Connection Between Line Integrals of Vector Fields

Section Connection Between Line Integrals of Vector Fields Definition: We start with a plane curve C given by the parametric equations x = x(t) a≤t≤b y = y(t) If f is defined on a smooth curve C given by the equation above, then the line integral of f along C is R Pn f (x, y) ds = limn→∞ i=1 f (x∗i , yi∗ ) △ si C if this limit exists. q R Rb dy 2 2 Line Integral Formula: C f (x, y) ds = a f (x(t), y(t)) ( dx dt ) + ( dt ) dt R Exercise 1. Evaluate C xy 2 ds if C has the parametrization x = cos t, 0 ≤ t ≤ π/2. y = sin t; (Swok Sec 18.2 Ex 1) R Exercise 2. Evaluate C (xy + 2z) ds on the following line segments. (a) The line segment from P (1, 0, 0) to Q(0, 1, 1) (b) The line segment from Q(0, 1, 1) to P (1, 0, 0) (Briggs Sec 14.2 Ex 3) Class R Exercise 1. Evaluate the line integral, where C is the given curve. (#2,4) (a) RC xy ds, C: x = t2 , y = 2t, 0 ≤ t ≤ 1 (b) C x sin y ds, C is the line segment from (0,3) to (4,6) Definition: The line integrals of f along C with respect to x and y: R C f (x, y) dx = limn→∞ Pn i=1 f (x∗i , yi∗ ) △ xi R C f (x, y) dy = limn→∞ Pn i=1 f (x∗i , yi∗ ) △ yi Formula: Suppose x = x(t) and y = y(t) R Rb f (x, y) dx = a f (x(t), y(t)) x′ (t) dt RC Rb f (x, y) dy = a f (x(t), y(t)) y ′ (t) dt C R R R Line Integral Abbreviation: C P (x, y) dx + C Q(x, y) dy = C P (x, y) dx + Q(x, y) dy. R Exercise 3. Evaluate C xy dx + x2 dy if (a) C consists of line segments from (2,1) to (4,1) and from (4,1) to (4,5) (b) C is the line segment from (2,1) to (4,5). (Swok Sec 18.2 Ex 4) Class R Exercise 2. Evaluate the line integral, where C is the given curve. (#6,8) (a) RC ex dx, C is the arc of the curve x = y 3 from (−1, −1) to (1, 1) (b) C x2 dx + y 2 dy, C consists of the arc of the circle x2 + y 2 = 4 from (2,0) to (0,2) followed by the line segment from (0,2) to (4,3) Definition: In general, a given parametrization x = x(t), y = y(t), a ≤ t ≤ b, determines an orientation of a curve C, with the positive direction corresponding to increasing values of the parameter t. Line Integrals in Space: We now suppose that C is a smooth space curve given by the parametric equations: x = x(t) y = y(t) z = z(t) a ≤ t ≤ b. Definition: We define the line integral of f along C in a manner similar to that for plane curves: R Pn f (x, y, z) ds = limn→∞ i=1 f (x∗i , yi∗ , zi∗ ) △ si . C Formula R C f (x, y, z) ds = Rb a f (x(t), y(t), z(t)) q dy 2 dz 2 2 ( dx dt ) + ( dt ) + ( dt ) dt. Line integrals along C with respect to x, y, and z can also be defined. For example, R Rb f (x, y, z) dz = limn→∞ f (x∗i , yi∗ , zi∗ ) △ zi = a f (x(t), y(t), z(t)) z ′ (t) dt. C Line Integral Abbreviation: We evaluate integrals of the form R P (x, y, z) dx + Q(x, y, z) dy + R(x, y, z) dz C by expressing everything (x, y, z, dx, dy, dz) in terms of the parameter t. R Exercise 4. Evaluate the line integral C −y dx + z dy + 2x dz, where C is the helix → − → − → − → − r (t) = (cos t) i + (sin t) j + t k , 0 ≤ t ≤ 2π. (Hass Sec 16.2 Ex 3) R Exercise 5. Evaluate C y sin z ds, where C is circular helix given by the equations x = cos t, y = sin t, z = t, 0 ≤ t ≤ 2π. (Stew Sec 16.2 Ex 5) R Exercise 6. Evaluate C yz dx + xz dy + xy dz if C is given by x = t, y = t2 , z = t3 ; 0 ≤ t ≤ 2. (Swok Sec 18.2 Ex 5) Class R Exercise 3. Evaluate the line integral, where C is the given curve. (#10-16 even) (a) RC xyz 2 ds, C is the line segment from (−1, 5, 0) to (1, 6, 4) (b) C (x2 + y 2 + z 2 ) ds, C: x = t, y = cos 2t, z = sin 2t, 0 ≤ t ≤ 2π √ R (c) RC y dx + z dy + x dz, C: x = t, y = t2 , z = t3 , 1 ≤ t ≤ 4 (d) C (y + z) dx + (x + z) dy + (x + y) dz, C consists of line segments from (0,0,0) to (1,0,1) and from (1,0,1) to (0,1,2). → − Definition: The work W done by the force field F is the limit of Riemann sums: R → R → − → − − → − W = C F (x, y, z) · T (x, y, z) ds = C F · T ds. Rb → − − − Here is the computational formula: W = a F (→ r (t)) · → r ′ (t) dt. → − Definition: Let F be a continuous vector field defined on a smooth curve C given by a vec→ − − tor function → r (t), a ≤ t ≤ b. Then, the line integral of F along C is Rb → R → R → − → − − − − − − F · d→ r = a F (→ r (t)) · → r ′ (t) dt = c F · T ds. C R → → − − → − → − → − − r , where F (x, y, z) = z i + xy j − y 2 k along the curve C given Exercise 7. Evaluate C F · d→ √→ − → − → − − by → r (t) = t2 i + t j + t k , 0 ≤ t ≤ 1. (Hass Sec 16.2 Ex 2) R → − − r , where C is given by the vector function Class Exercise 4. Evaluate the line integral C F · d→ → − r (t). (#20,22) → − − → − → − → − → − → − → − (a) F (x, y, z) = (x + y) i + (y − z) j + z 2 k , → r (t) = t2 i + t3 j + t2 k , 0 ≤ t ≤ 1 → − → → − → − → − → − → − → − (b) F (x, y, z) = x i + y j + xy k , − r (t) = cos t i + sin t j + t k , 0 ≤ t ≤ π → − → − → − → − Exercise 8. Find the work done by the force field F = (y − x2 ) i + (z − y 2 ) j + (x − z 2 ) k → − → − → − − along the curve → r (t) = t i + t2 j + t3 k , 0 ≤ t ≤ 1, from (0,0,0) to (1,1,1). (Hass Sec 16.2 Ex 4) → − → − → − → − Exercise 9. Find the work done by the force field F = x i + y j + z k in moving an object → − → − → − − along the curve C parametrized by → r (t) = cos(πt) i + t2 j + sin(πt) k , 0 ≤ t ≤ 1. (Hass Sec 16.2 Ex 5) → − → − → − Class Exercise 5. Find the work done by the force field F (x, y) = x2 i + yex j on a particle that moves along the parabola x = y 2 + 1 from (1,0) to (2,1). (#40) Class Exercise 6. The force exerted by an electric charge at the origin on a charged particle at a → − − − − − point (x, y, z) with position vector → r = < x, y, z > is F (→ r ) = K→ r /|→ r |3 where K is a constant. Find the work done as the particle moves from a straight line from (2,0,0) to (2,1,5). (#42) Connection Between Line Integrals of Vector Fields and Line Integrals of Scalar Fields: R → R → − − → − → − → − − F · d→ r = C P dx + Q dy + R dz, where F = P i + Q j + R k . C