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wA8_soln

# wA8_soln - Math 237 Assignment 8 Solutions 1 Consider the...

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Unformatted text preview: Math 237 Assignment 8 Solutions 1. Consider the map F : R2 → R2 deﬁned by (u, v ) = F (x, y ) = (y + e−x , y − e−x ). a) Show that F has an inverse map by ﬁnding F −1 explicitly. Solution: We have u = y + e−x and v = y − e−x , hence 1 u + v = 2 y ⇒ y = (u + v ) 2 1 u − v = 2e−x ⇒ x = − ln | (u − v )| 2 1 1 Hence (x, y ) = F −1 (u, v ) = (− ln | 2 (u − v )|, 2 (u + v )). b) Find the derivative matrices DF (x, y ) and DF −1 (u, v ) and verify that DF (x, y )DF −1(u, v ) = I . Solution: We have DF (x, y ) = Then DF (x, y )DF −1 −e−x 1 , e−x 1 DF −1 (u, v ) = 1 − u−v 1 2 1 u−v 1 2 . −e−x 1 (u, v ) = e−x 1 1 − u−v 1 2 1 u−v 1 2 = since u − v = 2e−x . c) Verify that the Jacobians satisfy ∂ (x,y ) ∂ (u,v) e −x +1 u−v 2 −x 1 e − u−v + 2 e 1 − u−v + 2 10 = , −x e 1 01 +2 u−v −x = ∂ (u,v) ∂ (x,y ) −1 . 1 2 1 u−v Solution: We have ∂ (x, y ) 1 1 1 − = det DF −1 (u, v ) = − ∂ (u, v ) u−v 2 u−v ∂ (u, v ) = det F (x, y ) = −e−x (1) − e−x (1) = −2e−x ∂ (x, y ) Hence ∂ (u,v) ∂ (x,y ) −1 =− = 1 −2e−x 1 = − u−v = ∂ (x,y ) . ∂ (u,v) 2 2. Find the Jacobian for the following mappings. a) (x, y, z ) = T (r, θ, z ) = (r cos θ, r sin θ, z ). cos θ −r sin θ 0 ) Solution: ∂ (x,y,z) = det sin θ r cos θ 0 = (r cos2 θ + r sin2 θ) = r ∂ (r,θ,z 0 0 1 b) (x, y, z ) = T (ρ, θ, φ) = (ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ). Solution: cos θ sin φ −ρ sin θ sin φ ρ cos θ cos φ ∂ (x, y, z ) = det sin φ sin θ ρ sin φ cos θ ρ sin θ cos φ ∂ (ρ, θ, φ) cos φ 0 −ρ sin φ = cos φ[−ρ2 sin2 θ sin φ cos φ − ρ2 cos2 θ sin φ cos φ] − ρ sin φ[ρ cos2 θ sin2 φ + ρ sin2 θ sin2 φ] = −ρ2 sin φ cos2 φ − ρ2 sin3 φ = −ρ2 sin φ(cos2 φ + sin2 φ) = −ρ2 sin φ c) (u, v ) = T (x, y ) = (2x + y 3 , x2 y ). Solution: ∂ (u,v) (x,y ) = det 2 3y 2 = 2x2 − 6xy 3. 2xy x2 3. Evaluate the following integrals. a) D 1 + xy dA where D is the region bounded by x = y , y = 0 and x + y = 2. Solution: Sketching the region gives: Drawing rectangles horizontally we get y ≤x ≤ 2 − y 0 ≤y ≤ 1 Thus, we get 1 (2−y ) 1 1 + xy dA = D 0 1 y 1 + xy dx dy = 0 1 x + x2 y 2 (2−y ) dy y 1 0 = 0 1 1 2 − y + (2 − y )2 y − (y + (y )2 y ) dy = 2 2 1 0 2 − 2y 2 dy 2 = 2y − y 3 3 4 = 3 3 b) D cos x dA where D is the region bounded by y = x2 − 1, y = 1 − x2 . Solution: Sketching the region gives: Drawing rectangles vertically we get x2 − 1 ≤y ≤ 1 − x2 −1 ≤x ≤ 1 Thus, we get 1 1−x2 1 1−x2 cos x dA = D −1 1 x 2 −1 cos x dy dx = −1 2 y cos x x 2 −1 2 dx 1 =2 −1 cos x − x cos x dx = 2 sin x − x sin x + 2x cos x − 2 sin x −1 = 8 sin 1 − 8 cos 1 c) 1 0 1 x cos(y 2 ) dy dx. Solution: We observe that we need to change the order of integration. The region is x ≤ y ≤ 1, 0 ≤ x ≤ 1, thus we draw x = y and y = 1 for 0 ≤ x ≤ 1 to get the region below. We now want to integrate with respect to x ﬁrst, so we draw rectangles horizontally to get 0 ≤ x ≤ y and 0 ≤ y ≤ 1. Hence we have 1 1 1 y cos(y ) dy dx = 0 x 0 1 0 2 cos(y 2 ) dx dy y = 0 1 x cos(y ) 0 2 dy = 0 y cos(y 2 ) dy 1 0 1 sin(y 2 ) 2 1 = sin 1 2 = 4 d) R √ x x 2 +y 2 dA where R is the region outside x2 + y 2 = 1 and inside x2 + y 2 = 2x. Solution: Since we have circles we see we want to use polar coordinates. Putting the equations into polar coordinates we get x2 + y 2 = 1 ⇒ r = 1 and x2 + y 2 = 2x ⇒ r 2 = 2r cos θ ⇒ r = 2 cos θ. The points of intersection are when 1 = 2 cos θ ⇒ θ = ± π . Thus 3 skething the region gives: Drawing sectors we see that 1 ≤r ≤ 2 cos θ π π − ≤θ ≤ 3 3 Hence, by the change of variables theorem we get x R x2 + y 2 dA = Drθ π /3 r cos θ √ · |r | dr dθ = r2 12 r cos θ 2 2 cos θ π /3 −π/3 π /3 1 2 cos θ r cos θ dr dθ 2 cos3 θ − 1 cos θ dθ 2 π /3 −π/3 = −π/3 π /3 dθ = 1 2 −π/3 2 1 1 = 2 cos θ(1 − sin θ) − cos θ dθ = 2 sin θ − sin3 θ − sin θ 2 3 2 −π/3 √ =3 e) Dxy x2 dA, where Dxy is bounded by the ellipse 5x2 + 8xy + 4y 2 = 4. Solution: Completing the square we get x2 + (2x + 2y )2 = 4 so we let u = x and v = 2x + 2y . Since u2 + v 2 = 5x2 + 8xy + 4y 2 = 4 we have Duv is a circle of radius 2. Also 10 ∂ (x,y ) ∂ (u,v) 1 = 2, so ∂ (u,v) = 2 . Now, using polar coordinates we have u = r cos θ and = ∂ (x,y ) 22 v = r sin θ so u2 = r 2 cos2 θ and we get x2 dA = Dxy Duv u2 2π 0 1 1 du dv = 2 2 2 3 2 2π 0 0 2 r 2 cos2 θ · |r | dr dθ 2π 1 = 2 r cos θ dr dθ = 0 0 2 cos θ dθ = 2 2 θ1 + sin 2θ 24 2π = 2π 0 5 4. Let R be the triangle bounded by x + y = 1, x = 0 and y = 0 and let (u, v ) = F (x, y ) = (x + y, x − y ). a) Sketch the image of R under F in the uv -plane. Solution: We ﬁnd the image of each line. LINE 1: x = 0, 0 ≤ y ≤ 1. We have u = x + y = y and v = x − y = −y so v = −y = −u, and 0 ≤ y ≤ 1 ⇒ 0 ≤ u ≤ 1. So the image of LINE 1 under F is the line v = −u, 0 ≤ u ≤ 1. LINE 2: y = 0, 0 ≤ x ≤ 1. We have u = x + y = x and v = x − y = x so v = x = u, and 0 ≤ x ≤ 1 ⇒ 0 ≤ u ≤ 1. So the image of LINE 2 under F is the line v = u, 0 ≤ u ≤ 1. LINE 3: x + y = 1, 0 ≤ x ≤ 1. We have u = x + y = 1 and v = x − y = x − (1 − x) = 2x − 1, so 0 ≤ x ≤ 1 ⇒ −1 ≤ 2x − 1 ≤ 1 ⇒ −1 ≤ v ≤ 1. hence the image of LINE 3 under F is the line u = 1, −1 ≤ v ≤ 1. Hence, we get the image below: b) Find the Jacobian of F and show that it is never 0 on R. Solution: ∂ (u,v) (x,y ) = det 11 = −2 = 0. 1 −1 6 c) Find the mapping F −1 and the Jacobian for F −1 . Solution: We have u = x + y and v = x − y so u + v = 2x ⇒ x = 1 u − v = 2y ⇒ y = 2 (u − v ). Also, ∂ (x,y ) (u,v) 1 (u 2 + v ) and = ∂ (u,v) ∂ (x,y ) −1 1 = −2. d) Use the mapping F to evaluate R cos x−y x+y dx dy . Solution: Drawing rectangles vertically we get −u ≤ v ≤ u, 0 ≤ u ≤ 1. Thus, by the change of variables theorem we get cos R x−y x+y 1 u dx dy = 0 −u 1 0 1 0 cos 1 v dv du ·− u 2 u 1 = 2 = 1 2 v u sin u du −u u sin(1) − u sin(−1) du 1 = sin(1) 0 u du = sin(1) 2 ...
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