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 ln 2 5 9 2 ln 5 Explanation Integrating the inner integral with respect to x keeping y fixed we see that Z 2 0 h 1 i2 1 dx x y 2 x y 0 1 1 y 2 y lewis scl876 HW 15 1 2 radin 55315 3 In this case 3 Z I 1 006 1 1 y 2 y h ln y ln 2 y Consequently dy i3 1 Determine the value I of the integral of the function f x y 3 1 2 9 I ln ln 2 3 5 005 10 0 points Evaluate the iterated integral Z ln 4 Z ln 3 I e2x y dx dy 0 10 0 points 2x y 1 8y y 2 over the rectangle n o A x y 1 x 3 0 y 3 1 I ln 34 correct 2 I ln 32 0 3 I 2 ln 34 1 I 6 4 I 34 2 I 4 5 I 32 3 I 5 6 I 2 ln 32 4 I 2 Explanation The integral of f over A can be written as the iterated integral 5 I 3 correct Explanation Integrating with respect to x with y fixed we see that Z ln 3 1 h 2x y iln 3 2x y e dx e 0 2 0 32 1 1 2 ln 3 y e e y e y 2 2 I 0 I 4 ln 4 e y 0 h dy 4 e y 4 e ln 4 1 Consequently I 4 1 4 1 iln 4 0 3Z 3 1 2x y dx dy 1 8y y 2 integrating first with respect to x Now Z 3 1 Thus Z Z h x2 xy i3 2x y dx 1 8y y 2 1 8y y 2 1 8 2y 1 8y y 2 Thus I Z 0 3 8 2y dy 1 8y y 2 To evaluate this last integral we use the substitution u 1 8y y 2 For then 3 du 8 2y dx lewis scl876 HW 15 1 2 radin 55315 while y 0 u 1 y 3 u 34 Thus 16 I 3 Consequently I Z 34 1 n A x y 0 x 1 008 over the rectangle o 0 y 1 A x y 0 x 2 0 y 1 1 I 2 I 3 I 5 3 I 6 4 I 4 I 5 I 7 5 I 6 6 I 0 y 1 o is a rectangle with sides parallel to the coordinate axes the double integral can be represented as the iterated integral Z 0 0 1Z 1 0 10 0 points A 4 2 I correct 3 Explanation Since n A x y 0 x 1 4 3 Evaluate the integral Z Z I xe2xy dxdy 3 1 I 2 1 0 i1 16 h 1 1 tan y dy 1 y2 3 0 I 10 0 points when Now Z 1 h i34 1 du ln u ln 34 u 1 Evaluate the double integral Z Z 5 x2 I dxdy 2 A 1 y I Z Consequently 007 4 5 x2 dxdy 1 y2 1 h 1 3 i1 5 x2 dx 5x x 0 1 y2 1 y2 3 1 4 e 5 correct 4 1 4 e 4 4 1 4 e 5 8 1 4 e 4 8 1 4 e 3 8 1 4 e 3 4 Explanation Since the integral with respect to y in Z Z I xe2xy dxdy A can be evaluated easily using substitution or directly making the substitution in one s head while the integral with respect to x requires integration by parts this suggests that we should represent the double integral as the repeated integral Z 2Z 1 I xe2xy dy dx 0 0 lewis scl876 HW 15 1 2 radin 55315 Now after integration the inner integral becomes h1 i1 1 2x e 1 e2xy 0 2 2 5 Consequently volume 7 cu units Thus 1 I 2 Z 2 2x e 0 i2 1 h e2x 1 dx x 0 2 2 keywords double integral volume of …

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