exam2 - Math 152 Au 11 K Kwa Name SO N0056 Signature...

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Math 152 Au. ’11 – K. Kwa Name: SO N0056 Signature: October 27, 2011 Name.nnn: Exam 2 Form A, 7 pages General Instructions: Do all the problems. Answer each part thoroughly. Show all of your work! Incorrect answers with work shown may receive partial credit but unsubstantiated answers will receive NO credit. You do not need to simplify numerical answers, e.g. 3 + 2 / 5 - 7 / 6. Calculators are permitted EXCEPT those calculators that have symbolic algebra or calculus capabilities. In particular , the following calculators and their upgrades are not permitted: TI-89, TI-92,and TI-NSPIRE CAS. In addition, neither tablets, laptops nor cell phones are permitted. permitted. Special Instructions: The Exam Duration is 48 minutes This exam consists of 5 problems on 5 pages (excluding this cover sheet and a formula sheet in the back). Make sure that your exam paper is not missing any pages before you start. Answers without supporting work will receive no credit. A random sample of graded exams will be xeroxed before being returned. Good Luck ! Problem # Points Score 1 20 2 20 3 20 4 20 5 20 Total 100 1
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1) (20 points) A large bowl is constructed by rotating the graph of x = p 2 y + 4 y 3 , 0 y 13 around the y -axis (with dimensions measured in feet). (a) Find the volume of water in the bowl when it is filled to height h (i.e. 0 y h ) (Hint: use the disk method, not the shell method.) Solution. The volume of the water is when the bowl is filled to height h is V ( h ) = Z h 0 π p 2 y + 4 y 3 2 dy = π Z h 0 ( 2 y + 4 y 3 ) dy = π ( h 2 + h 4 ) ft 3 . (b) If the bowl contains 6 π ft 3 of water, how high is the water level in the bowl? (Hint: the relevant equation factors in the form ( h 2 - a )( h 2 + b ) for suitable numbers a and b .) Solution. By part (a), V ( h ) = 6 π . Thus h 2 + h 4 = 6 h 4 + h 2 - 6 = 0 ( h 2 - 2)( h 2 + 3) = 0 h 2 - 2 = 0 h = 2 ft . (c) Write down, but do NOT simplify or evaluate, a definite integral for computing the work needed to pump out the 6 π ft 3 of water out over the top of the bowl. (The density of water is 62.5lb/ft 3 .) Solution. The work required is W = Z 2 0 62 . 5 π p 2 y + 4 y 3 2 (13 - y ) dy = 62 . 5 π Z 2 0 ( 2 y + 4 y 3 ) (13 - y ) dy = 62 . 5 π Z 2 0 ( - 4 y 4 + 52 y 3 - 2 y 2 + 26 y ) dy = 62 . 5 π 78 - 68 2 15 ! 14056 . 4 ft-lb .
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