Question

Find the value of the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ (Hint: If \(\mathbf{F}\) is conservative, the integration may be easier on an alternative path.) $$ \mathbf{F}(x, y)=y e^{x y} \mathbf{i}+x e^{x y} \mathbf{j} $$ (a) \(\mathbf{r}_{1}(t)=t \mathbf{i}-(t-3) \mathbf{j}, \quad 0 \leq t \leq 3\) (b) The closed path consisting of line segments from (0,3) to \((0,0),\) and then from (0,0) to (3,0)

Short answer

The line integration for part a is \(e^3 - 1\) and for part b is 0.

Step-by-step solution

Find the potential function

The function is conservative if it's a gradient of some scalar function \(\phi\). Take the partial derivatives of \(\phi\) with respect to x and y. Set these equal to the corresponding components of \(\mathbf{F}(x, y)\) to form the following system of equations and solve for \(\phi(x, y)\):\n\(\frac{\partial \phi}{\partial x}=y e^{x y}\) and \(\frac{\partial \phi}{\partial y}=x e^{x y}\)\nThe solution to this system gives \(\phi(x, y)=e^{x y}\).

Parameterize the given path for part (a)

First parameterize the given path \(\mathbf{r}_{1}(t)=t \mathbf{i}-(t-3) \mathbf{j}\) to get x(t) = t and y(t) = 3 - t.

Compute the line integral for part (a)

The line integral for this part is:\n\n\(\int_{C} \mathbf{F} \cdot d \mathbf{r} = \int_{0}^{3} \mathbf{F}(r(t)) \cdot r'(t) dt\)\n\nSubstituting the parameterized path and the potential function into the line integral gives:\n\n\(\int_{0}^{3} e^{t(3-t)}(1-(3-t)) dt = e^3 - 1\) .

Use Fundamental Theorem for part (b)

The line integral for path (b) can be computed using the Fundamental Theorem of Line Integrals, as the path goes from (0,3) to (0,0) and then from (0,0) to (3,0). According to the theorem, the line integral over a conservative field between two points a and b is \(\phi(b)-\phi(a)\). Thus, for (b) we have the integral is \(\phi(3,0)-\phi(0,3)\), which gives us \(1 - e^0 = 1 - 1 = 0\).

Key concepts

Understanding Conservative Vector Fields

In calculus, when dealing with vector fields, an important concept for simplifying calculations is to determine whether a vector field is conservative. A conservative vector field is one where the line integral between any two points only depends on the endpoints and not on the path taken.

This means that if you move a particle from point A to point B in a conservative field, the work done by the force field along any path will be the same. In the context of the exercise, the vector field \(\mathbf{F}(x, y)=y e^{x y} \mathbf{i}+x e^{x y} \mathbf{j}\) was found to be conservative since it could be expressed as the gradient of a potential function \(\phi(x, y)\).

To identify conservative fields, it's essential to check for a function whose gradient matches the vector field. It turns identifiying and calculating line integrals into a much simpler task, as we will see with the Fundamental Theorem of Line Integrals.

Parameterization of Paths

Line integrals often require the parameterization of paths to be calculated. Parameterization gives us a way to describe a curve or path using a single variable, which is extremely useful when it comes to integration.

For example, the path given in part (a) of the exercise is parameterized by \(\mathbf{r}_{1}(t)=t \mathbf{i}-(t-3) \mathbf{j}\), defining a line segment between two points as t varies from 0 to 3. The parameter t represents the position on the path in terms of time or another continuous parameter. By substituting the parameterized path into the integral, line integrals become a matter of straightforward calculus.

Fundamental Theorem of Line Integrals

The Fundamental Theorem of Line Integrals is a remarkable tool that simplifies computing line integrals remarkably if the vector field is conservative. This theorem states that if you have a scalar potential function \(\phi\), then the line integral of a gradient of \(\phi\) along a path from point A to point B only depends on the values of \(\phi\) at those endpoints.

Following this theorem, complex line integrals reduce to a simple subtraction: \(\phi\) at the end of the path minus \(\phi\) at the start. In part (b) of the exercise, the line integral over a closed path in a conservative field was found to be zero, emphasising the importance of recognizing conservative fields and applying this powerful theorem.