Ex.%2051-53%20in%2016.2 - 886 C H A P T E R 16 M U LTI P L...

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886 C H A P T E R 16 MULTIPLE INTEGRATION (ET CHAPTER 15) The double integral of f over D 2 is D 2 x y 2 d A = 5 3 x + 8 x + 1 2 x y 2 dy dx = 5 3 x y x + 8 y = x + 1 2 dx = 5 3 x x + 8 + x 1 2 x + 1 2 dx = 5 3 x x + 8 dx + 2 5 3 x x + 1 dx = 5 3 ( x + 8 ) 8 x + 8 dx + 2 5 3 ( x + 1 ) 1 x + 1 dx = 5 3 1 8 x + 8 dx + 2 5 3 1 1 x + 1 dx = 5 3 dx 8 5 3 dx 8 x + 2 5 3 dx 2 5 3 dx x + 1 = 2 + 8 ln ( 8 x ) 5 3 + 4 2 ln ( x + 1 ) 5 3 = 6 + 8 ( ln 3 ln 5 ) 2 ( ln 6 ln 4 ) = 6 + 8 ln 3 8 ln 5 2 ln 6 + 2 ln 4 = 6 + 6 ln 3 8 ln 5 + 2 ln 2 (3) Finally we substitute (2) and (3) in (1) to obtain the following solution: D x y 2 d A = 3 1 4 ln 5 2 ln 2 + 6 + 6 ln 3 8 ln 5 + 2 ln 2 = 9 8 . 25 ln 5 + 6 ln 3 2 . 314 51. Calculate the double integral f ( x , y ) = sin y y over the region D in Figure 23. D y y = x 2 1 x x 2 y = FIGURE 23 SOLUTION To describe D as a horizontally simple region, we rewrite the equations of the lines with x as a function of y , that is, x = y and x = 2 y . The inequalities for D are 1 y 2 , y x 2 y x y D x = y x = 2 y 1 2 We now compute the double integral of f ( x , y ) = sin y y over D by the following iterated integral: D sin y y d A = 2 1 2 y y sin y y dx dy = 2 1 sin y y x 2 y x = y dy = 2 1 sin y y ( 2 y y ) dy = 2 1 sin y y · y dy = 2 1 sin y dy = − cos y 2
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  • Spring '08
  • Rogawski
  • D1, dx

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