Line Integrals: Scalar Integrals & Arc Length Fundamentals

We first parameterize the curve C. Since C is a straight line from (0,0) to (3,4), we can use the parameterization: x = 3t, y = 4t, with the parameter t ranging from 0 to 1. Next, we compute the differential arc length ds. We find the derivatives dx/dt = 3 and dy/dt = 4. Then ds is given by ds = √((dx/dt)² + (dy/dt)²) dt = √(3² + 4²) dt = √(9+16) dt = √(25) dt = 5 dt. Now, we express the integrand in terms of t: 2(x+y) = 2(3t+4t) = 2*(7t) = 14t. The line integral becomes ∫ from t=0 to 1 of (14t) * (5 dt) = ∫₀¹ 70t dt. We evaluate this integral: 70 * [ (1/2) t² ] from 0 to 1 = 70 * (1/2) * (1² - 0²) = 70 * 1/2 = 35. Therefore, the value of the line integral is 35.

We parameterize the curve C. Let x = 1-t and y = t, with t ranging from 0 to 1. Then compute the derivatives: dx/dt = -1 and dy/dt = 1. The differential arc length ds is given by ds = √((dx/dt)² + (dy/dt)²) dt = √((-1)² + (1)²) dt = √(2) dt. The integrand is 2(x+y) = 2((1-t) + t) = 2(1) = 2. The line integral becomes ∫ from t=0 to 1 of 2 * √(2) dt = 2√2 ∫₀¹ dt = 2√2 * (1-0) = 2√2.

We parameterize the curve by letting x = t, which makes y = t². The parameter t ranges from 0 to 1. The derivatives are dx/dt = 1 and dy/dt = 2t. The arc length element is ds = √((dx/dt)² + (dy/dt)²) dt = √(1² + (2t)²) dt = √(1 + 4t²) dt. The function to integrate is x, which becomes t. The line integral is ∫ from 0 to 1 of t * √(1 + 4t²) dt. We can solve this integral using a u-substitution. Let u = 1 + 4t², then du = 8t dt, which means t dt = du/8. We must also change the limits of integration: when t=0, u = 1 + 4(0)² = 1. When t=1, u = 1 + 4(1)² = 5. The integral becomes ∫ from 1 to 5 of √u * (du/8) = (1/8) ∫ u^(1/2) du. The antiderivative is (1/8) * [ (2/3)u^(3/2) ]. Evaluating this from 1 to 5 gives (1/12) * [ u^(3/2) ] from 1 to 5 = (1/12) * (5^(3/2) - 1^(3/2)) = (5√5 - 1) / 12.

To set up the line integral ∫_C f(x,y) ds, we parameterize the curve using x as the parameter. Let x = t, then y = t³, with t from 0 to 2. Then f(x,y) = xy = t * t³ = t⁴. Compute ds: dx/dt = 1, dy/dt = 3t², so ds = √((dx/dt)² + (dy/dt)²) dt = √(1² + (3t²)²) dt = √(1+9t⁴) dt. Therefore, the line integral becomes ∫ from t=0 to 2 of t⁴ * √(1+9t⁴) dt. Replacing t with x, we get ∫₀² x⁴ √(1+9x⁴) dx.

Parameterize the line segment from (0,0) to (2,3) by letting x = 2t and y = 3t, where t ranges from 0 to 1. Compute the derivatives: dx/dt = 2, dy/dt = 3. Then the differential arc length ds is given by ds = √((dx/dt)² + (dy/dt)²) dt = √(2² + 3²) dt = √(4+9) dt = √13 dt. The integrand is x² + y² = (2t)² + (3t)² = 4t² + 9t² = 13t². The line integral becomes ∫ from t=0 to 1 of 13t² * √13 dt = 13√13 ∫₀¹ t² dt. Evaluate the integral: ∫ t² dt from 0 to 1 is [t³/3] from 0 to 1 = 1/3. Therefore, the line integral equals 13√13 * (⅓) = (13√13)/3.

We parameterize the curve by letting x = t, which makes y = t². The parameter t ranges from 0 to 1. The derivatives are dx/dt = 1 and dy/dt = 2t. The arc length element is ds = √((dx/dt)² + (dy/dt)²) dt = √(1² + (2t)²) dt = √(1 + 4t²) dt. The function to integrate is 8x, which becomes 8t. The line integral is ∫ from 0 to 1 of 8t * √(1 + 4t²) dt. We can solve this integral using a u-substitution. Let u = 1 + 4t², then du = 8t dt. This substitution matches the integrand perfectly. We must also change the limits of integration: when t=0, u = 1 + 4(0)² = 1. When t=1, u = 1 + 4(1)² = 5. The integral becomes ∫ from 1 to 5 of √u du = ∫ u^(1/2) du. The antiderivative is [ (2/3)u^(3/2) ]. Evaluating this from 1 to 5 gives (2/3) * (5^(3/2) - 1^(3/2)) = (2/3) * (5√5 - 1).

Parameterize the upper half of the unit circle by x = cos t, y = sin t, with t from 0 to π. Then dx/dt = -sin t, dy/dt = cos t, so ds = √((-sin t)² + (cos t)²) dt = √(sin² t+cos² t) dt = dt. The integrand is y = sin t. The line integral becomes ∫ from t=0 to π of sin t dt. The integral of sin t from 0 to π is [-cos t]_0^π = -cos π - (-cos 0) = -(-1) + 1 = 1+1=2.

The triangle consists of three line segments: C₁ from (0,0) to (1,0), C₂ from (1,0) to (0,1), and C₃ from (0,1) to (0,0). Compute the integral over each and sum. For C₁, parameterize as x=t, y=0, so that t is from 0 to 1 and ds = √(1²+0) dt = dt. So we have the integral ∫₀¹ t dt = 1/2. For C₂ from (1,0) to (0,1), parameterize as x=1-t, y=t so that t is from 0 to 1 and ds = √((-1)²+1²) dt = √2 dt. The integrand is (1-t)+t = 1. So we have ∫₀¹ 1 * √2 dt = √2. For C₃: from (0,1) to (0,0), parameterize as x=0, y=1-t so that t from 0 to 1 and ds = √(0²+(-1)²) dt = 1 dt. The integrand is 0+(1-t) = 1-t. So obtain ∫₀¹ (1-t) dt = [t - t²/2]₀¹ = 1 - 1/2 = 1/2. Taking the sum, we get 1/2 + √2 + 1/2 = 1 + √2.

Parameterize C: x = t, y = 2t, with t from 0 to 1. Then dx = dt, dy = 2 dt. The integrand is y dx + x dy = (2t)(dt) + (t)(2 dt) = 2t dt + 2t dt = 4t dt. So the integral becomes ∫ from t=0 to 1 of 4t dt = 4 * (1/2) = 2.

Parameterize the curve C. Let x = t and y = 1+t, with t ranging from 0 to 1. Then dx = dt and dy = dt. Substitute into the integrand: x² dx + y² dy = t² dt + (1+t)² dt = [t² + (1 + 2t + t²)] dt = (2t² + 2t + 1) dt. Now evaluate the line integral: ∫ from t=0 to 1 (2t² + 2t + 1) dt. Compute the antiderivative: (⅔)t³ + t² + t. Evaluate at t=1: (⅔) + 1 + 1 = 2/3 + 2 = 8/3. Evaluate at t=0: 0. So the integral equals 8/3 - 0 = 8/3.

The work done by a force field F(x,y) = (y, x) along a curve C is given by the line integral ∫_C y dx + x dy. We parameterize the curve C: y = x² from (0,0) to (1,1) by letting x = t, y = t², with t from 0 to 1. Then dx = dt (since dx/dt = 1) and dy = 2t dt (since dy/dt = 2t). Substitute into the integral: ∫_C y dx + x dy = ∫ from t=0 to 1 [ (t²)(dt) + (t)(2t dt) ] = ∫₀¹ (t² * 1 + t * 2t) dt. This matches option C.

Parameterize C: x = t, y = 2t, z = 3t, t from 0 to 1. Then dx = dt, dy = 2 dt, dz = 3 dt. The integrand is x dx + y dy + z dz = t dt + (2t)(2 dt) + (3t)(3 dt) = t dt + 4t dt + 9t dt = 14t dt. The integral is ∫ from t=0 to 1 of 14t dt = 14 * [t²/2]₀¹ = 14*(1/2)=7.

Parameterize C as given: x = cos t, y = sin t, z = t, with t from 0 to π/2. Compute the derivatives: dx/dt = -sin t, dy/dt = cos t, dz/dt = 1. Then ds = √( (-sin t)² + (cos t)² + 1² ) dt = √( sin² t + cos² t + 1 ) dt = √(1+1) dt = √2 dt. The scalar function is f(x,y,z) = xy = cos t sin t. So the line integral becomes ∫ from t=0 to π/2 of (cos t sin t) * √2 dt = √2 ∫_0^{π/2} cos t sin t dt. Use the identity cos t sin t = (1/2) sin(2t). Then the integral becomes √2 ∫_0^{π/2} (1/2) sin(2t) dt = (√2/2) ∫_0^{π/2} sin(2t) dt. Compute ∫ sin(2t) dt = -1/2 cos(2t). Evaluate from 0 to π/2: at π/2: -1/2 cos(π) = -1/2 * (-1) = 1/2; at 0: -1/2 cos(0) = -1/2. So the definite integral is 1/2 - (-½) = 1. Therefore, the line integral is (√2/2) * 1 = √2/2.

The line integral of a scalar function with respect to arc length is defined in a way that does not depend on the orientation or the specific parameterization used for the curve. It only depends on the curve and the function. Option A is false because reversing the direction does not change ds (arc length is always positive). Option B is false because if f is negative, the integral can be negative. Option D is false in general; it is only true if f is constant. Option C is correct: any smooth parameterization of the curve will yield the same value for the line integral.

Parameterize the line segment from (1,0,0) to (0,1,1) using t from 0 to 1: let x = 1-t, y = t, z = t. Then compute derivatives: dx/dt = -1, dy/dt = 1, dz/dt = 1. The differential arc length ds = √( (-1)² + 1² + 1² ) dt = √3 dt. The scalar function is f(x,y,z) = x²+y²+z² = (1-t)² + t² + t² = 1 - 2t + 3t². The line integral is ∫ from t=0 to 1 of (1 - 2t + 3t²) * √3 dt = √3∫₀¹ (1 - 2t + 3t²) dt. Evaluate the integral:∫₀¹ (1 - 2t + 3t²) dt = [t - t² + t³]₀¹ = (1 - 1 + 1) - 0 = 1. So the line integral equals √3 * 1 = √3.