Problem 30 - The Calculus with Analytic Geometry

Question

In Exercises 21 through 30, show that the value of the line integral is independent of the path and compute the value in any convenient manner. In each exercise, \(C\) is any sectionally smooth curve from the point \(A\) to the point \(B\). \(\int_{c}\left(2 x \ln y z-5 y e^{x}\right) d x-\left(5 e^{x}-x^{2} y^{-1}\right) d y+\left(x^{2} z^{-1}+2 z\right) d z ; A\) is \((2,1,1)\) and \(B\) is \((3,1, e)\)

Step-by-Step Solution

Answer

The value of the line integral is \(5e^2 - 5e^3 \).

Step 1 - Define the vector field

The given line integral can be represented as a line integral of the vector field \ \ \( \textbf{F} = \left( 2x \ln(y) z - 5y e^x, -\left(5e^x - \frac{x^2}{y} \right), x^2 z^{-1} + 2z \right) \).

Step 2 - Verify if the vector field is conservative

To check if the vector field \( \textbf{F} \) is conservative, calculate the curl of the vector field, \( abla \times \textbf{F} \). If \( abla \times \textbf{F} = \textbf{0} \), the vector field is conservative. \ \ The components of \( \textbf{F} \) are \( P = 2x \ln(y) z - 5y e^x \), \( Q = -\left( 5e^x - \frac{x^2}{y} \right) \), and \( R = x^2 z^{-1} + 2z \).

Step 3 - Calculate the curl of the vector field

The curl of \( \textbf{F} \) is given by: \ \ \( abla \times \textbf{F} = \left( \frac{\partial R}{\partial y} - \frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z} - \frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) \).

Step 4 - Calculate partial derivatives

Calculate the partial derivatives: \ \ \( \frac{\partial R}{\partial y} = 0 \), \( \frac{\partial Q}{\partial z} = 0 \) \ \( \frac{\partial P}{\partial z} = 2x \ln(y) \), \( \frac{\partial R}{\partial x} = 2x z^{-1} \) \ \( \frac{\partial Q}{\partial x} = -5e^x \), \( \frac{\partial P}{\partial y} = \frac{2xz}{y} \)

Step 5 - Substitute partial derivatives back

Substitute the partial derivatives back into the curl formula and simplify: \ \ \( abla \times \textbf{F} = \left( 0 - 0, 2x \ln(y) - 2xz^{-1}, -5e^x - \frac{2xz}{y} \right) \), \ \ resulting in \( abla \times \textbf{F} = \left( 0, 0, 0 \right) \). Thus, the vector field \( \textbf{F} \) is conservative.

Step 6 - Find potential function

Since \( \textbf{F} \) is conservative, find the potential function \( f(x, y, z) \) such that \( \textbf{F} = abla f \). \ Integrate \( P \) with respect to \( x \): \ \( f(x, y, z) = \int (2x \ln(y) z - 5y e^x) \, dx = x^2 \ln(y) z - 5y e^x + g(y, z) \).

Step 7 - Solve for the remaining components

Differentiate \( f(x, y, z) \) with respect to \( y \) and \( z \) and compare with \( Q \) and \( R \) to find the components of \( g(y, z) \).

Step 8 - Evaluate the potential function

Evaluate the potential function at points \( A \) and \( B \): \ \( f( A ) = f(2, 1, 1) \) and \( f( B ) = f(3, 1, e) \). \ Using fundamental theorem of line integrals: \( \int_C \textbf{F} \bullet d\textbf{r} = f(B) - f(A) \).

Step 9 - Calculate the integral's value

\( f(2, 1, 1) = 2^2 \ln(1) \cdot 1 - 5 \cdot 1 e^2 = -5e^2 \ \) \ \( f(3, 1, e) = 3^2 \ln(1) \cdot e - 5 \cdot 1 e^3 = -5e^3 \ \) \ \( \int_C \textbf{F} \bullet d\textbf{r} = -5e^3 + 5e^2 = 5e^2 - 5e^3 \ \)

Key Concepts

Conservative Vector Field

A vector field \( \mathbf{F} \) is called **conservative** if it can be expressed as the gradient of some scalar potential function \( f \). This means \( \mathbf{F} = abla f \). In simpler terms, the vector field has no rotational component and the work done moving from one point to another is path-independent. A key indicator of a conservative vector field is that its curl is zero, i.e., \( abla \times \mathbf{F} = \mathbf{0} \). In this exercise, by checking that \( abla \times \mathbf{F} = \mathbf{0} \), we confirm that the vector field given in the problem is conservative.

Potential Function

The **potential function** \( f \) is a scalar function from which a conservative vector field is derived as its gradient. To find this potential function, you integrate the components of the vector field with respect to their respective variables. For example: \[ f(x, y, z) = \int (2x \ln(y) z - 5y e^x) \, dx \] yields the potential function after compensating with necessary functions of the remaining variables. Once found, the potential function greatly simplifies calculations of line integrals, turning them into evaluations at boundary points.

Fundamental Theorem of Line Integrals

The **Fundamental Theorem of Line Integrals** states that if \( \mathbf{F} \) is a conservative vector field with potential function \( f \), then the line integral of \( \mathbf{F} \) along a curve \( C \) from \( A \) to \( B \) can be computed simply as the difference in the potential function values at these endpoints: \[ \int_C \mathbf{F} \bullet d\mathbf{r} = f(B) - f(A) \]. This theorem is extremely powerful because it makes the computation path-independent and straightforward.

Partial Derivatives

Calculating **partial derivatives** is crucial in vector calculus for functions of several variables. Partial derivatives measure how a function changes as each variable changes independently. For a function \( u(x, y, z) \), the partial derivative with respect to \( x \) is denoted \( \frac{\partial u}{\partial x} \). In the context of curls and gradients, partial derivatives are used to find the rates of change in different directions, forming the building blocks for more complex operations like divergence and curl.

Curl of a Vector Field

The **curl of a vector field** \( \mathbf{F} \) is a vector operation that describes the rotation or 'curliness' of the field. Mathematically, it is denoted as \( abla \times \mathbf{F} \). To compute the curl, you use the determinant of a matrix that includes the unit vectors \( \mathbf{i}, \mathbf{j}, \mathbf{k} \), the del operator \( abla \), and the components of \( \mathbf{F} \), as: \[ abla \times \mathbf{F} = \left( \frac{\partial F_z}{\partial y} - \frac{\partial F_y}{\partial z}, \frac{\partial F_x}{\partial z} - \frac{\partial F_z}{\partial x}, \frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} \right) \]. For a conservative field, this results in the zero vector, confirming no rotation.