Line Integrals: Conservative Fields, Path Independence & Closed Curves

The average value of a scalar function f(x, y) along a curve C is defined as (1 / Length of C) ∫C f(x, y) ds. Here, we are given that the length of C is 10 and the value of the integral is 50. Plugging these into the definition, the average value is 50 / 10 = 5.

Parametrize the path as x = t, y = t where t goes from 0 to 1. Then dy = dt and x = t, so x dy = t dt. The integral is ∫₀¹ t dt. The antiderivative of t is t²/2. Evaluating from 0 to 1 gives 1²/2 - 0²/2 = 1/2.

The line integral of a vector field F =

is always written as P dx + Q dy + R dz, where the components are taken in order. Here the first component yz multiplies dx, the second component xz multiplies dy, and the third component xy multiplies dz, giving yz dx + xz dy + xy dz.

The path lies on the y-axis where x = 0 the entire time. The function xy = 0 · y = 0 for every point on the path. The integral of the constant function 0 with respect to arc length ds is 0, regardless of the length of the path.

Parametrize as x = t, y = t where t goes from 0 to 2. Then dr =

The work is ∫C F · dr = ∫C x dx + y dy. Parametrize as x = t, y = t, t from 0 to 1. Then dx = dt, dy = dt, so t dt + t dt = 2t dt. The integral is ∫₀¹ 2t dt. The antiderivative is t². Evaluating from 0 to 1 gives 1 - 0 = 1. This is the work done by the force on the particle along the path.

Parametrize as x = cos t, y = sin t, t from 0 to 2π. Then dx = -sin t dt and dy = cos t dt. Substitute: -y dx = -sin t (-sin t dt) = sin² t dt and x dy = cos t (cos t dt) = cos² t dt. Adding gives sin² t dt + cos² t dt = (sin² t + cos² t) dt = dt. The integral is ∫ from 0 to 2π dt = 2π.

Parametrize as x = t, y = t, z = t, t from 0 to 1. Then dx = dt, dy = dt, dz = dt. Substitute: z dt + t dt + t dt = t dt + t dt + t dt = 3t dt. The integral is ∫₀¹ 3t dt = 3 · (t²/2) from 0 to 1 = 3/2.

Parametrize as x = cos t, y = sin t, t from 0 to π/2. Then dx = -sin t dt, dy = cos t dt. The integrand y dx + x dy = sin t (-sin t dt) + cos t (cos t dt) = -sin² t dt + cos² t dt = (cos² t - sin² t) dt = cos 2t dt. The integral is ∫ from 0 to π/2 cos 2t dt = (1/2) sin 2t from 0 to π/2 = (1/2)(sin π - sin 0) = 0.

Parametrize as x = 3t, y = 4t, t from 0 to 1. Then ds = 5 dt (speed = √(9+16) = 5). The function x² + y² = 9t² + 16t² = 25t². The integral is ∫₀¹ 25t² · 5 dt = 125 ∫ t² dt = 125 · (t³/3) from 0 to 1 = 125/3.

Parametrize as x = 1 + 3t, y = 2 + 3t, z = 3 + 3t, t from 0 to 1. Then dr = and F = . The dot product is (3+3t)3dt + (2+3t)3dt + (1+3t)3dt = (18 + 27t) dt. Taking the integral, ∫01 (18 + 27t) dt = [18t + 13.5t²]01 = 31.5.

The form 2x y dx + x² dy = d(x² y). It is exact. Evaluate x² y at (1,1) minus at (0,0): 1·1 - 0·0 = 1.

We must split the integral into three segments.

Segment 1 (0,0) to (1,0): y=0, x=t from 0 to 1. ds=dt. Integrand x+y = t. Integral = 1/2.

Segment 2 (1,0) to (0,1): Line y = 1-x. Param x=1-t, y=t from 0 to 1. dx=-dt, dy=dt. ds = √(1+1)dt = √(2)dt. Integrand x+y = (1-t)+t = 1. Integral = 1 * √(2) = √(2).

Segment 3 (0,1) to (0,0): x=0, y=1-t from 0 to 1. ds=dt. Integrand x+y = 1-t. Integral = [t - t²/2]01 = 1/2.

Total Sum: 1/2 + √(2) + 1/2 = 1 + √(2).

The field has curl ∂Q/∂x - ∂P/∂y = 1 - (-1) = 2. By Green’s theorem the integral equals ∬ 2 dA over the square of area 4, so 2 · 4 = 8.

In physics, the work done by a variable force F moving a particle along a path C is defined as the line integral of the tangential component of the force, which is the vector line integral ∫C F · dr. This represents the energy transferred to the particle by the force as it moves from A to B along C.