Problem 8 - Advanced Engineering Mathematics

Question

In Problems, show that the given line integral is independent of the path. Evaluate in two ways: (a) Find a potential function \(\phi\) and then use Theorem \(9.9 .1\), and (b) Use any convenient path between the endpoints of the path. $$ \int_{(-1,1)}^{(0,0)}(5 x+4 y) d x+\left(4 x-8 y^{3}\right) d y $$

Step-by-Step Solution

Answer

The line integral evaluates to \(\frac{7}{2}\) and is independent of the path.

Step 1: Confirm the Line Integral's Independence from the Path

To show that a line integral is independent of the path, verify that the vector field \(\mathbf{F} = (5x + 4y)\mathbf{i} + (4x - 8y^3)\mathbf{j}\) is conservative. Check if the partial derivatives as follows satisfy the condition for a vector field to be conservative:\(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\). Here, \(P = 5x + 4y\) and \(Q = 4x - 8y^3\).

Step 2: Calculate Partial Derivatives

Calculate \(\frac{\partial P}{\partial y} = \frac{\partial}{\partial y}(5x + 4y) = 4\). Calculate \(\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x}(4x - 8y^3) = 4\). Since \(\frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x}\), the vector field is conservative.

Step 3: Find the Potential Function \(\phi\)

To find the potential function \(\phi(x, y)\), integrate \(P\) with respect to \(x\): \(\phi(x, y) = \int (5x + 4y) \, dx = \frac{5x^2}{2} + 4yx + h(y)\). Then differentiate this with respect to \(y\) and set equal to \(Q\), which will give \(h'(y) = -8y^3\). Integrate to find \(h(y) = -2y^4 + C\), so \(\phi(x, y) = \frac{5x^2}{2} + 4xy - 2y^4 + C\).

Step 4: Use Potential Function for Integral

Using the potential function \(\phi(x, y)\), calculate \(\phi(0, 0)\) and \(\phi(-1, 1)\): \(\phi(0, 0) = C\), and \(\phi(-1, 1) = \frac{5(-1)^2}{2} + 4(-1)(1) - 2(1)^4 + C = \frac{5}{2} - 4 - 2 + C = \frac{-7}{2} + C\). Thus, the integral is \(\phi(0, 0) - \phi(-1, 1) = C - (\frac{-7}{2} + C) = \frac{7}{2}\).

Step 5: Evaluate Using a Convenient Path

Choose a straightforward path: first move vertically from \((-1, 1)\) to \((-1, 0)\), then horizontally from \((-1, 0)\) to \((0, 0)\). For the first path, only \(dy\) contributes: \(\int (4x - 8y^3) \ dy = \int_{1}^{0} (4(-1) - 8y^3) \, dy = \int_{1}^{0} (-4 - 8y^3) \, dy = 4 + 2 = 6,xy \text{ parts the integral as } 5 \). For the second path, only \(dx\) contributes: \(\int (5x + 4y) \, dx = \int_{-1}^{0} 5x \, dx = \frac{5(-1)^2}{2} = \frac{5}{2}\). Add them: \(\frac{7}{2}\).

Step 6: Conclusion

Both approaches confirm that the line integral is \(\frac{7}{2}\). Since the vector field is conservative and the integral evaluated is independent of path, both methods give consistent results.

Key Concepts

Conservative Vector Field

A vector field is conservative when its line integrals depend only on the endpoints of the path, not on the path taken. In simpler terms, it doesn't matter how you travel from point A to point B; the work done by a conservative vector field is consistent. This property is crucial in physics as it relates to energy conservation.

To determine if a vector field \( \mathbf{F} \) is conservative, we check whether there exists a potential function \( \phi \) such that \( \mathbf{F} = abla \phi \). In mathematical terms, \( \mathbf{F} \) is conservative if its curl is zero or its components satisfy \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \), where \( P \) and \( Q \) are the components of \( \mathbf{F} \). If this condition holds, the vector field has a potential function, and the line integral is path-independent.

Potential Function

A potential function \( \phi(x, y) \) for a vector field is a scalar function whose gradient gives the vector field itself. Discovering this function can be seen like finding the energy stored in a field. For example, in a gravitational field, the potential function corresponds to gravitational potential energy.

To find a potential function, start by integrating the component \( P \) with respect to \( x \) and then the component \( Q \) with respect to \( y \). Combine these to form the complete potential function \( \phi \). If the partial derivatives satisfy the condition for a conservative field, the integrated result is consistent, rendering the vector field path-independent. Once the potential function is known, evaluating the integral between two points is simply the difference \( \phi \) at these points.

Path Independence

For conservative vector fields, the line integral from one point to another is path-independent. This simply means the end-to-end journey matters, not the particular route taken. Think of it like flying a straight line between two cities versus taking several connecting flights – the change in altitude (or energy expended) would be the same either way.

Path independence assures that by finding a potential function, we can evaluate the integral simply using endpoint values. It assures that intricate path-specific calculations are bypassed. This quality of path independence, owing to its reliance on only endpoints, makes calculations both straightforward and efficient in practice.

Green's Theorem

Green's Theorem bridges the relationships between a line integral around a closed curve and a double integral over the region it encloses. In simple terms, it connects circulation around a path with a notion of 'flux' through the surface.

Mathematically, it states that a line integral of a vector field around a simple, closed curve is equal to the double integral of the curl of the field over the interior of the curve. Formulated as \( \int_{C} P \, dx + Q \, dy = \iint_{R} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA \), it provides an alternate way of understanding and solving line integrals.

Though in our original exercise we don't use Green's Theorem directly, understanding it highlights how different integral concepts interconnect, offering a broader perspective in vector analysis.