Solved exercises line integral
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This document contains exercises and solutions for line integrals from a chapter on the topic. It includes 6 exercises evaluating line integrals over various curves defined parametrically or through equations. It also contains exercises using Green's Theorem and evaluating line integrals for conservative vector fields. The solutions provide the parametrizations needed to set up and evaluate the line integrals.
Solved exercises line integral
- 1. Lebanese University - Faculty of Sciences Section º Chapter 3: Line integral Solved Problems Dr. Kamel ATTAR attar.kamel@gmail.com F 2020-2021 F
- 2. 2Ú19 Exercises Solutions 1 Exercises 2 Solutions Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 3. 3Ú19 Exercises Solutions . Exercise 1. Evaluate the following line integrals. a) Z C ydx + dy where C is the curve of equation x = t3 − t and y = t2 from the point (0, 0) to the point (6, 4). b) Z C (x + 2y)dx + (x − y)dy where C is the part of the ellipse of equation x = 2 cos(t) and y = 4 sin(t) for 0 ≤ t ≤ π 4 . c) Z C xydy where C is the portion of the ellipse of equation 4x2 + 9y2 = 36 lying in the first quadrant, oriented clockwise. d) Z C ydx − xdy where C is the portion of the curve y = 1 x from the point (1, 1) to the point (2, 1 2 ). e) Z C (ey + yex )dx + (xey + ex )dy where C is given parametrically x = sin πt 2 and y = ln(t) for 1 ≤ t ≤ 2. Go to Solution Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 4. 4Ú19 Exercises Solutions . Exercise 2. For each part below compute Z C − → F · d − → M a) C is the line segment from (2; 3) to (0; 3) and − → F = x~ i − y~ j b) C is the line segment from (5; 0; 2) to (5; 3; 4) and − → F = z~ i − y~ j + x~ k. c) C is the curve from y = ex from (2; e2 ) to (0; 1) and − → F = x2~ i − y~ j. d) C is the part of the circle of radius 3 in the first quadrant from (3; 0) to (0; 3) and − → F =~ i − y~ j e) C is the part of the curve x = cos(y) from (1; 2π) to (1; 0) and − → F = y~ i + 2x~ j. Go to Solution . Exercise 3. We consider the vector field − → F (x, y) = (x2 − y) ~ i + (y2 + x) ~ j . Evaluate Z AB − → F · d − → M along a) a straight line from (0, 1) to (1, 2), b) a straight line from (0, 1) to (1, 1) and then from (1, 1) to (1, 2), c) the parabola y = x2 + 1, Here A = (0, 1) and B = (1, 2). Go to Solution Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 5. 5Ú19 Exercises Solutions . Exercise 4. We consider the vector field − → F (x, y, z) = (3x2 − 6yz) ~ i + (2y + 3xz) ~ j + (1 − 4xyz2 )~ k . Evaluate Z C − → F · d − → M from (0, 0, 0) to (1, 1, 1) along the following paths C: a) x = t, y = t2 and z = t3 b) The straight lines from (0, 0, 0) to (0, 0, 1), then to (0, 1, 1), and then to (1, 1, 1), c) The straight line joining (0, 0, 0) and (1, 1, 1), Go to Solution . Exercise 5. Find the work done in moving a particle once around an ellipse C in the xy plane, if the ellipse has its center at the origin with equation x2 16 + y2 9 = 1 and if the force field is given by − → F (x, y, z) = (3x − 4y + 2z) ~ i + (4x + 2y − 3z2 ) ~ j + (2xz − 4y2 + z3 )~ k . Go to Solution Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 6. 6Ú19 Exercises Solutions . Exercise 6. Evaluate the following line integrals using Green’s Theorem. Unless otherwise stated, assume that all curves are oriented counterclockwise. a) I C 2xydx + y2 dy where C is the closed curve formed by y = x 2 and y = √ x. b) I C xydx + (x + y)dy where C is the triangle with vertices (0, 0), (2, 0) and (0, 1). c) I C (e3 x + 2y)dx + (x3 + sin y)dy where C is the rectangle with vertices (2, 1), (6, 1), (6, 4) and (2, 4). d) I C ln(1 + y)dx − xy 1 + y dy where C is the triangle with vertices (0, 0), (2, 0) and (0, 4). Go to Solution Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 7. 7Ú19 Exercises Solutions . Exercise 7. ¬ Let C be the curve joining the points A and B with parametrization ~ M(t) = 1 + (t − 1)et ~ i + 2t + cos πt 2 ~ j + sin πt 2 ~ k , 0 ≤ t ≤ 1 We consider the vector field − → F (x, y, z) = (2xy3 z4 + 2 cos x) ~ i + (sin y + 3x2 y2 z4 ) ~ j + (ez + 4x2 y3 z3 )~ k . a) Find the coordinates of A and B. b) Show that − → F is conservative (gradient field), then find a scalar potential f of − → F . c) Deduce the circulation of − → F along the curve C. Let Γ1 be the part of the parabola y = x2 − 3x from the point (0, 0) to the point (4, 4) and Γ2 be the line segment y = x from the point (4, 4) to the point (0, 0). We consider the vector field − → V = ex15 − 2y + cos(x3 ) ~ i + sin(y2018 ) + x2 − ln(y2 + 1) ~ j . a) Let Γ = Γ1 ∪ Γ2, draw the curve Γ. b) Using the formula of Green-Riemann, evaluate I Γ+ − → V · d − → M Go to Solution Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 8. 8Ú19 Exercises Solutions Solution 1. a) 308 15 b) − 9 2 − π c) 4 d) ln(4) e) −1 + ln 2 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 9. 9Ú19 Exercises Solutions Solution 2. a) Parameterization n x = −2t + 2 y = 3 , 0 ≤ t ≤ 1 then n dx = −2dt dy = 0 Integration Z C − → F · d − → M = Z (x~ i − y~ j) · (dx~ i + dy~ j) = Z 1 0 (4t − 4) dt = −2 b) Parameterization ( x = 5 y = 3t z = 2t + 2 , 0 ≤ t ≤ 1 then ( dx = 0 dy = 3dt dz = 2dt Integration Z C − → F · d − → M = Z (z~ i − y~ j + x~ k) · (dx~ i + dy~ j + dz~ k) = Z 1 0 (−9t + 10) dt = 11 2 c) Parameterization x = t y = et , 0 ≤ t ≤ 2 then dx = dt dy = et dt Integration Z C − → F · d − → M = Z (x2~ i − y~ j) · (dx~ i + dy~ j) = − Z 2 0 (t2 − e2t ) dt = − 19 6 + 1 2 e4 d) Parameterization x = 3 cos(t) y = 3 sin(t) , 0 ≤ t ≤ π 2 then dx = −3 sin(t)dt dy = 3 cos(t)dt Z C − → F · d − → M = Z ( ~ i − y~ j) · (dx~ i + dy~ j) = Z π 2 0 (−3 sin(t) − 9 sin(t) cos(t)) dt = − 15 2 e) Parameterization n x = cos(t) y = t , 0 ≤ t ≤ 2π then n dx = − sin(t)dt dy = dt Z C − → F · d − → M = Z (y~ i + 2x~ j) · (dx~ i + dy~ j) = − Z 2π 0 (−t sin(t) + 2 cos(t)) dt = −2π. Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 10. 10Ú19 Exercises Solutions Solution 3. a) An equation for the line joining (0, 1) and (1, 2) in the xy plane is y = x + 1, then the parametric equation is given by x = t and y = t + 1 and so dx = dt, dy = dt and the line integral Z − → F · d − → M = Z 1 t=0 (2t2 + 2t)dt = 5 3 b) Along the straight line from (0, 1) to (1, 1) x = t, y = 1, then dx = dt, dy = 0 and the line integral Z − → F · d − → M = Z 1 t=0 (t2 − 1)dt = − 2 3 Along the straight line from (1, 1) to (1, 2) x = 1, y = t, then dx = 0, dy = dt and the line integral Z − → F · d − → M = Z 2 t=1 (t2 + 1)dt = 10 3 Then the required value is 8/3 c) The parametric equation of the parabola is x = t and y = t2 + 1, with 0 ≤ t ≤ 1 the line integral Z − → F · d − → M = Z 1 t=0 (2t5 + 4t2 + 2t2 + 2t − 1)dt = 2 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 11. 11Ú19 Exercises Solutions Solution 4. Z C − → F · d − → M = Z C (3x 2 − 6yz)dx + (2y + 3xz)dy + (1 − 4xyz 2 )dz a) If x = t, y = t 2 and z = t 3 , then dx = dt, dy = 2tdt and dz = 3t 2 dt and points (0, 0, 0) and (1, 1, 1) correspond to t = 0 and t = 1 respectively. Then Z C − → F · d − → M = Z 1 0 (3t 2 − 6t 5 )dt + (4t 3 + 6t 5 )dt + (3t 2 − 12t 11 )dt = 2 b) • Along the straight line from (0, 0, 0) to (0, 0, 1) we have x = 0, y = 0 and z = t then dx = 0, dy = 0 and dz = dt with 0 ≤ t ≤ 1. Then the integral over this part of the path is Z 1 0 dt = 1. • Along the straight line from (0, 0, 0) to (0, 1, 1) we have x = 0, y = t and z = 1 then dx = 0, dy = dt and dz = 0 with 0 ≤ t ≤ 1. Then the integral over this part of the path is Z 1 0 2tdt = 1. • Along the straight line from (0, 1, 1) to (1, 1, 1) we have x = t, y = 1 and z = 1 then dx = dt, dy = 0 and dz = 0 with 0 ≤ t ≤ 1. Then the integral over this part of the path is Z 1 0 (3t 2 − 6)dt = −5. • Adding, Z C − → F · d − → M = 1 + 1 − 5 = −3 c) The straight line joining (0, 0, 0) and (1, 1, 1) is given in parametric form x = t, y = t, z = t with 0 ≤ t ≤ 1. Then Z 1 0 (3t 2 − 6t 2 )dt + (2t + 3t 2 )dt + (1 − 4t 4 )dt = 6/5 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 12. 12Ú19 Exercises Solutions Solution 5. In the plane z = 0, − → F (x, y, z) = (3x − 4y) ~ i + (4x + 2y) ~ j − 4y2~ k . and d − → M = dx~ i + dy~ j. The parametric equation is of form x = 4 cos θ, y = 3 sin θ and z = 0, then dx = −4 sin θdθ, dy = 3 cos θdθ and dz = 0 with 0 ≤ θ ≤ 2π. Then Z C − → F · d − → M = Z 2π 0 (12 cos θ − 12 sin θ)(−4 sin θdθ) + (16 cos θ + 6 sin θ)(3 cos θdθ) Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 13. 13Ú19 Exercises Solutions Solution 6. a) I C 2xydx + y2 dy = ZZ D ∂Q ∂x − ∂P ∂y dxdy = Z 4 0 dx Z √ x x/2 (−2x)dy = − 64 15 b) I C xydx + (x + y)dy = ZZ D ∂Q ∂x − ∂P ∂y dxdy = Z 2 0 dx Z − x 2 +1 0 (1 − x)dy = 1 3 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 14. 14Ú19 Exercises Solutions c) I C (e3 x + 2y)dx + (x3 + sin y)dy = ZZ D ∂Q ∂x − ∂P ∂y dxdy = Z 6 2 dx Z 4 1 (3x2 − 2)dy = 600 d) I C ln(1 + y)dx − xy 1 + y dy = ZZ D ∂Q ∂x − ∂P ∂y dxdy = Z 2 0 dx Z −2x+4 0 − y 1 + y − 1 1 + y dy = −4 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 15. 15Ú19 Exercises Solutions Solution 7. ¬ a) A = ~ r(0) = (0, 1, 0) and B = ~ r(1) = (1, 2, 1). b) − → ∇ × − → F =
- 23. ~ i ~ j ~ j ∂ ∂x ∂ ∂y ∂ ∂z 2xy3 z4 + 2 cos x sin y + 3x2 y2 z4 ez + 4x2 y3 z3
- 31. = (12x2 y2 z3 − 12x2 y2 z3 ) ~ i − (8xy3 x3 − 8xy3 x3 ) ~ j + (6xy2 z4 − 6xy2 z4 )~ k = ~ 0 . − → F · d − → M = df = ⇒ P(x, y, z)dx + Q(x, y, z)dy + R(x, y, z)dz = ∂f ∂x dx + ∂f ∂y dy + ∂f ∂z dz = ⇒ ∂f ∂x = P(x, y, z) ∂f ∂y = Q(x, y, z) ∂f ∂z = R(x, y, z) = ⇒ À Á Â ∂f ∂x = 2xy3 z4 + 2 cos x ∂f ∂y = sin y + 3x2 y2 z4 ∂f ∂z = ez + 4x2 y3 z3 Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 32. 16Ú19 Exercises Solutions • First integrate the first equation À with respect to x and obtain Z ∂f ∂x dx = Z (2xy3 z4 + 2 cos x)dx = ⇒ f(x, y, z) = x2 y3 z4 + 2 sin x + φ(y, z) • Then we differentiate f with respect to y and use equation Á to get ∂f ∂y = 3x2 y2 z4 + ∂ ∂y φ(y, z) = ⇒ sin y + 3x2 y2 z4 = 3x2 y2 z4 + ∂ ∂y φ(y, z) = ⇒ ∂ ∂y φ(y, z) = sin y = ⇒ φ(y, z) = − cos y + ψ(z) Replace φ(y, z) in f we get f(x, y, z) = x2 y3 z4 + 2 sin x − cos y + ψ(z) . • Finally we differentiate f with respect to z and use equation  to get ∂f ∂z = 4x2 y3 z3 + ψ0 (z) = ⇒ ez + 4x2 y3 z3 = 4x2 y3 z3 + ψ0 (z) = ⇒ ψ0 (z) = ez = ⇒ ψ(z) = ez + C . Therefore f(x, y, z) = x2 y3 z4 + 2 sin x − cos y + ez + C. Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 33. 17Ú19 Exercises Solutions c) Z C − → F · d~ r = Z AB df = f(B) − f(A) = 7 + 2 sin 1 − cos 2 + e + cos 1 . a) Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems
- 34. 18Ú19 Exercises Solutions b) Γ1 and Γ2 are both smooth so Γ is piecewise smooth, simple closed, positively orriented boundary, and ∂P ∂x = 15x 14 e x15 −3x 2 sin(x 3 ) , ∂P ∂y = −2 ∂Q ∂x = 2x and ∂Q ∂y = 2018y 2017 cos y 2018 − 2y y2 + 1 are continuous everywhere. So by Green’s Theorem, I Γ+ − → V · d − → r = ZZ D ∂Q ∂x − ∂P ∂y dA = Z 4 0 Z x x2−3x (2x + 2)dydx = Z 4 0 [(2x + 2)y]x x2−3x dx = Z 4 0 (−2x3 + 8x2 − 2x2 + 8x)dx = 64 . Go Back Dr. Kamel ATTAR | Chapter 4: Line integral | Solved Problems