Problem 128 - Calculus Volume 3

Question

For the following exercises, evaluate the integral using the Fundamental Theorem of Line Integrals. \([\mathrm{T}] \quad\) Evaluate \(\quad \int_{C} \nabla f \cdot d \mathbf{r},\) where \(f(x, y)=x^{2} y-x\) and \(C\) is any path in a plane from (1,2) to (3,2)

Step-by-Step Solution

Answer

The line integral evaluates to 14.

Step 1: Understand the Problem

We need to evaluate the line integral \( \int_{C} abla f \cdot d \mathbf{r} \) where \( C \) is a path from (1,2) to (3,2) and \( f(x, y) = x^{2} y - x \). The Fundamental Theorem of Line Integrals states that for a conservative vector field \( abla f \), the integral \( \int_{C} abla f \cdot d \mathbf{r} \) equals \( f(\mathbf{r}(b)) - f(\mathbf{r}(a)) \) where \( \mathbf{r}(a) \) and \( \mathbf{r}(b) \) are the start and end points of the path, respectively.

Step 2: Compute the Gradient of the Function

Calculate the gradient of \( f(x, y) = x^{2} y - x \). The gradient \( abla f \) is given by the vector of partial derivatives: \[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) = (2xy - 1, x^2) \]

Step 3: Apply the Fundamental Theorem of Line Integrals

Since \( abla f = (2xy - 1, x^2) \) is a conservative vector field, the line integral \( \int_{C} abla f \cdot d \mathbf{r} \) over any path \( C \) from (1, 2) to (3, 2) is simply \( f(3, 2) - f(1, 2) \). Calculate: \[ f(3, 2) = 3^2 \times 2 - 3 = 18 - 3 = 15 \]\[ f(1, 2) = 1^2 \times 2 - 1 = 2 - 1 = 1 \]

Step 4: Compute the Integral Value

Substitute the calculated values into the expression given by the theorem: \[ f(3, 2) - f(1, 2) = 15 - 1 = 14 \] Hence, the value of the line integral is 14.

Key Concepts

Conservative Vector Field

A conservative vector field is a key concept in vector calculus. It describes a vector field that is the gradient of some potential function. This means that the vector field can be expressed as the gradient of a scalar function, say \( f(x, y) \).

• A foundational property of conservative vector fields is that they are path independent. This implies that the line integral of a conservative field will only depend on the endpoints of the path.
• This feature makes them particularly easy to work with, as evaluating an integral over any path reduces to simply computing the difference in potential values at the endpoints.

In our example, we evaluated a line integral for a conservative vector field \( abla f = (2xy - 1, x^2) \). Because this vector field is conservative, the path between the endpoints (1, 2) and (3, 2) does not alter the outcome of the integral. This intrinsic property saves us from needing detailed information about the actual path taken.

Gradient of a Function

The gradient of a function is a vector that contains all its partial derivatives. It points in the direction of the greatest rate of increase of the function and its magnitude represents the rate of that increase.

• For any function \( f(x,y) \), the gradient \( abla f \) is found by computing the partial derivatives \( (\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}) \).

In the exercise given, the function is \( f(x, y) = x^2 y - x \). To find its gradient, we calculate:

• \( \frac{\partial f}{\partial x} = 2xy - 1 \)
• \( \frac{\partial f}{\partial y} = x^2 \)

Thus, the gradient is \( abla f = (2xy - 1, x^2) \). Knowing the gradient allows us to see how the function changes in different directions and is crucial for evaluating line integrals in conservative fields.

Line Integral Evaluation

Evaluating a line integral involves summing up a function's values along a path. In the context of the Fundamental Theorem of Line Integrals, this becomes much simpler with a conservative vector field.With conservative vector fields, the theorem states that:\[\int_{C} abla f \cdot d \mathbf{r} = f( \mathbf{r}(b) ) - f( \mathbf{r}(a) )\]Here's how it applies:

• Identify \( f(x, y) \) and calculate its gradient to verify it's conservative.
• Determine the function values at the endpoints of your path.
• Subtract the start point's function value from the endpoint's function value to find the integral.

For our problem, the endpoints were (1, 2) and (3, 2). By simply computing \( f(3, 2) - f(1, 2) \), we find that the line integral is 14. This approach offers a swift evaluation, bypassing the need to explicitly parametrize and integrate along the path.