hw5sol - Math 241 Solutions 5 5.1#6. 9 1 1 e √ ln x dxdy...

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Unformatted text preview: Math 241 Solutions 5 5.1#6. 9 1 1 e √ ln x dxdy = xy 9 1 1 (ln x)2 2y 2 e 9 dy = 1 1 1 dy = ln y/4|9 = ln 9/4. 1 4y 5.1#9. 1 2 −1 0 2x2 +y 4 sin(πx) 1 2 Volume = 0 1dzdydx = 0 1 −1 2x2 + y 4 sin(πx)dydx 2x2 y + y 5 sin(πx)/5 0 1 2 −1 = = 0 dx 6x2 + 33 sin(πx)/5dx 33 cos(πx) 5π 1 = 2x3 − 5.2#13. y 4 3 0,2 = 2 + 66/5π. 0 2 1,1 1 -1 -1 3, 1 -2 1 2 3 4 x -2 2 −1 2−y y 2 −2y 2 (x + y )dxdy = −1 2 x2 /2 + xy 2−y y 2 −2y dy = −1 2 (2 − y )2 /2 + (2 − y )y − (y 2 − 2y )2 /2 − (y 2 − 2y )ydy −y 4 /2 + y 3 − y 2 /2 + 2dy 2 −1 = −1 = −y 5 /10 + y 4 /4 − y 3 /6 + 2y 1 = 99/20. 5.2#22. Since there are two regions bounded by the given equations, here we give both solutions. y 2.5 -5 -4 -3 -2 -1 -2.5 -5 -7.5 2 5, 4 2 5 2 5, 4 2 5 1 2 x -10 -12.5 Area(smaller) = 0 √ −2+ 5 1−2x−x2 2x 1dydx = 0 √ −2+ 5 1 − 4x − x2 dx 3 √ −2+ 5 0 = x − 2x − x /3 0 2 √ 10 5 − 22 . = 3 Area(larger) = 1−2x−x2 √ 2x 0 1dydx = −2− 5 −2− 5 √ 1 − 4x − x2 dx 3 0 √ −2− 5 = x − 2x − x /3 2 √ 10 5 + 22 = . 3 5.2#23. Note that the equation of the ellipse is (x/a)2 + (y/b)2 = 1. √ √ 2 2 a b Area = −a −b √ 1−(x/a) a b 1dydx = 4 0 0 1−(x/a) a 1dydx = 4 0 b 1−(x/a)2 1 − (x/a)2 dydx. Set x/a = sin θ, then dx = a cos θdθ. Note θ = 0 when x = 0 and θ = π/2 when x = a. For 0 ≤ θ ≤ π/2, cos θ ≥ 0, so 1 − (x/a)2 = | cos θ| = cos θ. Now we have π /2 π /2 Area = 4b 0 cos θ(a cos θ)dθ =4ab 0 π /2 cos2 θdθ cos 2θ + 1 dθ = 4ab 2 sin 2θ θ + 4 2 π /2 =4ab 0 = πab. 0 2 5.3#4. 2 0 0 4−y 2 2 xdxdy = 0 2 x2 /2 4−y 2 0 dy = 0 (4 − y 2 )2 /2dy 12 (16 − 8y 2 + y 4 )dy = 20 1 = (16y − 8y 3 /3 + y 5 /5) 2 Here is the region. y 2 0,2 2 = 128/15. 0 1.5 1 0.5 4,0 1 2 3 4 x √ Note x = 4 − y 2 ⇒ y = ± 4 − x. Since y here is non-negative, if we reverse the order, then we get that 4 0 0 √ 4−x 4 xdydx = 0 xy |0 4−x dx = √ 4 0 √ x 4 − xdx. Let 4 − x = u2 , then u = 2 when x = 0 and u = 0 when x = 4. Note x = 4 − u2 and √ dx = −2udu, so x 4 − xdx = (4 − u2 )|u|(−2u)du = (2u4 − 8u2 )du. (Note 0 ≤ u ≤ 2). Hence the integral equals 0 2 (2u4 − 8u2 )du = 2u5 /5 − 8u3 /3 0 2 = 128/15. 3 5.3#10. y 3 2,2 2 1 x -2 -1.5 -1 -0.5 -1 0.5 1 1.5 1, 1 -2 0, 2 1 −2 −x x2 −2 1 (x − y )dydx = = −2 1 −2 1 xy − y 2 /2 −x x2 −2 dx −x2 − (−x)2 /2 − x(x2 − 2) + (x2 − 2)2 /2dx x4 /2 − x3 − 7x2 /2 + 2x + 2dx 1 −2 = −2 = x5 /10 − x4 /4 − 7x3 /6 + x2 + 2x = −9/20. If we reverse the order, we get two regions. One is above y = −1 and the other is below y = −1. Hence −1 −2 −1 −2 −1 −2 −1 −2 √ y +2 2 √ − y +2 2 (x − y )dxdy + √ y +2 √ − y +2 −y √ − y +2 −1 2 −1 (x − y )dxdy −y √ − y +2 = = = x /2 − xy dy + x2 /2 − xy dy 2 (y + 2)/2 − y −2y −1 −2 −1 −2 y + 2 − (y + 2)/2 − y 2 −1 y + 2dy + −1 y 2 /2 + y 2 − (y + 2)/2 − y y + 2dy y + 2dy + 3y 2 /2 − y/2 − 1 − y 2 −1 y + 2dy 2 =−2 =−2 y y y + 2dy + (y 3 /2 − y 2 /4 − y ) 2 − y −1 y + 2dy y + 2dy + 3/4 − y −1 y + 2dy 4 Let y + 2 = u2 , then u = 0 when y √ −2, u = 1 when y = −1, and u = 2 when y = 2. = 2 Note y = u − 2 and dy = 2udu, so y y + 2dy = (u2 − 2)|u|(2u)du = (2u4 − 4u2 )du. (Note 0 ≤ u). Hence the integral equals 1 2 −2 0 (2u4 − 4u2 )du − 3/2 − 5 3 1 0 1 (2u4 − 4u2 )du 2 1 =28/15 + 3/4 − 46/15 = −9/20. = − 2 (2u /5 − 4u /3) + 3/4 − (2u5 /5 − 4u3 /3) 5 ...
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This note was uploaded on 09/02/2010 for the course MATH 241 taught by Professor Any during the Spring '08 term at University of Illinois at Urbana–Champaign.

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