Line Integrals: Work, Vector Fields & Parametric Paths

To set up this scalar line integral, we first need to parameterize the curve C. Since C is the line segment connecting (0, 0) to (1, 1), we can define x(t) = t and y(t) = t for t ranging from 0 to 1. Next, we must find the arc length element ds. The formula for ds is √((dx/dt)² + (dy/dt)²) dt. Calculating the derivatives, we get dx/dt = 1 and dy/dt = 1. Substituting these into the formula gives us ds = √(1² + 1²) dt, which simplifies to √(2) dt. Then, we substitute the parameterization into the function f(x, y) = x + y, resulting in f(t, t) = t + t. Finally, we combine these parts to form the ∫₀¹ (t + t) * √(2) dt.

To parameterize a function given explicitly as y = f(x), it is most efficient to set x equal to the parameter t. By letting x = t, the equation y = x² naturally becomes y = t². Therefore, the position vector r(t) is <x(t), y(t)> = <t, t²>. Since the problem states that the curve runs from x = 0 to x = 1, and we defined x = t, the bounds for t are also from 0 to 1.

We begin by parameterizing the line segment from (0, 0) to (3, 4). We can set x(t) = 3t and y(t) = 4t for 0 ≤ t ≤ 1. Next, we calculate the arc length element ds. The derivatives are dx/dt = 3 and dy/dt = 4. Using the formula ds = √((dx/dt)² + (dy/dt)²) dt, we get ds = √(3² + 4²) dt = √(9 + 16) dt = √(25) dt = 5 dt. Now we substitute the parameterization into the function f(x, y) = 2x to get f(t) = 2(3t) = 6t. We can now set up the integral as the ∫₀¹ (6t) * 5 dt, which simplifies to the ∫₀¹ (30t dt. Evaluating this integral, we find the antiderivative is 15t² evaluated from 0 to 1. Plugging in the limits, we get 15(1)² - 15(0)², which equals 15.

To find the mass of a wire with variable density, we must compute the scalar line integral of the density function p(x, y) = y over the curve C. First, we parameterize the semicircle of radius 2 using polar coordinates: x(t) = 2cos(t) and y(t) = 2sin(t), with t ranging from 0 to pi. Next, we find the arc length element ds. The derivatives are dx/dt = -2sin(t) and dy/dt = 2cos(t). Calculating ds, we get √((-2sin(t))² + (2cos(t))²) dt = √(4sin²(t) + 4cos²(t)) dt = √(4) dt = 2 dt. We then substitute the density function y = 2sin(t) into the integral. The mass integral becomes the integral from 0 toπ of (2sin(t)) * (2) dt, which simplifies to the integral from 0 toπ of 4sin(t) dt. The antiderivative of 4sin(t) is -4cos(t). Evaluating this from 0 toπ gives -4cos(pi) - (-4cos(0)) = -4(-1) - (-4(1)) = 4 + 4 = 8.

The line integral of a vector field F along a curve C is defined as the integral of the tangential component of the field along the path. Physically, if F represents a force field, the dot product F · dr (or F · T ds) calculates the force applied in the direction of motion multiplied by the displacement. Summing these infinitesimal contributions along the entire path gives the total work done by the force field in moving an object along that specific curve.

First, we identify the components of the vector field and the path. Here, F = <y, x> and the path is x(t) = t, y(t) = t. Substituting the path into the vector field, we get F(r(t)) = <t, t>. Next, we calculate the derivative of the position vector, r'(t) = <1, 1>. We then compute the dot product F(r(t))·r'(t), which is <t, t> · <1, 1> = t(1) + t(1) = 2t. Finally, we integrate this result with respect to t from 0 to 1. The integral of 2t dt from 0 to 1 is t² evaluated from 0 to 1, which equals 1 - 0 = 1.

We start by parameterizing the path, which is a quarter circle of radius 1. We can use r(t) = <cos(t), sin(t)> for t from 0 to π/2. Next, we find the derivative vector r'(t) = < -sin(t), cos(t) >. We then substitute the parameterization into the force field F(x, y) = <x, y> to get F(r(t)) = <cos(t), sin(t)>. Now, we calculate the dot product F · r'(t) = <cos(t), sin(t)> · < -sin(t), cos(t) >. This results in -cos(t)sin(t) + sin(t)cos(t) = 0. Since the integrand is 0, the integral from 0 to π/2 of 0 dt is simply 0.

We can solve this by parameterizing the curve. Let x = t, so y = t². Then dx = dt and dy = 2t dt. The bounds for t are from 0 to 1. We substitute these into the integral expression 3x² dx + 2y dy. This gives us 3(t)² (dt) + 2(t²) (2t dt). Simplifying the terms, we get 3t² dt + 4t³ dt. Combining them, we integrate (3t² + 4t³) dt from 0 to 1. The antiderivative is t³ + t⁴. Evaluating from 0 to 1 yields (1³ + 1⁴) - (0 + 0) = 1 + 1 = 2. Alternatively, we could recognize that the field is conservative with potential function f(x,y) = x³ + y², and evaluate f(1,1) - f(0,0) = (1+1) - 0 = 2.

First, we find the derivatives of the components of r(t). We have x'(t) = -sin(t), y'(t) = cos(t), and z'(t) = 1. Next, we calculate the magnitude of the derivative vector |r'(t)| to find ds. This is √((-sin(t))² + (cos(t))² + 1²) = √(sin²(t) + cos²(t) + 1). Since sin²(t) + cos²(t) = 1, this simplifies to √(1 + 1) = √(2). So, ds = √(2) dt. Now we substitute the function f(x, y, z) = z with the parameter z = t. The integral becomes the integral from 0 to 2π of t * √(2) dt. Pulling out the constant, we get √(2) times the integral of t dt. The antiderivative of t is t²/2. Evaluating this from 0 to 2π gives √(2) * [(2π)² / 2 - 0] = √(2) * [4π² / 2] = 2π² * √(2).

Since the curve C is piecewise smooth, we must split the integral into two parts: C1 and C₂.

For C1 (from (0,0) to (1,0)): y is constant at 0, so dy = 0. The term xy dx becomes x(0) dx = 0. The term x dy becomes x(0) = 0. So the integral over C1 is 0.

For C₂ (from (1,0) to (1,2)): x is constant at 1, so dx = 0. The term xy dx becomes 0. The term x dy becomes 1 dy. We integrate 1 dy from y = 0 to y = 2. The result is y evaluated from 0 to 2, which equals 2.

Finally, we add the results of the two integrals: 0 + 2 = 2.

A fundamental property of line integrals of vector fields is that they depend on the orientation of the curve. If you reverse the direction of the path (traversing from B to A instead of A to B), the direction of the tangent vector dr reverses the sign (dr becomes -dr). Therefore, the dot product F · dr changes sign. This means ∫-C F · dr = - (∫C F · dr). Since the original integral was 10, the integral over the reversed path is -10.

We parameterize the line segment from (0, 0) to (2, 2) by setting x(t) = t and y(t) = t for 0 ≤ t ≤ 2. Consequently, dx = dt and dy = dt. We substitute these into the line integral Integral of y² dx + x² dy. Substituting the parameters, we get Integral from 0 to 2 of (t²)(dt) + (t²)(dt), which simplifies to Integral from 0 to 2 of 2t² dt. The antiderivative of 2t² is (⅔)t³. Evaluating this at the upper limit gives (⅔)(2)³ = (⅔)(8) = 16/3. Evaluating at the lower limit gives 0. Thus, the total work is 16/3.

We parameterize the upper half of the unit circle using x(t) = cos(t) and y(t) = sin(t) for t ranging from 0 to pi. The derivative vector is r'(t) = <-sin(t), cos(t)>. We substitute the path into the vector field F(x, y) = <x, y> to get F(r(t)) = <cos(t), sin(t)>. We then compute the dot product F(r(t)) · r'(t) = <cos(t), sin(t)> · <-sin(t), cos(t)>. This equals -cos(t)sin(t) + sin(t)cos(t), which simplifies to 0. Integrating 0 with respect to t from 0 toπ results in 0.

First, we parameterize the line segment connecting (1, 0) to (0, 1). A standard parameterization is r(t) = <1-t, t> for 0 ≤ t ≤ 1. From this, we find the derivatives: dx = -1 dt and dy = 1 dt. We substitute these into the work integral Integral of F · dr = Integral of (y dx - x dy). Substituting x = 1-t and y = t, the integral becomes ∫₀¹ ([t(-1) - (1-t)(1)] dt. Simplifying the integrand gives -t - (1 - t) = -t - 1 + t = -1. We integrate the constant -1 from 0 to 1. The result is [-t] evaluated from 0 to 1, which equals -1 - 0 = -1.

We start by finding the arc length element ds. The position vector is r(t) = <t, t, t>, so the derivative vector is r'(t) = <1, 1, 1>. The magnitude |r'(t)| is √(1² + 1² + 1²) = √(3). Therefore, ds = √(3) dt. Next, we substitute the parameterization into the scalar function f(x, y, z) = x + y + z. This gives f(t) = t + t + t = 3t. The integral becomes the ∫₀¹ (3t) * √(3) dt. Pulling out the constants, we have 3 * √(3) times ∫₀¹ (t dt. The integral of t is t²/2. Evaluating from 0 to 1 gives 1/2. Multiplying by the constant term, we get 3 * √(3) * (1/2) = 3 * √(3) / 2.