Problem 9 - 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. $$ \int_{10,0)}^{(2,8)}\left(y^{3}+3 x^{2} y\right) d x+\left(x^{3}+3 y^{2} x+1\right) d y $$

Step-by-Step Solution

Answer

The line integral is path-independent with a value of 1096.

Step 1: Determine if the vector field is conservative

Identify the components of the given vector field. The vector field \( \mathbf{F} = (M,N) = (y^3 + 3x^2y, x^3 + 3y^2x + 1) \). A vector field is conservative if the partial derivative \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \). Compute these partial derivatives:- \( \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(y^3 + 3x^2y) = 3y^2 + 3x^2 \).- \( \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^3 + 3y^2x + 1) = 3x^2 + 3y^2 \).Since \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the field is conservative, implying the integral is path-independent.

Step 2: Find the potential function \(\phi\)

To find a potential function \(\phi(x, y)\), integrate \(M\) with respect to \(x\), and \(N\) with respect to \(y\).1. \( \phi(x, y) = \int (y^3 + 3x^2y) \, dx = xy^3 + x^3y + g(y) \), where \(g(y)\) is a function of \(y\) only.2. Differentiate \(\phi(x, y)\) with respect to \(y\): \( \frac{\partial \phi}{\partial y} = 3xy^2 + x^3 + g'(y) \).3. Set this equal to \(N\): \( 3xy^2 + x^3 + g'(y) = x^3 + 3y^2x + 1 \).4. Solve for \(g'(y)\) to find that \( g'(y) = 1 \), resulting in \( g(y) = y \).Thus, the potential function is \( \phi(x, y) = xy^3 + x^3y + y \).

Step 3: Apply Theorem 9.9.1

Theorem 9.9.1 states that for a conservative vector field, the line integral is given by the difference in the potential function at the endpoints.Calculate \( \phi(2,8) \):\[ \phi(2,8) = 2 \times 8^3 + 2^3 \times 8 + 8 = 1024 + 64 + 8 = 1096 \].Calculate \( \phi(10,0) \):\[ \phi(10,0) = 10 \times 0^3 + 10^3 \times 0 + 0 = 0 \].The integral is \( \phi(2,8) - \phi(10,0) = 1096 - 0 = 1096 \).

Step 4: Evaluate the integral along a straight path

Choose the straight-line path from \((10,0)\) to \((2,8)\), described by the parametric equations \(x(t) = 10 - 8t\) and \(y(t) = 8t\) for \(t \in [0,1]\).Calculate \( dx \) and \( dy \):- \( dx = -8 \, dt \)- \( dy = 8 \, dt \)Substitute in the integral:\[\int_0^1 \left((8t)^3 + 3(10-8t)^2(8t)\right)(-8) \, dt + \left((10-8t)^3 + 3(8t)^2(10-8t) + 1\right)(8) \, dt.\]After solving, this integral evaluates to 1096, matching the result from using the potential function.

Key Concepts

Conservative Vector Field

A vector field is termed as **conservative** when it satisfies specific conditions, indicating that the work done by the field along any path depends only on the end points, not the path taken. To determine if a vector field \( \mathbf{F} = (M, N) \) is conservative, we verify if the mixed partial derivatives are equal: \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \).
This condition implies that the vector field has a scalar potential function \( \phi \) such that \( \mathbf{F} = abla \phi \). For the given vector field \( M = y^3 + 3x^2y \) and \( N = x^3 + 3y^2x + 1 \), the partial derivatives are \( 3y^2 + 3x^2 \) and \( 3x^2 + 3y^2 \) respectively, which are equal.
This confirms the vector field is conservative, providing the foundation for path-independent line integrals.

Line Integral

A **line integral** is a way to integrate functions over a curve in the field, embodying essential applications in physics and engineering. For a function in a conservative vector field, the line integral between two points can be computed simply using the potential function \( \phi \).
Given our vector field \( \mathbf{F} \), the line integral along a path \( C \) can be found using:

• The formula \( \int_C \mathbf{F} \cdot d\mathbf{r} \) reflects the accumulation of field vectors along the path.
• In a conservative field, this equates to \( \phi(B) - \phi(A) \), where \( A \) and \( B \) are the endpoints of path \( C \).

This links closely to our scenario where the evaluated integral has endpoints at \((10,0)\) and \((2,8)\). By finding the potential values at these endpoints, we discovered the line integral is path-independent and equals \( 1096 \).

Potential Function

A **potential function** \( \phi(x, y) \) is a scalar function whose gradient yields the original vector field. Finding \( \phi \) involves integrating the components of the vector field.
For our exercise, integrating the component \( M \) yields \( \phi(x, y) = xy^3 + x^3y + g(y) \), where \( g(y) \) is a function only of \( y \).

• Differentiating \( \phi(x, y) \) with respect to \( y \) gives \( \frac{\partial \phi}{\partial y} = 3xy^2 + x^3 + g'(y) \).
• Equating this with \( N \) gives us \( g'(y) = 1 \), leading to \( g(y) = y \).
• This constructs the potential function: \( \phi(x, y) = xy^3 + x^3y + y \).

The potential function plays a crucial role in confirming the vector field’s conservativeness and simplifying the evaluation of line integrals by replacing complex path calculations with endpoint evaluations.

Path Independence

The notion of **path independence** is a profound characteristic of conservative vector fields. For line integrals within such fields, the integral is the same irrespective of the path chosen between two points.
In other words, the work done by the field from point \( A \) to point \( B \) is identical regardless of whether the path is direct, circular, or convoluted. This simplifies calculations significantly. It enables us to substitute complex paths with any convenient or simple path when computing integrals.

Example from Exercise

The exercise illustrated this with the line integral from \( (10,0) \) to \( (2,8) \). Calculations using a straight-line path matched those using the potential function derived, both yielding \( 1096 \). This confirms that in a conservative field, the specific trajectory of the path makes no difference as long as the start and end points remain the same.