hw8sol - Math 53 Homework 8 – Solutions 1 2 1 3 x(x x2 +...

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Unformatted text preview: Math 53 Homework 8 – Solutions 1 2 1 3 x(x x2 + y 2 dy = xy 15.2 # 12: 0 1 1 xy So 0 x2 + y 2 dy dx = 0 1 1 3 0 R 3 y3 x x2 + 1 3 Inner: 1 18 Outer: 0 15.2 # 21: = −3 3 1 xy 2 dA = x2 + 1 −3 0 x x2 + 1 dA = √ 1 4 2−2 5/2 = . (2 − 1) − (1 − 0) = 15 15 1 0 xy 2 dy dx. x2 + 1 27 − (−27) 3 x 1 dx = 18 ln(x2 + 1) 2+1 x 2 x2 y R xye 1 = 1 x(x2 + 1)3/2 − 3 x4 . 3 0 x(x2 + 1)3/2 − x4 dx = 1 = (x2 + 1)5/2 − x5 15 15.2 # 17: 1 + y 2 )3/2 21 x2 y 0 0 xye = 18 x . x2 + 1 1 = 9 ln 2. 0 x=1 2 1 x2 y dy = 1 0 (ey − 1) dy 2e 2 x=0 1 1 y ]2 = 2 ((e2 − 2) − (e0 − 0)) = 2 (e2 − 0 2 0 dx dy = = 1 [ey − 2 (Note: integrating in the other order is quite a bit harder.) 1x 0 x2 (1 15.3 # 3: 1 0 [y + 2y ) dy dx = 1 0 (x 1v 00 15.3 # 8: D √ 1 0 1 − v 2 du dv = y dA = 5+1 x x2 0 y dy dx = 5+1 x 1 x2 y 2 /2 x5 + 1 y (0,2) x=2−y 2 (3,2) (1,1) 0 1 x = 2y − 1 2y −1 Outer: − 1 0 0 1 10 ln(x5 1 5 = 1 0 3 10 . = 1. 3 1 x4 dx 2 x5 + 1 + 1) 1 0 = 1 10 ln 2. 2−y 1 0 x 0 2y −1 3 3 2−y y dx = y x 2 4 3 1 3y − 3y dy = 1 2 y 3 dx dy 2 (or also: Inner: = dx = = 15.3 # 15: 1 0 1 = − 3 (1 − v 2 )3/2 1 0 + x2 ) − (x2 + x4 )) dx 1 − x4 ) dx = 1 x2 − 5 x5 2 √ u= v 1√ u 1 − v 2 u=0 dv = 0 v 1 − v 2 dv = 15.3 # 6: 1 0 ((x + y 2 ]y=x2 dx = y =x 3). 3 y 3 dy dx + 1 2−x x=2y −1 = y 3 ((2y − 1) − (2 − y )) = 3y 4 x=2−y 35 3 42 3 3 147 96 5 y − 4 y 1 = 5 − 12 − 5 + 4 = 20 . 2 x+1 2 y 3 dy dx) − 3y 3 . 15.3 # 23: The plane 3x + 2y + z = 6 (or z = 6 − 3x − 2y ) intersects the xy -plane (z = 0) along the line 3x +2y = 6 (with x-intercept 2 and y -intercept 3), or y = 3 − 3 x. So the region 2 3 of integration is the triangle with vertices (0,0), (2,0) and (0,3): x ≥ 0, 0 ≤ y ≤ 3 − 2 x, and the integrand is 6 − 3x − 2y . 3 3− 2 x 2 0 0 2 (6 − 3x − 2y ) dy dx = 0 6y − 3xy − y 2 = 1 2 0 (3 3 3− 2 x 0 2 dx = 0 3 3− 2 x [(6 − 3x − y )y ]0 3 − 3 x)2 dx = − 2 (3 − 2 x)3 2 9 2 0 = 2 9 dx = 33 = 6. 15.3 # 39: y √ x y= 2 Region of integration: 0 ≤ y ≤ √ x, 0 ≤ x ≤ 4 or equivalently: y 2 ≤ x ≤ 4, 0 ≤ y ≤ 2 0 15.3 # 47: y 4x y=0 √ x 4 f (x, y ) dy dx 00 So: y=2 2 √ y= x 2 4 4x So √ 0 Inner: y3 x +1 = 0 2 y2 y3 +1 . Outer: 0 y 3 15.3 # 58: D 0 y3 2 +1 x 1 3 dy = 1 dy dx = y3 + 1 ln(y 3 + 1) 2 0 3 2y 1 0 2 (2,1) f (x, y ) dA = = x = 2y 0 x ≤ y ≤ 2, 0 ≤ x ≤ 4 0 3−x y2 2 0 0 1 dx dy . y3 + 1 1 ln 9. 3 f (x, y ) dx dy f (x, y ) dy dx. x/2 0 D = 3−y f (x, y ) dx dy + x=3−y π /2 y2 1 1 f (x, y ) dx dy . or equivalently: 0 ≤ y 2 ≤ x, 0 ≤ y ≤ 2 0 y2 √ Region of integration: 24 0 y2 = x 4 cos θ r dr dθ represents the area of the region 0 ≤ r ≤ 4 cos θ, 0 ≤ θ ≤ π/2. 15.4 # 6: 0 0 Since r = 4 cos θ is the polar equation of the circle of radius 2 with center at (2, 0), R is the upper half of the disk enclosed by this circle; and the area is 2π . Evaluating the integral: Inner: Outer: 1 2 4 cos θ = 8 cos2 θ. 2r 0 π /2 π /2 8 cos2 θ dθ = 0 4(1 0 Outer: D xy dA = 15.4 # 17: By symmetry, A = 2 Inner: Outer: = 2π . 2π 3 0 (r cos θ )(r sin θ ) r dr dθ . 0 33 34 143 0 r sin θ cos θ dr = sin θ cos θ 4 r 0 = 4 sin θ cos θ . 2π 34 1 34 2π 2 4 0 sin θ cos θ dθ = 4 2 sin θ 0 = 0. (This was obvious 15.4 # 7: Inner: π/2 + cos 2θ) dθ = [4θ + 2 sin 2θ]0 π /4 sin θ 0 0 by symmetry...) y r dr dθ. 1 1 2 sin θ = 2 sin2 θ. 2r 0 π /4 π /4 1 sin2 θ dθ = 0 2 (1 − cos 2θ) dθ = 0 π /4 1 = 2 θ − 1 sin 2θ 0 = 1 ( π − 1 sin π ) 2 24 2 2 = r = sin θ θ= π 4 π 8 0 − 1. 4 x r = cos θ 15.4 # 25: The cone z = x2 + y 2 intersects the sphere x2 + y 2 + z 2 = 1 when x2 + y 2 + 1 ( x2 + y 2 )2 = 1, or x2 + y 2 = 2 . So √ 1/ 2 2π V= 1 − x2 − y 2 − x2 +y 2 ≤1/2 √ 1/ 2 √ = 2π 0 (r 1 √ = π (2 − 2). 3 x2 + y 2 dA = 0 0 1 − r2 − r2 ) dr = 2π − 1 (1 − r2 )3/2 − 3 r3 3 2 1 − r2 − r r dr dθ = √ 1/ 2 0 1 = 2π (− 1 )( 2√2 + 3 1 √ 22 − 1) √ 2−y 2 1 (x + y ) dx dy : the region of integration is the portion of the disk 15.4 # 31: 0 y 2 + y 2 ≤ 2 where x ≥ y and y ≥ 0, i.e. between the x-axis and the line y = x. x √ √ π /4 2−y 2 2 1 (x + y ) dx dy = 0 So 0 y 0 (r cos θ + r sin θ ) r dr dθ . Inner: Outer: √ √ 2 13 r (cos θ + sin θ) 0 = 2 3 2 (cos θ + sin θ). 3 √ √ π/4 2 2 π /4 (cos θ + sin θ) dθ = 2 3 2 [sin θ − cos θ]0 3 0 15.4 # 36: a) Using polar coordinates: = 2π ∞ 0 re −r 2 −r 2 1 dr = 2π − 2 e ∞ 0 R2 e−(x 2 +y 2 ) = √ 22 3 (0 dA = − (−1)) = 2π ∞ −r 2 0e 0 √ 22 3. r dr dθ = = 2π (0 + 1 ) = π . 2 Or, to be more rigorous, we integrate over the disk Da and take the limit as a → ∞: Da e−(x 2 +y 2 ) 2π a −r 2 0e 0 dA = The result then follows, since lim π (1 − e−(x b) 2 +y 2 ) a→∞ dA = Sa r dr dθ = 2π − 1 e−r 2 2 e−a ) a −x 2 −y 2 a e dy dx −a −a e = 2 a 0 2 = π (1 − e−a ). = π. a −x 2 dx −a e a −y 2 dy −a e . Taking the limit as a → ∞ on both sides, we deduce: ∞ −y 2 ∞ −x 2 −(x2 +y 2 ) dA = dy . dx −∞ e −∞ e R2 e ∞ −y 2 ∞ −x 2 c) Since −∞ e dy = −∞ e dx (the name of the integration variable is irrelevant), the 2 ∞ −x 2 dx = π . Taking the square root (and observing that result of (b) becomes: −∞ e √ 2 2 ∞ e−x > 0 for all x so the integral is positive), we get: −∞ e−x dx = π . √ √ 2 2 2 ∞ ∞1 ∞ 1 d) Letting x = t/ 2, we have −∞ e−x dx = −∞ √2 e−t /2 dt. Hence, π = √2 −∞ e−t /2 dt. √ 2 ∞ Equivalently, −∞ e−t /2 dt = 2π . √ √ 1 1 1 1 x y= x 2 15.5 # 8: m = D ρ dA = 0 0 x dy dx = 0 [xy ]y=0 dx = 0 x3/2 dx = 2 x5/2 0 = 5 . 5 √ √ 1 1 x y= x 51 2 5 1 51 x = m D xρ dA = 5 0 0 x2 dy dx = 2 0 [x2 y ]y=0 dx = 2 0 x5/2 dx = 5 7 x7/2 0 = 7 . ¯ 2 2 √ √ 1 x y= x 51 1 5 11 5 51 5 y = m D yρ dA = 2 0 0 xy dy dx = 2 0 [ 1 xy 2 ]y=0 dx = 2 0 2 x2 dx = 2 1 x3 0 = 12 . ¯ 2 6 π= π /2 1 π π1 3 2 0 kr r dr dθ = 2 0 kr dr = 8 k . 0 r =1 1 π /2 1 1 π /2 1 2 5 x ρ dA = m 0 5 kr cos θ r =0 dθ = 0 (r cos θ ) (kr ) r dr dθ = m 0 π/2 π /2 k/5 cos θ dθ = kπ/8 [sin θ]0 = 58 . π 0 1 π /2 1 2 ) r dr dθ = 1 π /2 1 kr 5 sin θ r =1 dθ = y ρ dA = m 0 m0 5 0 (r sin θ ) (kr r =0 π/2 π /2 k/5 8 sin θ dθ = kπ/8 [− cos θ]0 = 5π . 0 15.5 # 12: ρ(x, y ) = k (x2 + y 2 ) = kr2 , m = x= ¯ = y= ¯ = 1 m 1k m5 1 m 1k m5 (Note: x = y by symmetry: the lamina is symmetric about the axis y = x, and so is the ¯ ¯ density, so the center of mass lies on the symmetry axis.) 1 π /2 1 π /2 1 2 2 2 6 2 6 kr sin θ 0 dθ = 0 (r sin θ )(kr ) r dr dθ = 0 0 π /2 π /2 π /2 1 θ 1 π = 1 k 0 sin2 θ dθ = 1 k 0 1 (1 − cos 2θ) dθ = 6 k 2 − 4 sin 2θ 0 = 24 k . 6 6 2 1 π /2 1 π /2 1 2 6 2 2 2 2 6 kr cos θ 0 dθ = 0 (r cos θ )(kr ) r dr dθ = 0 D x ρ dA = 0 π /2 π /2 π /2 π 1 1 θ 1 = 1 k 0 cos2 θ dθ = 6 k 0 1 (1 + cos 2θ) dθ = 6 k 2 + 4 sin 2θ 0 = 24 k . 6 2 π /2 1 2 π1 π 2 2 61 0 r (kr ) r dr dθ = 2 6 kr 0 = 12 k . D r ρ dA = 0 15.5 # 18: Ix = Iy = I0 = D y 2 ρ dA = 3 (Note: by symmetry, Iy = Ix ; and as a general fact, I0 = Ix + Iy ; so it was enough to compute one of the three moments of inertia). 15.5 # 28: a) f (x, y ) ≥ 0, so f is a joint density function if R2 f (x, y ) dA = 1. Here 11 f (x, y ) = 0 outside of the unit square, so we just need to compute 0 0 f (x, y ) dy dx = 1 1 11 2 y =1 21 0 0 4xy dy dx = 0 2xy y =0 dx = 0 2x dx = x 0 = 1. 1 b) (i) The region where x ≥ 2 corresponds to the right half of the unit square (recall that X and Y only take values between 0 and 1). 1 So P (X ≥ 2 ) = 1 1 1/2 0 4xy dy dx 1 (ii) P (X ≥ 2 , Y ≤ 1 ) = 2 1 1/2 2xy 2 1/2 1 4xy dy dx 1/2 0 c) E (X ) = R2 E (Y ) = y f (x, y ) dA = R2 = = y =1 y =0 dx = 1 1/2 2xy 2 1 1/2 2x dx 1/2 0 dx = = x2 1x 1/2 2 1 1/2 = 3. 4 dx = 1 1 11 2 21 2 0 0 x(4xy ) dy dx = 0 [2x y ]0 dx = 0 2x dx 14 14 11 2 31 0 0 y (4xy ) dy dx = 0 [ 3 xy ]0 dx = 0 3 x dx = 3 . x f (x, y ) dA = 1 x2 4 1/2 = 2. 3 = 3 16 . Problem 1: a 2π 1 1a2 1 1 r r dr dθ = r dA = 2π r3 = a. Area πa2 0 πa2 30 3 0 b) (Using the setup suggested by the hint, so the circle has polar equation r = 2a cos θ): 2a cos θ π /2 π /2 1 1 1 1 Average distance = r2 dr dθ = dA = (2a cos θ)3 dθ 2 2 Area r πa −π/2 0 πa −π/2 3 a) Average distance = = = π /2 8a3 3πa2 8a 3π 1 cos3 θ dθ = −π/2 −1 (1 − u2 ) du = 8a 3π π /2 −π/2 (1 − sin2 θ) cos θ dθ = (substituting u = sin θ) 8a 1 u − u3 3π 3 1 = −1 8a 2 2 32a ( − (− )) = . 3π 3 3 9π Problem 2: sin 2θ π /2 r dr dθ. a) Area = 0 Inner: 0 Outer: sin 2θ 1 12 = 2 sin2 2θ. 2r 0 π /2 1 1 2 2 (1 − cos 4θ ) dθ 0 1 = 4θ − 1 16 sin 4θ π /2 0 = π. 8 b) By symmetry the centroid must be on the diagonal line y = x, so calculating x is enough. ¯ sin 2θ π /2 1 r cos θ r dr dθ . x= ¯ Area 0 0 Inner: = Outer: sin 2θ 13 1 = 3 sin3 2θ cos θ = 1 (2 sin θ cos θ)3 cos θ = 8 sin3 θ cos4 θ 3 r cos θ 0 3 3 8 4 θ (1 − cos2 θ ) = 8 sin θ (cos4 θ − cos6 θ ). 3 sin θ cos 3 π /2 8 16 8 1 1 1 16 5 7 = 8 5 − 1 = 105 . Therefore x = y = ¯¯ 3 − 5 cos θ + 7 cos θ 0 3 7 π 105 4 = 128 . 105π ...
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