HW13-solutions - liang(xl5432 HW13 berg(55120 This...

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Unformatted text preview: liang (xl5432) – HW13 – berg – (55120) This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points 1 is the volume of the solid below the graph of f having the rectangle A = (x, y) : 0 ≤ x ≤ 7, 0 ≤ y ≤ 6 for its base. Thus the solid is the wedge The graph of the function z z = f (x, y) = 7 − x 7 is the plane shown in y z (7, 6) 7 y x and so its volume is the area of triangular face multiplied by the thickness of the wedge. Consequently, 7 x I = 147 cu. units . Determine the value of the double integral f (x, y) dxdy I = A over the region A = (x, y) : 0 ≤ x ≤ 7, 0 ≤ y ≤ 6 in the xy-plane by first identifying it as the volume of a solid below the graph of f . keywords: double integral, linear function, volume under graph, volume, rectangular region, prism, triangle 002 Determine the value of the iterated integral 4 3 I = (1 + 2xy) dx dy . 0 1. I = 145 cu. units 2. I = 148 cu. units 10.0 points 1 1. I = 70 3. I = 147 cu. units correct 2. I = 74 4. I = 146 cu. units 3. I = 72 correct 5. I = 144 cu. units Explanation: The double integral I = 5. I = 76 f (x, y) dxdy A 4. I = 78 Explanation: liang (xl5432) – HW13 – berg – (55120) Integrating with respect to x and holding y fixed, we see that 3 (1 + 2xy) dx = 2 x+x y 1 x=3 x=1 In this case 3 I = (2 + 8y) dy = 2y + 4y 2 0 1 1 − y 3+y I = 2 1 . 4 3 0 1 . Consequently, . (3)(1 + 3) (3 + 3) I = 2 ln Consequently, 004 I = 8 + 64 = 72 . = 2 ln (2) . 10.0 points Evaluate the iterated integral 4 4 I = keywords: 1 003 dy = 2 ln(y) − ln(3 + y) Thus 4 2 1. I = 10.0 points 1 x y dydx . + y x 15 ln (15) 2 Evaluate the iterated integral 3 3 I = 1 0 2. I = 15 ln (4) correct 2 dx dy . (x + y)2 15 2 3. I = 4 ln 1. I = ln (2) 15 2 4. I = 15 ln 2. I = ln 6 5 5. I = 4 ln (15) 6 5 3. I = 1 ln 2 4. I = 1 ln (2) 2 5. I = 2 ln 6. I = Explanation: Integrating with respect to y keeping x fixed, we see that 6 5 4 1 6. I = 2 ln (2) correct Explanation: Integrating the inner integral with respect to x keeping y fixed, we see that 3 0 15 ln (4) 2 x y + y x dy = = (ln(4)) x + 2 − x+y 1 1 . = 2 − y 3+y 3 0 15 2 1 x 4 y2 2x 1 . Thus 4 2 dx = (x + y)2 x ln(y) + I = (ln(4)) x + 1 = x2 2 15 2 1 x 15 ln(4) + ln(x) 2 dx 4 . 1 liang (xl5432) – HW13 – berg – (55120) 3 when Consequently, A = I = 15 ln(4) . (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 . 1. I = 7 3 2. I = 1 ln 6 7 3 3. I = 1 ln (3) 3 1 ln 4 5. I = 1 ln (3) 6 6. I = 10.0 points 1 ln 3 4. I = 005 1 ln (3) 4 Evaluate the iterated integral ln(3) ln(4) I = 0 e2x−y dx dy . 0 1. I = 9 2. I = 8 3. I = 7 4. I = 5 correct Explanation: Integrating with respect to x with y fixed, we see that e2x−y dx = 0 ln(4) 1 2x−y e 2 0 A = (x, y) : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1 is a rectangle with sides parallel to the coordinate axes, the double integal can be interpreted as the iterated integral 1 Thus 0 2 xy 2 dx dy . x2 + 3 2 42 − 1 −y e . 2 1 2 ln(4)−y e − e−y = = 2 I = 7 3 Explanation: Since 5. I = 6 ln(4) correct xy 2 dx x2 + 3 0 But to integrate 15 2 ln(3) 0 e−y dy = − 15 −y e 2 15 − ln(3) = − e −1 . 2 006 0 0 with respect to x with y fixed we use the substitution u = x2 + 3. For then Consequently, 15 1 −1 I = − 2 3 ln(3) du = 2x dx = 5 . while Evaluate the double integral I = A xy 2 dx dy x2 + 3 =⇒ u = 3, x = 2 10.0 points x = 0 =⇒ u = 7. In this case 2 0 xy 2 1 dx = 2+3 x 2 7 3 y2 y2 ln u du = u 2 7 3 . liang (xl5432) – HW13 – berg – (55120) Thus 1 ln 2 7 3 1 = ln 6 7 3 I = 1 Thus π/2 y 2 dy 2x cos(x + y) dx = π sin 0 y3 4 0 1 0 . + 2 cos Consequently, π +y 2 π + y − 2 cos(y) . 2 In this case, 1 I = ln 6 7 3 I = . −π cos π +y 2 + 2 sin 007 10.0 points 2x cos(x + y) dxdy 008 . 10.0 points Find the volume of the solid under the graph of when A is the rectangle π , 2 0 I = −(4 − π) . A (x, y) : 0 ≤ x ≤ π/2 Consequently, Calculate the value of the double integral I = π + y − 2 sin(y) 2 0 ≤ y ≤ π 2 f (x, y) = 5 + 3x2 + 4y . 1. I = −2π and above the rectangle A = (x, y) : 1 ≤ x ≤ 2 , 0 ≤ y ≤ 2 . 2. I = (4 − π) 1. volume = 32 cu.units correct 3. I = 2π 2. volume = 39 cu.units 3. volume = 31 cu.units 4. I = π 4. volume = 37 cu.units 5. I = −2(4 − π) 5. volume = 30 cu.units 6. I = −(4 − π) correct Explanation: By treating I as an iterated integral, integrating first with respect to x, we see that 2x cos(x + y) dx = 2x sin(x + y) Explanation: The volume is given by the double integral 2 V = 0 sin(x + y) dx = 2 x sin(x + y) + cos(x + y) . (5 + 3x2 + 4y) dxdy 1 of f (x, y) over the rectangular region A. Integrating each term separately, we see that 2 −2 2 2 V = 2 2 5 dxdy + 0 1 0 2 3x2 dxdy 1 2 + 4y dxdy = 10 + 14 + 8. 0 1 liang (xl5432) – HW13 – berg – (55120) Consequently, volume = 32 cu.units . 009 √ 3y dx dy . I = y 0 16 15 Explanation: Treating I as an iterated integral, integrating first with respect to x with y fixed, we see that 1 1. I = correct 5 2. I = 19 15 5. I = − y 13 15 4. I = − Evaluate the iterated integral 1 2 2. I = − correct 3 3. I = − 10.0 points 4 5 1 y −1 y2 I = 2 3. I = 5 (5x − 2y) dx dy 1 = 4. I = 1 5. I = 5 −1 5 2 x − 2xy 2 y dy . y2 Thus 3 5 1 Explanation: As an interated integral, integrating first with respect to x, we see that 1 √ [ 3xy ]y y dy I = 0 1 = 3 0 I = −1 = 5 2 (y − y 4 ) dy − 2 y3 y5 − 3 5 5 2 011 1 = 0 1 . 5 10.0 points Evaluate the double integral y −1 y2 I = 1. I = − 7 15 2(y 2 − y 3 ) dy y3 y4 − 3 4 (5x − 2y) dx dy . 2 . 3 10.0 points Evaluate the iterated integral 3π/2 1 −2 I = − Consequently, 010 −1 Consequently, y 3/2 − y 2 dy . 2 5/2 1 3 y − y I = 3 5 3 1 cos(θ) I = 0 1. I = e − 2 2. I = 1 −2 e 3. I = 2(e − 1) 0 2 esin(θ) dr dθ . 1 . −1 liang (xl5432) – HW13 – berg – (55120) 1 −1 e 4. I = 2 6 The integral can be written as the repeated integral correct √ 1 y I = 5. I = 2e 0 (8x − y) dx y dy. Now 6. I = 0 √ Explanation: After simple integration cos(θ) sin(θ) 2e y 2re sin(θ) = 2 cos(θ) e . 1 In this case, 0 I = sin(θ) 2 cos(θ) e sin(θ) dθ = 2e 0 y y But then I = 3π/2 4x − xy √ = 4y − y 3/2 − 3y 2 . 0 0 2 (8x − y) dx = cos(θ) sin(θ) dr = y 3π/2 0 (4y − y 3/2 − 3y 2) dx = . 2 2y 2 − y 5/2 − y 3 5 1 0 . Consequently, Consequently, I = 2 1 −1 e . 013 012 10.0 points 3 . 5 I = 10.0 points Evaluate the double integral Find the value of the double integral I = I = A (8x − y) dxdy D 4y dxdy +1 x2 when D is the region when A is the region (x, y) : y ≤ x ≤ √ y, 0≤y≤1 . (x, y) : 0 ≤ x ≤ 1, in the xy-plane. 1. I = 4 5 1. I = 2 ln(2) 2. I = 3 correct 5 2. I = ln(2) correct 3. I = 9 10 3. I = 2 4. I = 7 10 4. I = 4 ln(2) 5. I = 1 Explanation: 5. I = 1 6. I = 4 0≤y≤ √ x liang (xl5432) – HW13 – berg – (55120) Explanation: As an iterated integral, integrating first with respect to y, we see that √ 1 x I = 0 0 4y dy 2+1 x x dx . Explanation: After integration with respect to y we see that 1 4y dy x2 + 1 0 I = √ 0 x x = 2 2 . x +1 In this case, x dx = 2+1 x I = 2 0 4x sin(y) 0 y2 = 2 2 x +1 1 7. I = 4 (1 − cos(1)) 8. I = 2 (cos(1) − 1) Now √ 7 ln(x2 + 1) 1 0 0 dx 4x sin(x2 ) dx = 0 −2 cos(x2 ) 1 0 , using substitution in the second integral. Consequently, . Consequently, I = ln(2) 1 = x2 . I = 2 (1 − cos(1)) . 015 10.0 points The graph of f (x, y) = 6xy keywords: 014 10.0 points over the bounded region A in the first quadrant enclosed by Evaluate the double integral I = 4x cos(y) dxdy D y = 9 − x2 and the x, y-axes is the surface when D is the bounded region enclosed by the graphs of y = 0, y = x2 , x = 1. 1. I = 2 (1 − sin(1)) 2. I = 4 (1 − sin(1)) 3. I = 4 (sin(1) − 1) 4. I = 4 (cos(1) − 1) 5. I = 2 (sin(1) − 1) 6. I = 2 (1 − cos(1)) correct Find the volume of the solid under this graph over the region A. 1. Volume = 81 cu. units 2 liang (xl5432) – HW13 – berg – (55120) 243 cu. units 8 4. Volume = 3. I = 243 cu. units correct 4 3. Volume = 2. I = 243 cu. units 2 2. Volume = 4. I = 1 0 1 2y 2 1 2 f (x, y) dx dy which in turn can be written as the repeated integral 6xy dy dx. Now the inner integral is equal to 3xy √ 9−x2 0 1 2 6. I = f (x, y) dx dy 0 y (x, y) : x/2 ≤ y ≤ 1 , 0 ≤ x ≤ 2 in the plane bounded by the y-axis and the graphs of x y = 1. y = , 2 This is the shaded region in 9−x2 0 2 0 Explanation: The region of integration is the set of all points A 0 y/2 5. I = f (x, y) dxdy, √ y f (x, y) dx dy 0 Explanation: The volume of the solid under the graph of f is given by the double integral 3 2 f (x, y) dx dy 0 0 V = 2y f (x, y) dx dy correct 0 243 cu. units 16 5. Volume = 8 = 3x(9 − x2 ). y Thus 1 3 V =3 0 3 x(9 − x2 ) dx = − (9 − x2 )2 4 3 0 . x Consequently, 2 243 cu. units . Volume = 4 016 10.0 points Reverse the order of integration in the integral 2 y 1 I = f (x, y) dy dx , 0 Integration is taken first with respect to y for fixed x along the dashed vertical line. To change the order of integration, now fix y and let x vary along the solid horizontal line in x/2 1 but make no attempt to evaluate either integral. 2 1 1. I = f (x, y) dx dy 0 y/2 x 2 liang (xl5432) – HW13 – berg – (55120) Since the equation of the slant line is x = 2y, integration in x is along the line from (0, y) to (2y, y) for fixed y, and then from y = 0 to y = 1. Consequently, after changing the order of integration, 1 f (x, y) dx dy √ 3 2 . 0 3 keywords: double integral, reverse order integration, linear function, 017 1 2y I = 0 10.0 points Reverse the order of integration in the integral I = f (x, y) dx dy 3/2 018 3/2 10.0 points Reverse the order of integration in the double integral f (x, y) dx dy I = f (x, y) dx dy , 0 9y 2 4y 2 1 f (x, y) dx dy cor3/2 y2 4 2 0 √ 0 but make no attempt to evaluate either integral. 3 √ 16 rect f (x, y) dy dx 0 f (x, y) dx dy x2 2. I = f (x, y) dy dx 0 0 3 0 4 16 4x2 3/2 3. I = 16 2y 2 1 4. I = √ f (x, y) dx dy 3/2 1 5. I = x 1. I = √ √ 4 3. I = f (x, y) dy dx 0 3 x2 16 4. I = 3 2y 2 Explanation: The region of integration is similar to the one in the figure 4 f (x, y) dy dx correct √ 0 f (x, y) dx dy 3/2 . 3 f (x, y) dy dx, 1. I = 2. I = √ x/2 but make no attempt to evaluate either integral. √ 4y 2 1 1 √ 3 4 This shaded region is enclosed by the graphs of 4y 2 = x, y = 1 , and x = 3. To change the order of integration, first fix y. Then, as the graph shows, x varies from 3 to 4y 2 . To cover the region√ integration, therefore, of y must vary from 3/2 to 1. Hence, after changing the order of integration, I = 4 9 x √ 16 5. I = x f (x, y) dy dx 0 0 4 6. I = 0 16 √ f (x, y) dy dx x liang (xl5432) – HW13 – berg – (55120) Explanation: The region of integration is the set of all points (x, y) : y 2 ≤ x ≤ 16 , 0 ≤ y ≤ 4 in the plane bounded by the y-axis and the graphs of √ y = x, y = 4. This is the shaded region in y 4 x 16 (not drawn to scale). Integration is taken first with respect to x for fixed y along the solid horizontal line. To change the order of integration, first fix x in the interval [0, 16] and let y vary along the vertical dashed line. Then y varies from √ 0 to x. Hence, after changing the order of integration, 16 I = 0 4 √ f (x, y) dy dx . x keywords: double integral, reverse order integration, parabola, 10 ...
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