Math 132 Exam II Spring 2008 1 This exam contains 15 multiple choice questions and 2 hand graded questions The multiple choice questions are worth 5 points each and the hand graded questions are worth a total of 25 points The latter questions will be evaluated not only for having the correct solutions but also for clarity Points may be taken for confusing and disorganized writing even when the answer is correct 1 Find the area enclosed by the graphs of f x x and g x 2 x2 A 5 2 B 3 C 7 2 D 4 E 9 2 F 5 G 11 2 H 6 I 13 2 J 7 The 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 1 and x 2 The area is Z 1 Z 1 g x f x dx 2 x2 x dx 2 2 1 x2 x3 2x 3 2 2 8 1 1 2 4 2 3 2 3 1 8 3 2 9 2 Math 132 Exam II Spring 2008 2 2 Calculate the volume of a cylinder inclined at an angle 6 whose height is 10 and whose base is a circle of radius 4 A 320 B 160 C 120 D 80 E 80 3 F 160 3 G 4 3 H 96 I 96 3 J 320 3 The cross section of the cylinder at height x for 0 x 10 is a disc of radius 4 hence the cross sectional area is constant equal to A x 42 16 The volume is given by the integral Z 10 Z 10 A x dx 16 dx 160 0 0 Math 132 Exam II Spring 2008 3 3 Find the volume of the solid obtained by rotating the region enclosed by the graphs of x y and x 0 about the y axis over the interval 1 y 3 A 4 B 3 C 2 D E 2 F 3 G 4 H 8 I 10 J 12 By the disc method the volume is Z 1 3 2 3 y y 2 dy 2 1 9 1 2 4 Math 132 Exam II Spring 2008 4 4 Compute the volume of the solid obtained by rotating the region enclosed by the graphs of y x2 y 2 x2 and x 0 about the y axis A 8 B 7 C 6 D 5 E 4 F 3 G 2 H I 2 J 4 The intersection of the graphs of y x2 and y 2 x2 happens for x such that x2 2 x2 This equation has solutions x 1 The region bounded by these two graphs and x 0 lies over the interval 0 x 1 The volume obtained by the shell method is then Z 1 Z 1 2 x 2 x2 x2 dx 2 x 2 2x2 dx 0 0 Z 4 2 x x3 dx 0 x4 x2 2 4 1 1 4 2 4 1 4 0 Math 132 Exam II Spring 2008 5 5 Calculate the volume of the solid obtained by rotating the region under the graph of f x x 2 over the interval 2 1 about the axis of rotation x 4 A 2 ln 2 2 B 1 ln 2 C 2 1 ln 2 D 2 2 ln 2 E 4 ln 2 5 F 3 1 ln 5 G 2 2 3 ln 2 H 5 2 3 ln 2 I 5 1 ln 2 3 J 7 2 ln 2 Using the shell method we get Z 1 Z 1 2 4 x x 2 dx 2 4 x x 2 dx 2 2 2 Z 4 u u 2 du doing a substitution u x 2 1 Z 2 2 4u 2 u 1 du 1 2 2 4u 1 log u 1 2 2 ln 2 4 0 2 2 ln 2 Math 132 Exam II Spring 2008 6 6 How much work is done raising a 4 kg mass to a height of 1 m above ground Recall the acceleration of gravity is 9 8 meters per second squared A 39 2 J B 38 2 J C 37 2 J D 36 2 J E 35 2 J F 34 2 J G 33 2 J H 32 2 J I 31 2 J J 30 2 J The work is given by the product of 4 kg by 9 8 m s2 the weight by 1 m So W 4 9 8 1 39 2 J Math 132 Exam II Spring 2008 7 7 Compute the work required to stretch a spring from 1 to 2 centimeters past equilibrium assuming that the spring constant is 200 kg per second squared A 0 36 J B 0 07 J C 0 15 J D 0 02 J E 0 03 J F 0 43 J G 1 53 J H 2 03 J I 3 42 J J 4 50 J The force function is F x 200x x is stretched from 0 01 m to 0 02 m Thus the work is given by Z 0 02 0 02 200xdx 100x2 0 01 0 01 0 03 J Math 132 Exam II Spring 2008 8 8 Calculate the work in Joules required to pump all of the water out of the horizontal cylindrical tank shown in the figure The cylinder has radius r and length l The acceleration of gravity is g 9 8 m per second squared and the density ofRwater is 1000 kg per meter cubed Hint You may r need the integral r r2 u2 du r2 2 A B C D E F G H I J 9 8 r3 l 98 r3 l 980r2 l2 9800r2 l2 9 8r2 l2 9800 rl3 9800 r2 l2 9800 r3 l 9800 r3 l 3 9800 r3 l 6 Let x be the height measured from the ground Then the cross section area is p the area of a rectangle with sides l and 2 r2 r x 2 so p A x 2l r2 r x 2 The work to pump out the water a distance 2r x from level x is then Z 2r Z 2r p g A x 2r x dx g 2l r2 r x 2 2r x dx 0 0 Z r p 2lg r u r2 u2 du using u r x r Z r p Z r p 2 2 2 2 2lg r r u du u r u du r 2lg r r3 0 2 lg r3 In the second to last line the second integral is zero since the integrand is an odd function and the interval is symmetric about 0 The first integral is half the area of a disc of radius r Thus the work is W 9800 r3 l Math 132 Exam II Spring 2008 9 Evaluate the integral R 0 sin 3x cos 4x dx A 3 2 B 2 3 C 5 4 D 3 5 E 3 …
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