Problem 24 - Advanced Engineering Mathematics

Question

In Problems, show that the given integral is independent of the path. Evaluate. $$ \int_{(-2,3,1)}^{(0,0,0)} 2 x z d x+2 y z d y+\left(x^{2}+y^{2}\right) d z $$

Step-by-Step Solution

Answer

The integral evaluates to \(-13\).

Step 1: Understand vector field notation

The line integral given is \( \int_{(-2,3,1)}^{(0,0,0)} \, 2 x z \, dx + 2 y z \, dy + (x^2 + y^2) \, dz \). This means the vector field \( \mathbf{F} \) is \( \mathbf{F} = (2xz, 2yz, x^2 + y^2) \).

Step 2: Check if field is conservative

A vector field is conservative if it is the gradient of a scalar potential function. To check, calculate \( abla \times \mathbf{F} \). If \( abla \times \mathbf{F} = \mathbf{0} \), the field is conservative.

Step 3: Calculate curl of the vector field

Compute \( abla \times \mathbf{F} = \left( \frac{\partial}{\partial y}(x^2 + y^2) - \frac{\partial}{\partial z}(2yz), \frac{\partial}{\partial z}(2xz) - \frac{\partial}{\partial x}(x^2 + y^2), \frac{\partial}{\partial x}(2yz) - \frac{\partial}{\partial y}(2xz) \right) \).

Step 4: Calculate individual partial derivatives

Perform the calculations: \( \frac{\partial}{\partial y}(x^2 + y^2) = 2y \), \( \frac{\partial}{\partial z}(2yz) = 2y \), \( \frac{\partial}{\partial z}(2xz) = 2x \), \( \frac{\partial}{\partial x}(x^2 + y^2) = 2x \), \( \frac{\partial}{\partial x}(2yz) = 0 \), and \( \frac{\partial}{\partial y}(2xz) = 0 \).

Step 5: Verify the curl is zero

Each component of the curl is zero: \( 2y - 2y = 0 \), \( 2x - 2x = 0 \), and \( 0 - 0 = 0 \). Thus, \( abla \times \mathbf{F} = \mathbf{0} \), confirming \( \mathbf{F} \) is conservative.

Step 6: Determine potential function if field is conservative

Since \( \mathbf{F} = abla f \), find \( f(x,y,z) \) such that \( abla f = (2xz, 2yz, x^2 + y^2) \). Integrate each component to find a scalar function \( f \).

Step 7: Integrate to find potential function

Integrate: \( \frac{\partial f}{\partial x} = 2xz \) gives \( f(x,y,z) = x^2 z + g(y,z) \). \( \frac{\partial f}{\partial y} = 2yz \) gives \( f(x,y,z) = y^2 z + h(x,z) \). \( \frac{\partial f}{\partial z} = x^2 + y^2 \) gives \( f(x,y,z) = x^2 z + y^2 z + C \).

Step 8: Evaluate the potential difference

Since \( f(x,y,z) = x^2 z + y^2 z \), evaluate \( f(0,0,0) - f(-2,3,1) \). This is \( 0 - ( (-2)^2 \cdot 1 + 3^2 \cdot 1 ) = 0 - (4 + 9) = -13 \).

Key Concepts

Line Integrals

Line integrals are a crucial concept in vector calculus. They allow us to integrate a function along a curve or path in the space. Essentially, this involves summing up a function's values across small segments that make up the path.
Line integrals are used extensively in physics and engineering, especially when dealing with fields like electromagnetism and fluid dynamics. For a vector field \( \mathbf{F} = ( M, N, P ) \), the line integral from a point \( A \) to a point \( B \) along a path \( C \) is given by:\[ \int_C ( M\,dx + N\,dy + P\,dz ) \].
This formula involves breaking down the path into tiny parts, evaluating the field at these parts, and then adding all these tiny contributions together to get the total integral value.

Conservative Vector Fields

Conservative vector fields are those where the line integral between two points depends only on those points, not on the path taken. This characteristic means that the integral of a conservative field over a closed loop is zero. A conservative field is always the gradient of some scalar function known as the potential function.

Key characteristics of conservative vector fields include:

• The field can be expressed as the gradient of a scalar potential function, \( \mathbf{F} = abla f \).
• The curl of the field must be zero (\( abla \times \mathbf{F} = \mathbf{0} \)).
• There is no energy loss or gain when moving between two points in the field, similar to conservative forces like gravity.

To determine if a field is conservative, it is often essential to check if its curl is zero.

Curl of a Vector Field

The curl of a vector field provides a measure of the field's rotational tendency at a point, visualized as the swirling motion around a point. The vector field \( \mathbf{F} \) has a curl defined by the vector \( abla \times \mathbf{F} \). This is calculated by evaluating the cross product of the del operator, \( abla \), and the vector field:
\[ abla \times \mathbf{F} = \left( \frac{\partial P}{\partial y} - \frac{\partial N}{\partial z}, \frac{\partial M}{\partial z} - \frac{\partial P}{\partial x}, \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) \].
For a conservative field, this curl is zero. This characteristic allows us to deduce the existence of a potential function whose gradient will produce the original vector field. Verifying the curl is particularly useful in exercises that require proving the independence of the path, ensuring the field is conservative.

Potential Function

A potential function is a scalar function \( f(x, y, z) \) such that its gradient \( abla f \) equals a given vector field \( \mathbf{F} \). This function is crucial for conservative vector fields as it offers a simpler representation of the field.

Steps to find the potential function involve:

• Identifying the components of the vector field \( \mathbf{F} = ( M, N, P ) \).
• Integrating each component: For example, \( \frac{\partial f}{\partial x} = M \) implies integrating with respect to \( x \).
• Ensuring that the resultant function satisfies the partial derivatives for all components.

A potential function is highly beneficial as it allows for the calculation of line integrals directly through the difference of potentials at endpoints. This substantially simplifies evaluation when dealing with conservative fields.