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Unformatted text preview: 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: TI89, TI92,and TINSPIRE 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 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 yaxis (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 π p 2 y + 4 y 3 2 dy = π Z h ( 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 62 . 5 π p 2 y + 4 y 3 2 (13 y ) dy = 62 . 5 π Z √ 2 ( 2 y + 4 y 3 ) (13 y ) dy = 62 . 5 π Z √ 2 ( 4 y 4 + 52 y 3 2 y 2 + 26 y ) dy = 62 . 5 π 78 68 √ 2 15 !...
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This note was uploaded on 11/11/2011 for the course MATH 152.01 taught by Professor Geline during the Fall '09 term at Ohio State.
 Fall '09
 GELINE
 Math, Calculus, Geometry

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