Chapter 9: Problem 22
Show that the given integral is independent of the path. Evaluate. $$ \int_{(1,2,1)}^{(3,4,1)}(2 x+1) d x+3 y^{2} d y+\frac{1}{z} d z $$
Short Answer
Expert verified
The provided integral is independent of the path, and its value is 66.
Step by step solution
01
Identify the Vector Field
The given integral is a line integral over a vector field, which can be represented as \( \mathbf{F} = (2x + 1)\mathbf{i} + 3y^2 \mathbf{j} + \frac{1}{z}\mathbf{k} \) where \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) are unit vectors along the \(x, y, z \) axes respectively.
02
Check for Path Independence Conditions
For the integral to be path independent, the vector field \( \mathbf{F} \) must be conservative, i.e., it should have no curl. We compute the curl of \( \mathbf{F} \) using \( abla \times \mathbf{F} \). The vector field is conservative if \( abla \times \mathbf{F} = \mathbf{0} \).
03
Calculate the Curl
Compute the curl: \[ abla \times \mathbf{F} = \left(\frac{\partial}{\partial y} \left(\frac{1}{z}\right) - \frac{\partial}{\partial z} (3y^2) \right) \mathbf{i} - \left(\frac{\partial}{\partial x} \left(\frac{1}{z}\right) - \frac{\partial}{\partial z} (2x+1) \right) \mathbf{j} + \left(\frac{\partial}{\partial x} (3y^2) - \frac{\partial}{\partial y} (2x+1) \right) \mathbf{k} \]. Compute individual components: - For \(\mathbf{i}\): \( \frac{\partial}{\partial y} (\frac{1}{z}) = 0\text{ and } \frac{\partial}{\partial z} (3y^2) = 0\) so it is 0.- For \(\mathbf{j}\): \( \frac{\partial}{\partial x} (\frac{1}{z}) = 0\text{ and } \frac{\partial}{\partial z} (2x+1)=0\) so it is 0.- For \(\mathbf{k}\): \( \frac{\partial}{\partial x} (3y^2) = 0\text{ and } \frac{\partial}{\partial y} (2x+1)=0\) so it is 0.Thus, \( abla \times \mathbf{F} = \mathbf{0}\), indicating the field is conservative.
04
Find a Potential Function
Since the vector field is conservative, there exists a potential function, \( f(x, y, z) \), such that \( \mathbf{F} = abla f \). This means \( \frac{\partial f}{\partial x} = 2x + 1, \frac{\partial f}{\partial y} = 3y^2, \frac{\partial f}{\partial z} = \frac{1}{z} \). Integrate each partial derivative separately to find \( f(x, y, z) \). - Integrate \( \frac{\partial f}{\partial x} = 2x + 1 \) with respect to \( x \) to get \( f(x, y, z) = x^2 + x + C(y, z) \).- Integrate \( \frac{\partial f}{\partial y} = 3y^2 \) with respect to \( y \) to get \( f(x, y, z) = y^3 + D(x, z) \).- Integrate \( \frac{\partial f}{\partial z} = \frac{1}{z} \) with respect to \( z \) to get \( f(x, y, z) = \ln|z| + E(x, y) \).Combine results: \( f(x, y, z) = x^2 + x + y^3 + \ln|z| + C \) (where \( C \) is a constant).
05
Evaluate the Line Integral Using the Potential Function
The line integral over a conservative field from point \( \mathbf{a} = (1, 2, 1) \) to point \( \mathbf{b} = (3, 4, 1) \) can be evaluated using the potential function as \( f(\mathbf{b}) - f(\mathbf{a}) \).- Compute \( f(3, 4, 1) = 3^2 + 3 + 4^3 + \ln|1| + C = 9 + 3 + 64 + 0 + C = 76 + C \).- Compute \( f(1, 2, 1) = 1^2 + 1 + 2^3 + \ln|1| + C = 1 + 1 + 8 + 0 + C = 10 + C \).- Therefore, \( f(3, 4, 1) - f(1, 2, 1) = (76 + C) - (10 + C) = 66 \).
06
Conclusion
The line integral is independent of the path, and its value is \( \boxed{66} \). The path independence was confirmed by showing that the vector field is conservative, allowing us to use the potential function to evaluate the integral directly.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservative Vector Field
A conservative vector field is a type of field that is significant in vector calculus. It is characterized by specific properties that simplify certain computations, particularly line integrals. Let's break down what it means for a field to be conservative and why it's useful in calculus.
A vector field, represented by \( \mathbf{F} \), is said to be conservative if the line integral of \( \mathbf{F} \) around any closed path is zero. This implies that there exists a single-valued potential function \( f \) such that \( \mathbf{F} = abla f \). In simpler terms, a conservative vector field is the gradient of some scalar function.
Moreover, one of the most practical characteristics of a conservative vector field is that it has zero curl. The curl determines the rotational tendency of a field. Mathematically, this is expressed as \( abla \times \mathbf{F} = \mathbf{0} \).
A vector field, represented by \( \mathbf{F} \), is said to be conservative if the line integral of \( \mathbf{F} \) around any closed path is zero. This implies that there exists a single-valued potential function \( f \) such that \( \mathbf{F} = abla f \). In simpler terms, a conservative vector field is the gradient of some scalar function.
Moreover, one of the most practical characteristics of a conservative vector field is that it has zero curl. The curl determines the rotational tendency of a field. Mathematically, this is expressed as \( abla \times \mathbf{F} = \mathbf{0} \).
- Zero curl implies no net rotation, indicating path independence for integrals over the field.
- A conservative vector field has a path-independent integral, meaning the integral's value depends only on the endpoints of the path, not the specific trajectory.
Line Integral
Line integrals are a crucial concept in both physics and mathematics, providing insights into various properties of fields. In vector calculus, a line integral involves integrating a vector field along a curve or path. Essentially, it's a way to sum up field values along a specific trajectory.
Given a vector field \( \mathbf{F} \) and a path \( C \), the line integral of \( \mathbf{F} \) along \( C \) is denoted as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This considers both the magnitude and direction of \( \mathbf{F} \) along the path.
When calculating line integrals, you must:
Given a vector field \( \mathbf{F} \) and a path \( C \), the line integral of \( \mathbf{F} \) along \( C \) is denoted as \( \int_C \mathbf{F} \cdot d\mathbf{r} \). This considers both the magnitude and direction of \( \mathbf{F} \) along the path.
When calculating line integrals, you must:
- Parameterize the path \( C \), converting it into a vector function.
- Substitute the path into the expression for \( \mathbf{F} \) and evaluate the dot product.
- Integrate with respect to the parameter of the path.
Potential Function
A potential function is a fundamental concept in vector calculus related to conservative fields. This function provides a simple way to compute line integrals over a conservative vector field.
For a vector field \( \mathbf{F} \) to be conservative, there must exist a scalar potential function \( f \) where \( \mathbf{F} = abla f \). You can think of \( f \) as the antiderivative of \( \mathbf{F} \).
Finding a potential function involves the following steps:
For a vector field \( \mathbf{F} \) to be conservative, there must exist a scalar potential function \( f \) where \( \mathbf{F} = abla f \). You can think of \( f \) as the antiderivative of \( \mathbf{F} \).
Finding a potential function involves the following steps:
- Calculate the partial derivatives of \( f \) that correspond to each component of the vector field \( \mathbf{F} \).
- Integrate these expressions separately to recover \( f \).
- Verify consistency across different variables by checking compatibility conditions.
Path Independence
Path independence is a significant property associated with conservative vector fields. It states that the line integral of a vector field between two points is independent of the path taken.
This property is a direct consequence of the field's conservativeness. Since the curl of a conservative field is zero, the integral over any closed loop is zero. Thus, for any starting point \( A \) and endpoint \( B \), the integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) is the same for any path \( C \) connecting \( A \) and \( B \).
Path independence is integral in simplifying problems involving work done by forces, as it confirms that:
This property is a direct consequence of the field's conservativeness. Since the curl of a conservative field is zero, the integral over any closed loop is zero. Thus, for any starting point \( A \) and endpoint \( B \), the integral \( \int_C \mathbf{F} \cdot d\mathbf{r} \) is the same for any path \( C \) connecting \( A \) and \( B \).
Path independence is integral in simplifying problems involving work done by forces, as it confirms that:
- The work done only depends on the initial and final states, irrelevant of the path.
- You can compute the integral efficiently using the potential function, by simply evaluating \( f(B) - f(A) \).