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271s3

# 271s3 - MATH 271 Test#3T Solutions(1(20 points A tank in...

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MATH 271 Test #3T Solutions (1) (20 points) A tank in the shape of a sphere with a radius of 5 meters is filled halfway with water (density: 1000 kg/m 3 ). The center of the tank is 20 meters underground. How much work does it take to pump the water to the surface? Evaluate the integral. (Note g = 9 . 8 m/s 2 .) Solution: The amount of work needed to pump the water to the surface is the integral of the weight of the water (in a thin slice) times the distance it has to be moved. The weight of the water is the density of water times g , times the cross-section area at a given depth, times dx . If the depth of the water is x , then the cross-section area of the tank is πr 2 , where r 2 + ( x - 20) 2 = 5 2 (because the tank is 20 meters below the surface and is spherical with radius 5 ). The area is only counted for the bottom half of the tank, so 20 x 25 . The integral which represents the work needed to pump out all the water is 25 Z 20 (1000)(9 . 8)( πr 2 ) x dx = 9800 π 25 Z 20 (5 2 - ( x - 20) 2 ) x dx. To evaluate the integral, expand ( x - 20) 2 = x 2 - 40 x + 400 and multiply out the polynomial to get 9800 π 25 Z 20 25 x - x 3 + 40 x 2 - 400 x = 9800 π - x 4 4 + 40 x 3 - 375 2 x 2 25 20 = 56 , 123 , 243 . 79 J The substitution u = x - 20 could also be made. Grading: This problem was graded on a 0 5 10 15 20 basis. Grading for common mistakes: - 5 points if the integral wasn’t evaluated; +7 points (total) if the mass times a depth of 20 m or 25 m was used; +10 points (or more) if the expression involved an integra; +15 points (or more) if the integral of an area was found.

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271s3 - MATH 271 Test#3T Solutions(1(20 points A tank in...

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