Section 16.2

Section 16.2

Line Integrals

4. Parametric equations for C are x = 4t, y = 3 + 3t, 0 ≤ t ≤ 1. Then ∫

Cx sin yds =

∫1

0(4t) sin(3 + 3t)√

4²+ 3²dt = 20∫1

0 t sin(3 + 3t)dt.

Intergrating by parts with u = t⇒ du = dt, dv = sin(3+3t)dt ⇒ v = −¹₃ cos(3+3t) gives ∫

Cx sin yds = 20[−¹₃cos(3 + 3t) + ¹₉sin(3 + 3t)]¹₀ = 20[−¹₃cos 6 +¹₉sin 6 + 0−

1

9sin 3] = ²⁰₉(sin 6− 3 cos 6 − sin 3)

5. If we choose x as the parameter, parametric equations for C are x = x, y =√ x for 1≤ x ≤ 4 and ∫

C(x²y³ −√

x)dy = ∫4

1[x²· (√

x)³ −√ x]₂√¹

xdx = ¹₂∫4

1(x³ − 1)dx =

1

2[¹₄x⁴− x]⁴₁ = ¹₂(64− 4 −¹₄ + 1) = ²⁴³₈ .

8. On C₁ : x = cos t⇒ dx = − sin tdt, y = sin t ⇒ dy = cos tdt, 0 ≤ t ≤ ^π₂. On C₂ : x = 2t− 2 ⇒ dx = 2dt, y = t ⇒ dy = dt, 1 ≤ t ≤ 3.

Then∫

Cx√ydx + 2y√

xdy =∫

C1x√ydx + 2y√

xdy +∫

C2x√ydx + 2y√ xdy

=∫^π

2

0 cos t√

sin t(− sin t)dt+2 sin t√

cos t cos tdt+∫₃

1(2t−2)√

t(2dt)+2t√

2t− 2dt =

∫ ^π

2

0 −(sin t)³²d sin t−2(cos t)³²d cos t+4∫₃

1 t³²−t¹²dt+¹₄∫₄

0 y³²+2y¹²dy (y = 2t−2) =

309√ 3−1 30

16. On C₁ : x = t⇒ dx = dt, y = 2t ⇒ dy = 2dt, z = −t ⇒ dz = −dt, 0 ≤ t ≤ 1.

On C₂ : x = 1 + 2t⇒ dx = 2dt, y = 2 ⇒ dy = 0dt, z = −1+t ⇒ dz = dt, 0 ≤ t ≤ 1.

Then∫

Cx²dx + y²dy + z²dz =∫

C1x²dx + y²dy + z²dz +∫

C2x²dx + y²dy + z²dz

= ∫₁

0 t²dt + (2t)² · 2dt + (−t)²(−dt) +∫₁

0(1 + 2t)²· 2dt + 2² · 0dt + (−1 + t)²dt =

∫₁

0 8t²dt +∫₁

0(9t²+ 6t + 3)dt = [⁸₃t³]¹₀+ [3t³+ 3t²+ 3t]¹₀ = ³⁵₃ 22. ∫

CF·dr =∫_π

0 hcos t, sin t, −ti·h1, cos t, − sin tidt =∫_π

0 (cos t + sin t cos t + t sin t)dt = [sin t + ¹₂sin²t + (sin t− t cos t)]^π0 = π

28. We graph F (x, y) = √ ^x

x²+y²i +√ ^y

x²+y²j and the curve C. In the first quadrant, each vector starting on C points in roughly the same direction as C, so the tangential component F · T is positive. In the second quadrant, each vector starting on C points in roughly the direction opposite to C, so F· T is negative. Here, it appears that the tangential components in the first and second quadrants counteract each other, so it seems reasonable to guess that∫

CF · dr = ∫

CF · T ds is zero. To verify, 1

we evaluate ∫

CF · dr. The curve C can be represented by r(t) = ti + (1 + t²)j,−1 ≤ t ≤ 1, so F (r(t)) = √ ^t

t²+(1+t²)²i +√ ^1+t2

t²+(1+t²)²j and r⁰(t) = i + 2tj. Then ∫

CF · dr = ∫1

−1F (r(t))· r⁰(t)dt = ∫1

−1(√ ^t

t²+(1+t²)² + √^2t(1+t2)

t²+(1+t²)²)dt =

∫1

−1 √t(3+2t²)

t⁴+3t²+1dt = 0 [since the integrand is an odd function].

43. ∫6π

0 (0, 0, 185)· (− sin θ, cos θ,³⁰_2π)dθ = ∫6π

0 185· ³⁰_2πdθ = 185· 90 = 16650 44. ∫6π

0 (0, 0, 185− _6π⁹ θ)· (− sin θ, cos θ,³⁰_2π)dθ =∫6π

0 (185− _6π⁹ θ)· ³⁰_2πdθ = 16245

48. Use the orientation pictured in the figure. Then since B is tangent to any circle that lies in the plane perpendicular to the wire, B = |B|T where T is the unit tangent to the circle C : x = r cos θ, y = r sin θ. Thus B = |B| < − sin θ, cos θ >.

Then ∫

CB· dr = ∫_2π

0 |B| < − sin θ, cos θ > · < −r sin θ, r cos θ > dθ =∫_π

0 |B|rdθ = 2πr|B|.(Note that |B| here is the magnitude of the field at a distance r from the wire’s center.) But by Ampere,s Law∫

CB· dr = µ0I. Hence |B| = ^µ_2πr^0I.

2