Problem 20 - Advanced Engineering Mathematics

Question

Show that the given integral is independent of the path. Evaluate. $$ \int_{(0,0,0)}^{(1,1,1)} 2 x d x+3 y^{2} d y+4 z^{3} d z $$

Step-by-Step Solution

Answer

The integral is path-independent and evaluates to 3.

Step 1: Check if the vector field is conservative

For the integral \( \int 2x \, dx + 3y^2 \, dy + 4z^3 \, dz \), identify the components of the vector field as \( \mathbf{F} = (2x, 3y^2, 4z^3) \). A vector field is conservative if it can be expressed as the gradient of a scalar function. Check if the mixed partial derivatives are equal: \( \frac{\partial (3y^2)}{\partial x} = \frac{\partial (2x)}{\partial y} = 0 \), \( \frac{\partial (4z^3)}{\partial y} = \frac{\partial (3y^2)}{\partial z} = 0 \), and \( \frac{\partial (4z^3)}{\partial x} = \frac{\partial (2x)}{\partial z} = 0 \). All conditions are satisfied, so the vector field is conservative, and the integral is path-independent.

Step 2: Find the potential function

Since the vector field is conservative, there exists a potential function \( f(x, y, z) \) such that \( abla f = \mathbf{F} \). To find \( f \), integrate each component with respect to its variable: - Integrate \( 2x \) with respect to \( x \) to get \( f(x, y, z) = x^2 + g(y, z) \).- Integrate \( 3y^2 \) with respect to \( y \) to get \( f(x, y, z) = x^2 + y^3 + h(z) \).- Integrate \( 4z^3 \) with respect to \( z \) to get \( f(x, y, z) = x^2 + y^3 + z^4 + C \).

Step 3: Evaluate the potential function at the endpoints

The value of the integral is given by the difference of the potential function values at the endpoints. Calculate \( f(0, 0, 0) = 0^2 + 0^3 + 0^4 = 0 \) and \( f(1, 1, 1) = 1^2 + 1^3 + 1^4 = 1 + 1 + 1 = 3 \).

Step 4: Compute the integral using the potential function values

Since the integral is path-independent and determined by the potential function, calculate the integral as \( f(1, 1, 1) - f(0, 0, 0) = 3 - 0 = 3 \). Thus, the integral evaluates to 3.

Key Concepts

Conservative Vector Fields

In vector calculus, a vector field is considered conservative if it originates from a potential function. This means that the vector field can be expressed as the gradient of a scalar function.

One of the key characteristics of conservative vector fields is the fact that they are "path-independent." This means the value of an integral of a conservative vector field between two points only depends on those end points, not on the actual path taken.

• To determine whether a vector field is conservative, we check that the mixed partial derivatives of its components are equal to each other.
• If all conditions for mixed partial derivatives being equal are satisfied, then the vector field is confirmed to be conservative.

For example, in the original exercise, the vector field described by components \( \mathbf{F} = (2x, 3y^2, 4z^3) \) has all its mixed partial derivatives equal, thereby showing it's conservative. This makes understanding and solving path-independent integrals simpler and more straightforward.

Potential Function

When a vector field is conservative, there exists a potential function \( f(x, y, z) \) such that its gradient yields the original vector field. Finding this potential function relies on integrating each component of the vector field with respect to its respective variable.

• Start by integrating the first component with respect to one variable, often called \( x \), and include other variables as constants. This helps to construct a portion of the potential function.
• Similarly, integrate the second component with respect to the second variable, here it is \( y \), adding the result to the potential function.
• Finally, integrate the third component with respect to the third variable, \( z \), and combine all results.

In the provided solution, the potential function \( f(x, y, z) \) was found to be \( x^2 + y^3 + z^4 + C \), after integrating each component accordingly. This potential function captures all the characteristics of the vector field, encapsulating the way it forms across space.

Integration of Vector Fields

Integration within the realm of conservative vector fields is streamlined due to its path-independent nature. This integral's evaluation process becomes straightforward by using the potential function values at determined endpoints.

• First, calculate the value of the potential function at the starting point, usually where all components are zero. This tends to simplify the calculation.
• Second, find the potential function's value at the endpoint.
• The final result of the integral is simply the potential function difference between the endpoint and the starting point.

For instance, using the potential function from the exercise, calculated values at the endpoints \( f(0, 0, 0) = 0 \) and \( f(1, 1, 1) = 3 \), we easily find \( 3 - 0 = 3 \). This shows how the potential function simplifies calculating integrals and aids in understanding the essence of vector fields and their interactions at various points.