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Math 116 / Exam 1 (February 8, 2016) DO NOT WRITE YOUR NAME ON THIS PAGE page 9 7 . [10points] MaizeandBlueJewelryCompanyistryingtodecideonadesignfortheirsignature aMaize-ing bracelet. There are two possible designs: type W and type J . The company has done research and the two bracelet designs are equally pleasing to customers. The design for both rings starts with the function C ( x )=cos ( π 2 x ) where all units are in millimeters. Let R be the region enclosed by the graph of C ( x ) and the graph of - C ( x ) for - 1 x 1. a . [5points] Thetype W braceletisintheshapeofthesolidformedbyrotating R aroundthe line x =50. Writeanintegralthatgivesthevolumeofthetype W bracelet. Include units . Solution: The volume of the type W bracelet, in mm 3 , using the shell method, is integraldisplay 1 - 1 2 π (50 - x ) · 2 C ( x ) dx. b . [5 points] The type J bracelet is in the shape of the solid formed by rotating R around the line y = - 50. Write an integral that gives the volume of the type J bracelet. Include units . Solution: The volume of the type J bracelet, in mm 3 , using the washer method, is integraldisplay 1 - 1 π (50+ C ( x )) 2 - π (50 - C ( x )) 2 dx. University of Michigan Department of Mathematics Winter, 2016 Math 116 Exam 1 Problem 7 (jewelry) Solution
Math 116 / Exam 1 (February 11, 2019) page 7 6 . [12 points] Ryan Rabbitt is making a smoothie with his new electric drink mixer. Mathemat- ically, the container of the mixer has a shape that can be modeled as the surface obtained by rotating the region in the first quadrant bounded by the curves y = 27 and y = x 3 / 2 about the y -axis, where all lengths are measured in centimeters. a . [7 points] Write, but do not evaluate, two integrals representing the total volume, in cm 3 , the mixer can hold: one with respect to x , and one with respect to y . Answer (with respect to x ): integraldisplay 9 0 2 πx parenleftBig 27 - x 3 / 2 parenrightBig dx Answer (with respect to y ): integraldisplay 27 0 π parenleftBig y 2 / 3 parenrightBig 2 dy b . [5 points] Ryan adds 1600 cubic centimeters of liquid to his mixer. The container spins around the y -axis at a very high speed, causing the liquid to move away from the center of the container. The result is the solid made by rotating the shaded region around the y -axis in the diagram below. Note that this means that there is an empty space inside the liquid that has the shape of a cylinder. r y = 27 y = x 3 / 2 x y Let r be the radius of this cylinder of empty space. Set up an equation involving one or more integrals that you would use to solve to find the value of r . Do not solve for r . Solution: integraldisplay 9 r 2 πx parenleftBig 27 - x 3 / 2 parenrightBig dx = 1600 , or integraldisplay 27 r 3 / 2 π parenleftBig y 2 / 3 parenrightBig 2 dy - πr 2 (27 - r 3 / 2 ) = 1600 . (There are other equations that would also work.) Answer: University of Michigan Department of Mathematics Winter, 2019 Math 116 Exam 1 Problem 6 (smoothies) Solution
Math 116 / Exam 1 (February 9, 2015) page 4 2 . [13 points] Fred is designing a plastic bowl for his dog, Fido. Fred makes the bowl in the shape of a solid formed by rotating a region in the xy -plane around the y -axis. The region, shaded in the figure below, is bounded by the x -axis, the y -axis, the line y = 1 for 0 x 4, and the curve y = - ( x - 5) 4 + 2 for 4 x 2 1 / 4 + 5. Assume the units of x and y are inches. x y a . [7 points] Write an expression involving one or more integrals which gives the volume of plastic needed to make Fido’s bowl. What are the units of your expression? Solution: Using the cylindrical shell method, we have that the volume of plastic needed to make Fido’s bowl is given by integraldisplay 4 0 2 πxdx + integraldisplay 5+2 1 4 4 2 πx (2 - ( x - 5) 4 ) dx . Using the washer method, we have that the volume of plastic needed to make Fido’s bowl is given by π integraldisplay 1 0 (5 + (2 - y ) 1 / 4 ) 2 dy + π integraldisplay 2 1 (5 + (2 - y ) 1 / 4 ) 2 - (5 - (2 - y ) 1 / 4 ) 2 dy. The units for either expression are in 3 . b . [6 points] Fred wants to wrap a ribbon around the bowl before he gives it to Fido as a gift. The figure below depicts the cross section of the bowl obtained by cutting it in half across its diameter. The thick solid curve is the ribbon running around this cross section, and the dotted curve is the outline of the cross section which is not in contact with the ribbon. Write an expression involving one or more integrals which gives the length of the thick solid curve in the figure (the length of ribbon Fred needs to wrap the bowl). Solution: The length of ribbon Fred needs to wrap the bowl is given by 10 + 2(2 1 4 + 5) + 2 integraldisplay 5+2 1 4 5 radicalbig 1 + 16( x - 5) 6 dx . x y University of Michigan Department of Mathematics Winter, 2015 Math 116 Exam 1 Problem 2 (dog bowl) Solution
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Math 116 / Exam 1 (October 8, 2014) page 8 7 . [13 points] Kazilla is designing a new board game. She is interested in using the region R in the xy -plane bounded by y = 2, y = x , x = 1 and x = 0. a . [4 points] The first part of the game is a spinning top formed by rotating the region R around the y -axis. Write an integral (or a sum of integrals) that gives the volume of the spinning top. Do not evaluate your integral(s). Solution: Shell method: integraldisplay 1 0 2 π (2 - x ) xdx Washer method: integraldisplay 1 0 πy 2 dy + integraldisplay 2 1 πdy b . [4 points] Another game piece has a base in the shape R , but with semicircular cross sections perpendicular to the x -axis. Write an integral which gives the volume of the game piece. Do not evaluate your integral. Solution: π 8 integraldisplay 1 0 (2 - x ) 2 dx c . [5 points] A third game piece has volume given by integraldisplay 2 0 π ( h ( x )) 2 dx where h ( x ) is a contin- uous function of x . Use MID(3) to approximate the volume of this third game piece. Be sure to write out all of the terms in your approximation. Your answer may contain the function h ( x ). Solution: MID (3) = 2 3 [ π ( h (1 / 3)) 2 + π ( h (1)) 2 + π ( h (5 / 3)) 2 ] University of Michigan Department of Mathematics Fall, 2014 Math 116 Exam 1 Problem 7 (board game) Solution