Math 132 - Exam II - Spring 2008 1This exam contains 15 multiple choice questions and 2 hand graded ques-tions. The multiple choice questions are worth 5 points each and the handgraded questions are worth a total of 25 points. The latter questions will beevaluated not only for having the correct solutions but also for clarity. Pointsmay be taken for confusing and disorganized writing, even when the answer iscorrect.1. Find the area enclosed by the graphs of f(x) = x and g(x) = 2 − x2.A) 5/2B) 3C) 7/2D) 4→E) 9/2F) 5G) 11/2H) 6I) 13/2J) 7The points of intersection of the two graphs are obtained by setting f(x) =g(x), or x = 2 − x2. Equivalently, x2+ x − 2 = 0. The roots are x = 1and x = −2. The area isZ1−2[g(x) − f(x)]dx =Z1−2[2 − x2− x]dx=2x −x33−x221−2= 2 −13−12−−4 +83− 2= 8 − 3 −12=92.Math 132 - Exam II - Spring 2008 22. Calculate the volume of a cylinder inclined at an angle π/6 whose heightis 10 and whose base is a circle of radius 4.A) 320ππ→B) 160πC) 120πD) 80πE) 80√3πF) 160√3πG) 4π/3H) 96πI) 96√3πJ) 320√3πThe cross section of the cylinder at height x, for 0 ≤ x ≤ 10, is a disc ofradius 4, hence the cross-sectional area is constant, equal to A(x) = π42=16π. The volume is given by the integralZ100A(x)dx = 16πZ100dx = 160π.Math 132 - Exam II - Spring 2008 33. Find the volume of the solid obtained by rotating the region enclosedby the graphs of x =√y and x = 0 about the y-axis over the interval1 ≤ y ≤ 3.A) π/4B) π/3C) π/2D) πE) 2πF) 3π→G) 4πH) 8πI) 10πJ) 12πBy the disc method, the volume isZ31π(√y)2dy =»πy22–31= π9 −12= 4π.Math 132 - Exam II - Spring 2008 44. Compute the volume of the solid obtained by rotating the region enclosedby the graphs of y = x2, y = 2 − x2, and x = 0 about the y-axis.A) 8πB) 7πC) 6πD) 5πE) 4πF) 3πG) 2π→H) πI) π/2J) π/4The intersection of the graphs of y = x2and y = 2 −x2happ e ns for x such thatx2= 2 − x2. This equation has solutions x = ±1. The region bounded by thesetwo graphs and x = 0 lies over the interval 0 ≤ x ≤ 1. The volume, obtained bythe shell method, is thenZ102πx[2 − x2− x2]dx = 2πZ10x(2 − 2x2)dx= 4πZ20(x − x3)dx= 4π»x22−x44–10= 4π„12−14«= π.Math 132 - Exam II - Spring 2008 55. Calculate the volume of the solid obtained by rotating the region underthe graph of f(x) = x−2over the interval [−2, −1], about the axis ofrotation x = 4.A) π(2 + ln 2)/2B) π(1 + ln 2)C) 2π(1 + ln 2)→D) 2π(2 + ln 2)E) π(4 + ln 2)/5F) 3π(1 + ln 5)G) 2π(2 + 3 ln 2)H) 5(2 + 3 ln 2)I) 5π(1 + ln 2)/3J) 7(2 + ln 2)Using the shell method we get:Z−1−22π(4 − x)x−2dx = 2πZ−1−2(4 − x)x−2dx= 2πZ21(4 + u)u−2du (doing a substitution u = −x)= 2πZ21(4u−2+ u−1)du= 2π−4u−1+ log u 21= 2π [−2 + ln 2 + 4 − 0]= 2π(2 + ln 2).Math 132 - Exam II - Spring 2008 66. How much work is done raising a 4 kg mass to a height of 1 m aboveground? (Recall: the acceleration of gravity is 9.8 meters per secondsquared.)→A) 39.2 JB) 38.2 JC) 37.2 JD) 36.2 JE) 35.2 JF) 34.2 JG) 33.2 JH) 32.2 JI) 31.2 JJ) 30.2 JThe work is given by the product of 4 kg by 9.8 m/s2(the weight) by 1 m. SoW = 4 × 9.8 × 1 = 39.2 J.Math 132 - Exam II - Spring 2008 77. Compute the work required to stretch a spring from 1 to 2 centimeterspast equilibrium, assuming that the spring constant is 200 kg per secondsquared.A) 0.36 JB) 0.07 JC) 0.15 JD) 0.02 J→E) 0.03 JF) 0.43 JG) 1.53 JH) 2.03 JI) 3.42 JJ) 4.50 JThe force function is F (x) = 200x, x is stretched from 0.01 m to 0.02 m. Thusthe work is given byZ0.020.01200xdx =ˆ100x2˜0.020.01= 0.03 J.Math 132 - Exam II - Spring 2008 88. Calculate the work in Joules required to pump all of the water out of thehorizontal cylindrical tank shown in the figure. The cylinder has radius rand length l. The acceleration of gravity is g = 9.8 m per second squaredand the density of water is ρ = 1000 kg per meter cubed. (Hint: You mayneed the integralRr−r√r2− u2du = πr2/2.)A) 9.8πr3lB) 98πr3lC) 980r2l2D) 9800r2l2E) 9.8r2l2F) 9800πrl3G) 9800πr2l2→H) 9800πr3lI) 9800πr3l/3J) 9800πr3l/6Let x be the height measured from the ground. Then the cross-section area isthe area of a rectangle with sides l and 2pr2− (r − x)2, soA(x) = 2lpr2− (r − x)2.The work to pump out the water (a distance 2r − x from level x) is thengρZ2r0A(x)(2r − x)dx = gρZ2r02lpr2− (r − x)2(2r − x)dx= 2lgρZr−r(r + u)pr2− u2du using u = r − x= 2lgρ„rZr−rpr2− u2du +Zr−rupr2− u2«du= 2lgρ„πr32+ 0«= lgρπr3.In the second to last line, the second integral is zero since the integrand is ano dd function and the interval is symmetric about 0. The first integral is halfthe area of a disc of radius r. Thus the work is W = 9800πr3l.Math 132 - Exam II - Spring 2008 99. Evaluate the integralRπ0sin(3x) cos(4x)dx.A) 3/2B) 2π/3C) 5/4D) 3π/5E) −3F) −2πG) −5/7H) 3/7I) π/5→J) −6/7We can use here the table of trigonometric integrals. It givesZπ0sin(3x) cos(4x)dx =»−cos(3 − 4)x2(3 − 4)−cos(3 + 4)x2(3 + 4)–π0=»cos x2−cos(7x)14–π0= −12+114−»12−114–= −67.Math 132 - Exam II - Spring 2008 1010. Use the error bound Error(SN) ≤ K4(b − a)5/(180N4) to find a value ofN such that the Simpson’s rule approximation SNforR51ln x dx has anerror of at most 10−6. (Recall that in Simpson’s rule N must be an eveninteger.)A) 605548B) 98C) 674D) 68E) 6F) 32G) 58→H) 78I) 74J) 78432N must be an even integer such thatK4(5 − 1)5180N4≤ 10−6.To obtain K4we take the fourth derivative of f(x) = ln x:f0(x) = 1/x, f00(x) = −1/x2, f(3)(x) = 2/x3, f(4)(x) = −6/x4.The maximum value of |f(4)(x)| over the interval [1, 5] is K4= 6/14= 6. So welo ok for N such that6(45)180N4≤ 10−6.Solving for N:N ≥„6(45)106180«1/4= 76.4354Rounding up to the next even integer givesN = 78.Math 132 - Exam II - Spring 2008 1111. Evaluate the integralR10(2x + 9)exdxA) 8B) 4C) e2D) 2e + 4E) e + 7F) 3e − 2G) 2e3+ 5H) 5e2− 9→I) 9e − 7J) 8e3− 5We can set u(x) = 2x + 9 and v0(x) = ex. Integrating by parts:Z10(2x + 9)exdx =(2x + 9)ex− 2Zexdx10= [(2x + 9)ex− 2ex]10= 11e − 9 − 2(e − 1)= 9e − 7.Math 132 - Exam II - Spring 2008 1212. Evaluate the integralRπ /20(x − 1) cos x dx.A)π2− 1B) π − 2C) 2π − 1→D)π2− 2E)π2+ 1F) 1G) 2H) 4I) 0J) −π2Zπ /20(x − 1) cos xdx =(x − 1) sin x −Zsin xdxπ /20= [(x − 1) sin x + cos
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