MATH 2011 Tutorial Problem Solutions-2.pdf - Tutorial...

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Tutorial Problem Solutions Problem 98. Problem 99. Note that ˆ 1 0 ˆ 3 1 x 3 y 2 dy dx = ˆ 1 0 1 3 x 3 y 3 3 1 dx = ˆ 1 0 9 x 3 - 1 3 x 3 dx = 9 4 x 4 - 1 12 x 4 2 0 = 104 3 . Note that ˆ 3 1 ˆ 3 2 ( x 2 - 2 xy + 2 y 2 ) dy dx = ˆ 3 1 x 2 y - xy 2 + 2 3 y 3 3 2 dx = ˆ 3 1 ( 3 x 2 - 9 x + 18 ) - 2 x 2 - 4 x + 16 3 dx = ˆ 3 1 x 2 - 5 x + 38 3 dx = 1 3 x 3 - 5 2 x 2 + 38 3 x 3 1 = 14 . Problem 100. For part (a) the area is ˆ 1 0 ˆ x 2 x 3 dy dx = 1 12 . The region of integration is the following. 1
2 - 0 . 5 0 . 5 1 1 . 5 - 0 . 5 0 . 5 1 1 . 5 x y Note that the curve y = x 3 is below the curve y = x 2 on the interval 0 x 1. Problem 101. For part (a) we write ¨ Ω xy dA = ˆ 2 a 0 ˆ 1 4 a x 2 0 xy dy dx = ˆ 2 a 0 1 2 xy 2 1 4 a x 2 0 dx = ˆ 2 a 0 1 2 x 1 4 a x 2 2 dx = 1 32 a 2 ˆ 2 a 0 x 5 dx = 1 32 a 2 1 6 x 6 2 a 0 = 1 32 a 2 · 64 a 6 6 = a 4 3 .
3 For part (b) the region of integration is a quarter circle of radius 1 in the first quadrant. So, we write ¨ Ω x 2 y + y 3 dA = ˆ 1 0 ˆ 1 - x 2 0 x 2 y + y 3 dy dx = ˆ 1 0 1 2 x 2 y 2 + 1 4 y 4 1 - x 2 0 dx = ˆ 1 0 1 2 x 2 (1 - x 2 ) + 1 4 (1 - x 2 ) 2 dx = 1 5 . . Problem 102. For part (a) we have ˆ 1 0 ˆ 1 y 2 2 xe x 2 dx dy = ˆ 1 0 ˆ x 0 2 xe x 2 dy dx = ˆ 1 0 h 2 xe x 2 y i x 0 dx = ˆ 1 0 2 xe x 2 dx = h e x 2 i 1 0 = e - 1 . For part (b) we have ˆ 1 0 ˆ 1 y e x 2 dx dy = ˆ 1 0 ˆ x 0 e x 2 dy dx = ˆ 1 0 h ye x 2 i x 0 = ˆ 1 0 xe x 2 dx = 1 2 e x 2 1 0 = 1 2 ( e - 1) . For part (c) we have ˆ 1 0 ˆ 1 y sin ( x 2 ) dx dy = ˆ 1 0 ˆ x 0 sin ( x 2 ) dy dx = ˆ 1 0 x sin ( x 2 ) dx = - 1 2 cos ( x 2 ) 1 0 = 1 2 (1 - cos 1) .
4 Finally, for part (d) we have ˆ 1 - 1 ˆ 2 - x 2 x 2 dy dx = ˆ 1 0 ˆ y - y dx dy + ˆ 2 1 ˆ 2 - y - 2 - y dx dy = 8 3 . Note that part (d) requires somewhat more work than parts (a)–(c). Here is a drawing of the region of integration. - 1 . 5 - 1 - 0 . 5 0 . 5 1 1 . 5 - 1 1 2 x y For the lower boundary we originally have y = x 2 so this leads to x = y or x = - y ; note that x = y forms the lower right boundary, whereas x = - y forms the lower left boundary. And for the upper boundary we originally have y = 2 - x 2 so this leads to x = 2 - y or x = - 2 - y ; note that x = 2 - y forms the upper right boundary, whereas x = - 2 - y forms the upper left boundary. To be honest, there really isn’t any mathematical reason to change the order of integration here; in fact, this new order is much more technical. Problem 103. We get ˆ 1 0 ˆ 1 x ( 1 - y 2 ) - 1 2 dy dx = ˆ 1 0 ˆ y 0 ( 1 - y 2 ) - 1 2 dx dy = lim a 1 - ˆ a 0 y p 1 - y 2 dy = lim a 1 - h - ( 1 - y 2 ) 1 2 i a 0 = 1 . Note that the integral was improper, and so, we had to rewrite it in terms of limits. Problem 104.
5 Problem 105. Problem 106. For part (a), notice that the region of integration is the right half of a circle of radius 2 centered at (0,0). We know this because the limits of integration show 1 us that - 4 - x 2 y 4 - x 2 and 0 x 2. In polar this is described by 0 r 2 and - π 2 θ π 2 . So, ˆ 2 0 ˆ 4 - x 2 - 4 - x 2 x 2 y 2 dy dx = ˆ 2 0 ˆ π 2 - π 2 ( r cos θ ) 2 ( r sin θ ) 2 · r dθ dr = ˆ 2 0 r 5 1 32 (4 θ - sin (4 θ )) π 2 - π 2 | {z } = π 8 = π 8 ˆ 2 0 r 5 dr = π 8 1 6 r 6 2 0 = π 8 · 64 6 = 4 π 3 . For part (b), to determine the region of integration note that from the x limits of integration - p 2 y - y 2 x p 2 y - y 2 . Consider the boundary of the region – i.e., when x = p 2 y - y 2 . Square both sides to get x 2 = 2 y - y 2 or, equivalently, x 2 + y 2 - 2 y = 0 .

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