(-y)) dy, 0, 1-x) dx, 0, 1) =approx. 0.799 About 80% that X+Y<=1 P(E)+P(E line)=1 1-0.799=0.200 #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q005. Find the average value of the function f(x,y,z) = x + 2y + 3z over the solid region S bounded by the tetrahedron with vertices (0,0,0), (1,0,0), (0,1,0), (0,0,1). YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** Tetrahedron: first octant that lies beneath the plane x+y+z=1 =[Int(Int(Int x+2y+3z dz, 0, 1-x-y), dy, 0, 1-x), dx, 0, 1] =Int (x+2y+3z dz, 0, 1-x-y) =[(x+y-1)*(x-y-3)]/2 =Int ([(x+y-1)*(x-y-3)]/2 dy, 0, 1-x) =[-(x-1)^2*(2x-5)]/6 =Int ([-(x-1)^2*(2x-5)]/6 dx, 0, 1) =1/4 #$&*
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: