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WUSTL MATH 132 - m132_E3sF07

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Math 132 Fall 2007 Exam III1. The region in the first quadrant that is bounded above by = = = = y − − − − 4 x x2 and bounded below by = = = = y x is rotated about the vertical line = = = = x −−−−1. What is the volume of the resulting solid of revolution?a) 15 ππππ2 b) 16 ππππ c) 18 ππππ d) 21 ππππ e) 45 ππππ2 f) 20 ππππ g) 27 ππππ2 h) 25 ππππ2 i) 14 ππππ j) 15 ππππ Solution: eFirst we plot the planar region that is to be rotated.> f := x -> 4*x-x^2; g := x -> x; := f → x − 4 x x2 := g → x xThe planar region bounded by = y x and = y − 4 x x2 is shown below: Problem 1, Figure 1When this region is rotated about the specified axis, the following solid results. Problem 1, Figure 2Calculation of the Volume by the Method of Cylindrical ShellsThe next figure shows a typical cylindrical shell. Problem 1, Figure 3Here is the calculation of the volume by means of the Method of Cylindrical Shells:> radius_of_shell := x -> x + 1; height_of_shell := x -> f(x) - g(x); := radius_of_shell → x + x 1 := height_of_shell → x − ( )f x ( )g x> Volume = Int(2*Pi*radius_of_shell(x)*height_of_shell(x), x = 0 .. 3); = Volume d⌠⌡032 π ( ) + x 1 ( ) − 3 x x2x> Volume = 2*Pi*Int(expand(radius_of_shell(x)*height_of_shell(x)), x = 0 .. 3); = Volume 2 π d⌠⌡03 − + 2 x2x33 x x> antiderivative := int(2*x^2-x^3+3*x,x); := antiderivative − + 2 x33x443 x22> Volume = simplify(subs(x=3,antiderivative)-subs(x=0,antiderivative)); = Volume454Calculation of the Volume by the Method of WashersIt is possible (but more difficult) to obtain the volume using washers. A glance at Figure 2 above indicates that we have to divide the integration into two pieces: one for ≤ 0 y ≤ 3 and one for ≤ 3 y ≤ 4. Figures 4 and 5 below show the two types of washer. For < y 3, the outer radius of the washer extends to the curve = y x. For < 3 y, the outer radius of the washer extends to the curve = y − 4 x x2.Problem 1, Figure 4 Problem 1, Figure 5> Volume = Int(Pi*( (1+y)^2-(1+2-sqrt(4-y))^2), y = 0 .. 3)+Int(Pi*( (1+2+sqrt(4-y))^2-(1+2-sqrt(4-y))^2), y = 3 .. 4); = Volume + d⌠⌡03π ( ) − ( ) + 1 y2( ) − 3 − 4 y2y d⌠⌡34π ( ) − ( ) + 3 − 4 y2( ) − 3 − 4 y2y> Volume = int(Pi*( (1+y)^2-(1+2-sqrt(4-y))^2), y = 0 .. 3)+int(Pi*( (1+2+sqrt(4-y))^2-(1+2-sqrt(4-y))^2), y = 3 .. 4); = Volume45 π22. The region in the first quadrant bounded above by = = = = y x and below by = = = = y x2 for ≤ ≤ ≤ ≤ 0 x ≤ ≤ ≤ ≤ 1 is rotated about the line y = 2. What is the volume of the solid of revolution that results?a) 8 ππππ15 b) 3 ππππ5 c) 2 ππππ3 d) 4 ππππ5 e) 14 ππππ15 f) ππππ2 g) 3 ππππ8 h) 5 ππππ8 i) 3 ππππ4 j) 7 ππππ8 Solution: a The region (shown with blue boundary) is rotated about = y 2.In the next figure, we show the solid that results from the rotation. The inner boundaty, in the shape of a conical frustum, is rendered with solid tan. The outer boundary, a paraboloid, is rendered using a brown wireframe. A washer is also shown. Calculation of the Volume by the Method of WashersThe outer radius is the distance from = y x2 to = y 2, namely − 2 x2. The inner radius is the distance from = y x to = y 2, namely − 2 x.> volume := Pi*int((2-x^2)^2-(2-x)^2, x = 0 .. 1); := volume8 π15Calculation of the Volume by the Method of Cylindrical ShellsFor a shell that has an edge at level y, the radius is − 2 y and the height of the shell is − y y.> int(2*Pi*(2-y)*(sqrt(y)-y), y = 0..1);8 π153. Calculate the arc length of the graph of = = = = y + + + + x3314 x for ≤ ≤ ≤ ≤ 12x ≤ ≤ ≤ ≤ 1. a) 512 b) 1124 c) 12 d) 1324 e) 712 f) 58 g) 23 h) 1724 i) 34 j) 56 Solution: d> f := x -> x^3/3 + 1/(4*x); := f → x + 13x3141x> diff(f(x),x)^2; − x214 x22> expand(1+diff(f(x),x)^2); + + 12x4116 x4> eqn := 1+diff(f(x),x)^2 = (x^2+1/(4*x^2))^2; := eqn = + 1 − x214 x22 + x214 x22> testeq( eqn );true> `arc length = `*Int((x^2+1/(4*x^2)), x=1/2..1) = int((x^2+1/(4*x^2)), x=1/2..1);= arc length = d⌠⌡ / 1 21 + x214 x2x13244. The line segment with end points (2,3) and (6,0) is rotated about the y-axis. What is the surface area of the resulting conical frustum?a) 12 ππππ b) 16 ππππ c) 20 ππππ d) 24 ππππ e) 25 ππππ f) 27 ππππ g) 28 ππππ h) 32 ππππ i) 36 ππππ j) 40 ππππ Solution: j The average circumference of the frustum is 2 π ( + 2 62 ), or 8 π. The slant length is + 3242, or 5. The surface area is the product of the average circumference and the slant length: 40 π.5. The graph of = = = = y 3 x for ≤ ≤ ≤ ≤ 4 x ≤ ≤ ≤ ≤ 10 is rotated about the x-axis. What is the surface area of the resulting figure? a) 54 π b) 60 π c) 64 π d) 72 π e) 80 π f) 84 π g) 94 π h) 102 π i) 109 π j) 115 π Solution: i > f := x -> 3*sqrt(x);:= f → x 3 x> `Surface area` = 2*Pi*Int(f(x)*sqrt(1+diff(f(x),x)^2), x = 4 .. 10); = Surface area 2 π d⌠⌡4103 x + 49x2x> Answer := Pi*Int(3*(4*x+9)^(1/2),x = 4 .. 10); := Answer π d⌠⌡4103 + 4 x 9 x> Answer := student[changevar](u = 4*x+9, Answer, u); := Answer π d⌠⌡25493 u4u> `Surface area` = value(Answer); = Surface area 109 π6. Let R be the trapezoidal region in the first quadrant that is bounded above by the graph of = = = = y 2 x, below by the x-axis, on the left by the vertical line = = = = x 1, and on the right by the vertical line = = = = x 2. What is the moment of R about the vertical line = = = = x −−−−1 ? (Assume that R has a constant density equal to 1.)a) 203 b) 7 c) 233 d) 8 e) 253 f) 9


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WUSTL MATH 132 - m132_E3sF07

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