#$&* course Phy 277 4/17 12 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): ------------------------------------------------ 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. 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: YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Tried not to look at or pay attention to solution as I was scrolling to my solution section, but it’s like being a kid and knowing before Christmas your presents are unwrapped in the closet. We have a continuous fn for the density( `rho = y) on a lamina corresponding to a plane region R, so we have the moments of mass with respect to the x and y axis given as…. M_x = Int(Int( y*`rho(x,y) `dA) M_y = Int(Int( x*`rho(x,y) `dA) Knowing this, if m is the mass of the lamina, the center of mass is (x_bar, y_bar) where x_bar = M_y/m y_bar = M_x/m We first find the mass of the object, which is easily done because we are given boundries and `rho m = Int(Int( y dy,0,e^(-x)) dx,0,3) Int(y dy) = y^2/2, evaluated e^(-2x)/2 Int((1/2)e^(-2x) dx,) = (1/2)Int(e^(-2x) dx) = (1/2)*(e^(-2x)/-2), evaluated (1/-4)e^(-2*3) - ((1/-4)e^(-2*0)) = approx -.0006 + .25 = approx .2494 For `rho(x,y) = y over the region bounded by y = e^-x, x = 0, x = 3, y = 0 we find the moments M_x = Int(Int( y*(`rho) dA)) =Int(Int(y^2 dy,0,e^(-x)) dx,0,3) Int(y^2 dy) = y^3/3, evaluated (e^(-x))^3/3 - (0)^3/0 = (1/3)e^-3x Int(e^-3x/3 dx)=(1/3)Int(e^(-3x)dx)=(1/3)(e^(-3x)/-3)=(1/-9)e^(-3x),evaluated (-1/9)e^(-3*3) - (-1/9)e^(-3*0) = (-1/9)e^(-9) M_x = (-1/9)e^(-9) = approx -.00001 M_y = Int(Int( x*(`rho) dA)) =Int(Int(xy dy,0,e^(-x)) dx,0,3) Int(xydy) = x*Int(y dy) = x*(y^2/2) = (xy^2)/2, evaluated (x*(e^(-x))^2)/2 - (x*(0)^2)/2 = (1/2)xe^-2x Int((1/2)xe^-2x dx) = (1/2)Int(xe^(-2x)dx), using int. by parts w/ u= x, u’=1,v’=e^(-2x), v=e^(-2x)/-2 = (1/2)[(x(e^(-2x)/-2) - (e^-2x)/4] = e^(-2x)/-4*(x + (1/2)),evaluated e^(-2*3)/-4*(3 + (1/2))] - [e^(-2*0)/-4*(0 + (1/2))] =(7/2)e^(-6)/-4 + (1/8) M_y = (7/2)e^(-6)/-4 + (1/8) = approx .1228 Center of mass is (x_bar, y_bar) where x_bar = M_y/m = .1228/.2494 = approx .4924 y_bar = M_x/m = -.00001/.2494 = approx -.00004 Center of mass then is at the approx pt (.4924, -.00004) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Got everything right I believe except for the fact that when calculating the first moment wrt x axis when evaluating the integral I had final expression (-1/9)e^(-9) when instead it was (-1/9)e^(-9) + (1/9) = approx .1111 Which corresponds to the solution you provided. 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: For y = 3-x^2 we have a parabola opening up downwards, which for each fixed x between -`sqrt(3) and `sqrt(3) varies from 0 to (3-x^2) . Because we know moment of inertia about the x axis is I_x = Int(Int( y^2*`rho(x,y) `dA)) For region defined as y = 3 - x^2 w/ `rho = x^2*y^2 we find I_x by… I_x = Int(Int(y^2*(x^2*y^2) dy,0,(3-x^2)) dx,-`sqrt(3),`sqrt(3)) Int(y^2(x^2*y^2) dy) = Int(x^2y^4 dy) = x^2*Int(y^4 dy) = x^2*(y^5/5), evaluated x^2((3-x^2)^5/5) - 0 = x^2*(1/5)*(243-405x^2+270x^4-90x^6+15x^8-x^10) - 0 =(1/5)*(243x^2-405x^4+270x^6-90x^8+15x^10-x^12) Int((1/5)*(243x^2-405x^4+270x^6-90x^8+15x^10-x^12) dx) =(1/5)Int(243x^2-405x^4+270x^6-90x^8+15x^10-x^12 dx), to same time and space I’ll calculate on sep sheet (1/5)*[81x^3- 81x^5+ (270/7)x^7- 10x^9+ (15/11)x^11- (1/13)x^13], evaluated (1/5)*[81(`sqrt(3))^3- 81(`sqrt(3))^5+ (270/7) (`sqrt(3))^7- 10(`sqrt(3))^9+ (15/11) (`sqrt(3))^11- (1/13) (`sqrt(3))^13] - (1/5)*[81(-`sqrt(3))^3- 81(-`sqrt(3))^5+ (270/7) (-`sqrt(3))^7- 10(-`sqrt(3))^9+ (15/11) (-`sqrt(3))^11- (1/13) (-`sqrt(3))^13] = approx 7.176 - (-7.176) = 14.352 I don’t know what I was thinking when I worked out (3-x^2)^5, I’m pretty sure this could have been done without taking the time to do that. Regardless we’ll see how close I was and the appropriate way to solve in a sec. 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): Got the correct answer but I’m pretty sure it wasn’t in the most efficient way. ------------------------------------------------ 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: For X + Y <= 1 we see in the first Quad a triangle w/ vertexs at (0,0), (0,1), (1,0) giving us boundries 0<=x<=1 and 0<=y<=1-x. So that… P[(X,Y) is in R] = Int(Int( f(x,y) dA) =Int(Int(4e^(-2x)*e^(-y) dy,0,1-x) dx,0,1) Int(4e^(-2x)*e^(-y) dy) = 4e^(-2x)*Int(e^(-y) dy) = 4e^(-2x)*(-e^(-y)), evaluated 4e^(-2x)*(-e^(-(1-x))) - 4e^(-2x)*(-e^(0)) = -4e^(-(x+1)) + 4e^(-2x) = -1.472e^(-x) + 4e^(-2x) Int(4e^(-2x) - 1.472e^-x dx) = Int(4e^(-2x) dx) - Int(1.472e^-x dx) = -2e^(-2x) + 1.472e^-x,evaluated [-2e^(-2*1) + 1.472e^-1] - [-2e^(-2*0) + 1.472e^-0] = [-2e^-2 + 1.472e^-1] + .528 = approx -.271 + .542 + .528 = aprox .8 or around an 80% chance 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: or Tetrahedron lies in first Quad where x+y+z = 1 so boundries are 0<=x<=1, 0<=y<=1-x and 0<=z<=1-x-y so we have integral Int(Int(Int(x + 2y + 3z dz,0,1-x-y) dy,0,1-x) dx,0,1) Int(x + 2y + 3z dz) = xz + 2yz + (3/2)z^2, evaluated [x(1-x-y)+ 2y(1-x-y) + (3/2)(1-x-y)^2] - [x*0 + 2y*0 + (3/2)0^2] = [x-x^2-yx] + [2y -2xy - 2y^2] + (3/2)[x^2+y^2+2xy-2x-2y+1] = (1/2)x^2 - (1/2)y^2 - 2x - y + (3/2) Int((1/2)x^2 - (1/2)y^2 - 2x - y + (3/2) dy) = (1/2)x^2y-(1/6)y^3-2xy-y^2/2+(3/2)y, evaluated [(1/2)x^2(1-x)-(1/6)(1-x)^3-2x(1-x)-(1-x)^2/2+(3/2)(1-x)] - [0] =(-1/3)x^3+(1/2)x^2-2x+(5/6) Int((-1/3)x^3+(1/2)x^2-2x+(5/6) dx) = (-1/12)x^4 + (1/6)x^3 - x^2 + (5/6)x,evaluated [(-1/12)1^4 + (1/6)1^3 - 1^2 + (5/6)] - 0 = -1/12 + 1/6 - 1 + 5/6 = -1/12
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Given Solution: GOOD STUDENT SOLUTION FOR INTEGRAL OF FUNCTION OVER REGION: 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 #$&* INSTRUCTOR'S SUPPLEMENT: That is the correct integral for the function. The average value of the function is the integral, divided by the volume. The volume is 1/6 (this could be obtained by integrating the volume increment over the given region). The average value of the function is therefore 1/4 / (1/6) = 3/2. There are a number of ways to think about this result. First, does it make common sense? x + 2y + 3z is equal to 0 at (0, 0, 0), to 1 at (1, 0, 0), to 2 at (0, 1, 0), and to 3 at (0, 0, 1). The largest cross-section is in the x-y plane; for the region 0 <= x <= 1, 0 <= y <= 1 - x we have x + y <= 1. The greatest values of f(x, y, z) occur near the 'tips' of the tetrahedron, which don't occupy a lot of volume. So an average value of 1.5 seems to be in the right general range. Second, how can we understand the meaning of the average value? If f(x, y, z) is interpreted as a mass density, our first integral tells us that the mass of the tetrahedron is 1/4 and our second integral gives us volume 1/6. Since average density = mass / volume, our result 1/4 / (1/6) = 3/2 = 1.5 gives us the average density of the object. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): My solution was not correct and I have spent lots of time working these solutions out where I should not factored everything out ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: