ch08 - 8.1 SOLUTIONS 523 CHAPTER EIGHT Solutions for...

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8.1 SOLUTIONS 523 CHAPTER EIGHT Solutions for Section 8.1 Exercises 1. Each strip is a rectangle of length 3 and width Δ x , so Area of strip = 3Δ x, so Area of region = ± 5 0 3 dx = 3 x ² ² ² ² 5 0 = 15 . Check: This area can also be computed using Length × Width = 5 · 3 = 15 . 2. Using similar triangles, the height, y , of the strip is given by y 3 = x 6 so y = x 2 . Thus, Area of strip y Δ x = x 2 Δ x, so Area of region = ± 6 0 x 2 dx = x 2 4 ² ² ² ² 6 0 = 9 . Check: This area can also be computed using the formula 1 2 Base · Height = 1 2 · 6 · 3 = 9 . 3. By similar triangles, if w is the length of the strip at height h , we have w 3 = 5 - h 5 so w = 3 ³ 1 - h 5 ´ . Thus, Area of strip w Δ h = 3 ³ 1 - h 5 ´ Δ h. Area of region = ± 5 0 3 ³ 1 - h 5 ´ dh = µ 3 h - 3 h 2 10 ¶ ² ² ² ² 5 0 = 15 2 . Check: This area can also be computed using the formula 1 2 Base · Height = 1 2 · 3 · 5 = 15 2 . 4. Suppose the length of the strip shown is w . Then the Pythagorean theorem gives h 2 + ³ w 2 ´ 2 = 3 2 so w = 2 · 3 2 - h 2 . Thus Area of strip w Δ h = 2 · 3 2 - h 2 Δ h, Area of region = ± 3 - 3 2 · 3 2 - h 2 dh. Using VI-30 in the Table of Integrals, we have Area = ³ h · 3 2 - h 2 + 3 2 arcsin ³ h 3 ´ ² ² ² 3 - 3 = 9(arcsin 1 - arcsin( - 1)) = 9 π. Check: This area can also be computed using the formula πr 2 = 9 π .
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524 Chapter Eight /SOLUTIONS 5. The strip has width Δ y , so the variable of integration is y . The length of the strip is x . Since x 2 + y 2 = 10 and the region is in the Frst quadrant, solving for x gives x = ± 10 - y 2 . Thus Area of strip x Δ y = ± 10 - y 2 dy. The region stretches from y = 0 to y = 10 , so Area of region = ² 10 0 ± 10 - y 2 dy. Evaluating using VI-30 from the Table of Integrals, we have Area = 1 2 ³ y ± 10 - y 2 + 10 arcsin ³ y 10 ´ µ µ µ µ 10 0 = 5(arcsin 1 - arcsin 0) = 5 2 π. Check: This area can also be computed using the formula 1 4 πr 2 = 1 4 π ( 10) 2 = 5 2 π . 6. The strip has width Δ y , so the variable of integration is y . The length of the strip is 2 x for x 0 . ±or positive x , we have x = y . Thus, Area of strip 2 x Δ y = 2 y Δ y. Since the region extends from y = 0 to y = 4 , Area of region = ² 4 0 2 y dy = y 2 µ µ µ µ 4 0 = 16 . Check: The area of the region can be computed by 1 2 Base · Height = 1 2 · 8 · 4 = 16 . 7. The width of the strip is Δ y , so the variable of integration is y . Since the graphs are x = y and x = y 2 , the length of the strip is y - y 2 , and Area of strip ( y - y 2 y. The curves cross at the points (0 , 0) and (1 , 1) , so Area of region = ² 1 0 ( y - y 2 ) dy = y 2 2 - y 3 3 µ µ µ µ 1 0 = 1 6 . 8. The width of the strip is Δ x , so the variable of integration is x . The line has equation y = 6 - 3 x . The length of the strip is 6 - 3 x - ( x 2 - 4) = 10 - 3 x - x 2 . (Since x 2 - 4 is negative where the graph is below the x -axis, subtracting x 2 - 4 there adds the length below the x -axis.) Thus Area of strip (10 - 3 x - x 2 x.
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This note was uploaded on 04/05/2010 for the course MATH 115 taught by Professor Blakelock during the Fall '08 term at University of Michigan.

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ch08 - 8.1 SOLUTIONS 523 CHAPTER EIGHT Solutions for...

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