# 1322FinalSolutionF17.pdf - MAT1322D Solution to Final...

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Chapter 13 / Exercise 15
Mathematical Applications for the Management, Life, and Social Sciences
Harshbarger Expert Verified
MAT1322D Solution to Final Examination December 2017 1 Solution to the Final Examination MAT1322D, Fall 2017 Part I. Multiple-choice Questions (2 12 = 24 marks) FADAFEECBBDD 1. The area of the region above the graph of y= x2and under the graph of y= 8 x2in the first quadrant is Solution. (F) Let x2= 8 x2. 2x2= 8, x= 2. The area is 232222200032(8)2(4)2 433xxxxdxxdxx. 2. Let Rbe the region above the parabola y= x2and under the line y= 2x. Solid Bis obtained by revolving Rabout the line y= 4. Then the volume of Bis calculated by the integral Answer. (A) rin= 4 2x rout = 4 − x2 y= 4 y= x2 y= 2x X Y O
##### We have textbook solutions for you!
The document you are viewing contains questions related to this textbook. The document you are viewing contains questions related to this textbook.
Chapter 13 / Exercise 15
Mathematical Applications for the Management, Life, and Social Sciences
Harshbarger Expert Verified
MAT1322D Solution to Final Examination December 2017 2 3. Suppose a pool has the shape of an inverted truncated pyramid. The top of the pool at the ground level is a square with side length 20 meters, and the bottom of the pool is a square with side length 10 meters. The depth of the pool is 5 meters. The pool is filled with water with density kg / m3. Let xbe the distance between a horizontal layer of water and the top of the pool. Denote the acceleration of gravity be gm / sec2. Then the work, in Joules, needed to pump the water in the pool to a point 2 meters above the ground is calculated by the integral Solution. (D) A horizontal layer of water in the pool is a square with side-length L(x) = 10 + 2 (5 x) = 20 2x. The volume of this layer with thickness dxis V(x) = (L(x))2= (20 2x)2dx. The weight of this layer of water is w(x) = gV(x) = g(20 2x)2dx. The work needed to pump this layer of water to a point 2 meters above the ground is W(x) = w(x)(x+ 2) = g(20 2x)2(x+ 2)dx. The total work is W= 520(202 ) (2)gxxdx.
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