UT M 408C - HW 15.1,2-solutions-1 (6 pages)

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HW 15.1,2-solutions-1



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HW 15.1,2-solutions-1

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Pages:
6
School:
University of Texas at Austin
Course:
M 408c - Differential and Integral Calculus

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lewis scl876 HW 15 1 2 radin 55315 This print out should have 11 questions Multiple choice questions may continue on the next column or page find all choices before answering 001 1 The graph of the function z f x y 8 x is the plane shown in 10 0 points Evaluate the double integral Z Z I 5 dxdy z A y with n o A x y 4 x 6 3 y 7 by first identifying it as the volume of a solid 8 x 1 I 34 Determine the value of the double integral Z Z I f x y dxdy 2 I 42 3 I 36 A 4 I 40 correct 5 I 38 Explanation The value of I is the volume of the solid below the graph of z f x y 5 and above the region n o A x y 4 x 6 3 y 7 Since A is a rectangle this solid is a box with base A and height 5 Its volume therefore is given by length width height 6 4 7 3 5 Consequently I 40 over the region n o A x y 0 x 3 0 y 6 in the xy plane by first identifying it as the volume of a solid below the graph of f 1 I 118 cu units 2 I 120 cu units 3 I 117 cu units correct 4 I 119 cu units 5 I 116 cu units Explanation The double integral Z Z I f x y dxxdy A keywords volume double integral rectangular region rectangular solid 002 10 0 points is the volume of the solid below the graph of f having the rectangle n o A x y 0 x 3 0 y 6 lewis scl876 HW 15 1 2 radin 55315 for its base Thus the solid is the wedge But then I z 1 Z h 9 3 i1 4x x2 dx 2x2 x3 2 2 0 0 8 2 Consequently y I 3 6 3 x and so its volume is the area of trapezoidal face multiplied by the thickness of the wedge Consequently I 117 cu units keywords keywords definite integral iterated integral polynomial function 004 Z 1 10 0 points Evaluate the integral Z 1Z 2 I 4x 3x2 y dydx 0 1 1 I 2 I 1 I 4 2 I 7 correct 2 3 I 3 I 9 2 4 I 4 I 5 5 I 5 I 11 2 6 I Explanation The integral can be written in iterated form Z 1Z 2 I 4x 3x2 y dy dx 0 Now Z 2 1 2 4x 3x y dy 1 h 9 4x x2 2 i2 3 4xy x2 y 2 1 2 10 0 points Evaluate the iterated integral I 003 7 2 3Z 2 0 1 dx dy x y 2 3 2 ln 4 3 ln 4 1 3 ln 2 4 9 correct ln 5 1 9



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