16.2.Ex60-64

16.2.Ex60-64 - S E C T I O N 16.2 y Double Integrals over...

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SECTION 16.2 Double Integrals over More General Regions (ET Section 15.2) 893 x D y y = x 2 1 Therefore, ZZ D ydA = Z 1 0 Z x 2 0 ydydx = Z 1 0 1 2 y 2 ¯ ¯ ¯ ¯ x 2 y = 0 dx = Z 1 0 1 2 ³ x 4 0 ´ = Z 1 0 1 2 x 4 = x 5 10 ¯ ¯ ¯ ¯ 1 0 = 1 10 (2) We compute the area of D using the formula for the area as a double integral: Area ( D ) = D 1 dA = Z 1 0 Z x 2 0 dydx = Z 1 0 y ¯ ¯ ¯ ¯ x 2 y = 0 = Z 1 0 x 2 = x 3 3 ¯ ¯ ¯ ¯ 1 0 = 1 3 (3) Substituting (2) and (3) in (1) we obtain H = 1 1 3 · 1 10 = 3 10 60. Find the average height of the “ceiling” in Figure 26 de±ned by z = y 2 sin x for 0 x π ,0 y 1. z y x 1 FIGURE 26 SOLUTION y x 1 0 D The average height is H = 1 Area ( D ) D y 2 sin xdA = 1 · 1 Z 1 0 Z 0 y 2 sin xdxdy = 1 Z 1 0 y 2 ( cos x ) ¯ ¯ ¯ ¯ x = 0 dy = 1 Z 1 0 y 2 ( cos + cos 0 ) = 1 Z 1 0 2 y 2 = 1 · 2 3 y 3 ¯ ¯ ¯ ¯ 1 0 = 2 3 61. Calculate the average value of the x -coordinate of a point on the semicircle x 2 + y 2 R 2 , x 0. What is the average value of the y -coordinate? The average value of the x -coordinates of a point on the semicircle D is x = 1 Area ( D ) D
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894 CHAPTER 16 MULTIPLE INTEGRATION (ET CHAPTER 15) x y The area of the semicircle is π R 2 2 . To compute the double integral, we identify the inequalities deFning D as a horizontally simple region: R y R , 0 x q R 2 y 2 x R R R y 0 x R 2 y 2 Therefore, x = 1 R 2 2 Z R R Z R 2 y 2 0 xdxdy = 2 R 2 Z R R x 2 2 ¯ ¯ ¯ ¯ R 2 y 2 x = 0 dy =
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16.2.Ex60-64 - S E C T I O N 16.2 y Double Integrals over...

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