Problem 31 - Thomas Calculus

Question

Evaluating a work integral two ways Let \(F=\nabla\left(x^{3} y^{2}\right)\) and let \(C\) be the path in the \(x y\) -plane from \((-1,1)\) to \((1,1)\) that consists of the line segment from \((-1,1)\) to \((0,0)\) followed by the line segment from \((0,0)\) to \((1,1)\) . Evaluate \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\) in two ways. \begin{equation}\begin{array}{l}{\text { a. Find parametrizations for the segments that make up } C \text { and }} \\ {\text { evaluate the integral. }} \\ {\text { b. Use } f(x, y)=x^{3} y^{2} \text { as a potential function for } \mathbf{F} \text { . }}\end{array}\end{equation}

Step-by-Step Solution

Answer

The integral \( \int_C F \cdot d\mathbf{r} \) along path \( C \) is 0.

Step 1: Identify the vector field

The given vector field is defined as the gradient of the function, i.e., \( F = abla(x^3 y^2) \). The gradient will be calculated as follows: \[ F = \left( \frac{\partial}{\partial x}(x^3 y^2), \frac{\partial}{\partial y}(x^3 y^2) \right) = (3x^2 y^2, 2x^3 y) \].

Step 2: Parametrize the first segment of path C

The path \( C \) consists of two segments. The first segment is from \((-1, 1)\) to \((0, 0)\). Parametrize this as \( r_1(t) = (-1 + t, 1 - t) \) where \( t \) ranges from 0 to 1.

Step 3: Compute the integral over the first segment

For the first segment, calculate \( F \cdot \frac{d\mathbf{r}}{dt} \). The derivative is \( \frac{d\mathbf{r_1}}{dt} = (1, -1) \) and \( F(r_1(t)) = (3(-1+t)^2(1-t)^2, 2(-1+t)^3(1-t)) \). Compute the dot product, integrate \( \int_0^1 F \cdot \frac{d\mathbf{r}}{dt} \, dt \).

Step 4: Parametrize the second segment of path C

The second segment is from \((0, 0)\) to \((1, 1)\). Parametrize this as \( r_2(t) = (t, t) \) where \( t \) ranges from 0 to 1.

Step 5: Compute the integral over the second segment

For the second segment, calculate \( F \cdot \frac{d\mathbf{r}}{dt} \). Derivative is \( \frac{d\mathbf{r_2}}{dt} = (1, 1) \) and \( F(r_2(t)) = (3t^2 t^2, 2t^3 t) \). Compute the dot product, integrate \( \int_0^1 F \cdot \frac{d\mathbf{r}}{dt} \, dt \).

Step 6: Combine results from both segments

Sum the integrals from both segments to find the total work done along path \( C \).

Step 7: Use potential function approach

Since \( F \) is the gradient of \( f(x,y) = x^3 y^2 \), the work integral \( \int_C F \cdot d\mathbf{r} \) can be simplified using the Fundamental Theorem for Line Integrals. Calculate \( f(1, 1) - f(-1, 1) \).

Step 8: Verify both methods give the same result

Confirm that the result from directly integrating along path \( C \) matches the result from evaluating the potential function difference.

Key Concepts

Gradient Vector Field

The concept of a gradient vector field is central to understanding how forces and movements interact in space. A gradient vector field, denoted as \( F = abla f \), represents the "direction" and "rate" of steepest ascent of a scalar function \( f \). The gradient \( abla f \) is a vector of partial derivatives of \( f \), describing how \( f \) changes with respect to each variable.

In the given exercise, \( F \) is the gradient of \( x^3 y^2 \), resulting in \( F = (3x^2 y^2, 2x^3 y) \). This gradient shows how the function \( x^3 y^2 \) would increase fastest at any point \((x, y)\) in its domain. Understanding the gradient helps in computing line integrals since it allows us to determine how the field behaves along a curve.

Parametrization

Parametrization involves expressing a mathematical object, like a curve or a surface, using parameters that simplify calculations. For curves in vector fields, especially in line integrals, parametrization maps the path into a form that can be easily integrated over.

In the exercise, the curve \( C \) is split into two segments. The first segment, from \((-1, 1)\) to \((0, 0)\), is parametrized as \( r_1(t) = (-1 + t, 1 - t) \), where \( t \) varies from 0 to 1. The second segment, from \((0, 0)\) to \((1, 1)\), is parametrized as \( r_2(t) = (t, t) \).

Parametrization enables us to convert the curve into a form where we can apply calculus operations, such as differentiation and integration. This makes the evaluation of integrals along paths straightforward and manageable.

Line Integral

A line integral of a vector field along a curve sums up the field's effect along the path of travel. In context, it gives us insight into the work done by a field as we move along a path. The line integral is expressed as \( \int_C F \cdot d\mathbf{r} \), which calculates the cumulative effect of the vector field \( F \) along the curve \( C \).

For our exercise, we calculate line integrals over two parametrized segments. For each segment, we evaluate \( F \cdot \frac{d\mathbf{r}}{dt} \) — the dot product of the vector field and the derivative of the parametrization with respect to \( t \). Integrating this over each segment from 0 to 1 gives the work done along that segment of the path.

Summing these integrals gives the total work done by the field along the entire path \( C \). This integral calculation incorporates both the magnitude and direction of the field effecting motion across the curve.

Potential Function

A potential function is a scalar function from which a vector field can be derived as a gradient. This concept simplifies the evaluation of work integrals by circumventing the necessity to compute directly using parametrizations.

In this problem, the vector field \( F \) is the gradient of \( f(x, y) = x^3 y^2 \). Because \( F \) is a conservative field (having a potential function), the work done between any two points is path-independent. Therefore, we can directly calculate the work integral using the Fundamental Theorem for Line Integrals:

• Calculate the potential function \( f(x, y) = x^3 y^2 \) at both endpoints of \( C \).
• Compute the difference \( f(1, 1) - f(-1, 1) \), which gives the work done along \( C \).

This method of using a potential function verifies and simplifies the integration process, highlighting the field's conservative nature, where the same work is achieved irrespective of the path taken across \( C \).