Problem 7
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. $$ \int_{(1,2)}^{(3,6)}\left(2 y^{2} x-3\right) d x+\left(2 y x^{2}+4\right) d y $$
Step-by-Step Solution
Verified Answer
The line integral is path-independent, and its value is \( 114 \).
1Step 1: Identify the Vector Field
The given line integral is \( \int_{(1,2)}^{(3,6)}(2y^2 x - 3) \, dx + (2yx^2 + 4) \, dy \). Identify the vector field \( \mathbf{F} = (P, Q) \), where \( P(x, y) = 2y^2 x - 3 \) and \( Q(x, y) = 2yx^2 + 4 \).
2Step 2: Check for Path Independence
The line integral is path-independent if the vector field \( \mathbf{F} \) is conservative. We check if \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). Calculate \( \frac{\partial P}{\partial y} = 4yx \) and \( \frac{\partial Q}{\partial x} = 4xy \). Since they're equal, the field is conservative.
3Step 3: Find the Potential Function \( \phi \)
Since the field is conservative, there exists a potential function \( \phi(x, y) \) such that \( abla \phi = \mathbf{F} \). Integrate \( P \) with respect to \( x \) to get \( \phi(x, y) = \int (2y^2 x - 3) \ dx = y^2 x^2 - 3x + g(y) \), where \( g(y) \) is a function of \( y \).
4Step 4: Integrate for \( g(y) \)
Differentiate \( \phi \) with respect to \( y \): \( \frac{d\phi}{dy} = 2xy^2 + g'(y) \). Set equal to \( Q(x, y) = 2yx^2 + 4 \) to solve for \( g'(y) \). We find \( 2xy^2 + g'(y) = 2yx^2 + 4 \). Therefore, \( g'(y) = 4 \), and \( g(y) = 4y \) after integration.
5Step 5: Complete the Potential Function
Substitute \( g(y) \) back into \( \phi \): \( \phi(x, y) = y^2 x^2 - 3x + 4y \). This is the potential function.
6Step 6: Evaluate the Line Integral Using the Potential Function
According to Theorem 9.9.1, \( \int_C \mathbf{F} \cdot d\mathbf{r} = \phi(3, 6) - \phi(1, 2) \). Compute \( \phi(3, 6) = 6^2 \cdot 3^2 - 3 \cdot 3 + 4 \cdot 6 = 108 + 24 - 9 = 123 \) and \( \phi(1, 2) = 2^2 \cdot 1^2 - 3 \cdot 1 + 4 \cdot 2 = 4 + 8 - 3 = 9 \). Thus, \( \phi(3, 6) - \phi(1, 2) = 123 - 9 = 114 \).
7Step 7: Choose Convenient Path for Direct Evaluation
Select any convenient path, such as a piecewise linear path from \((1,2)\) to \((3,6)\), for direct computation if needed. However, since path-independence is confirmed, the computed value from the potential function suffices.
Key Concepts
Vector FieldsConservative Vector FieldPotential FunctionTheorem 9.9.1
Vector Fields
Vector fields are a way to associate a vector to every point in a space. These vectors can represent things like forces, velocities, or any other directional quantities. In mathematical terms, a vector field in two dimensions is represented as \( \mathbf{F} = (P, Q) \), where \( P(x, y) \) and \( Q(x, y) \) are functions that describe the vector's components in the x and y directions, respectively.
For example, consider the vector field \( \mathbf{F} = (2y^2 x - 3, 2yx^2 + 4) \). Here \( P(x, y) = 2y^2 x - 3 \) is the x-component and \( Q(x, y) = 2yx^2 + 4 \) is the y-component. The behavior of a vector field can be visualized by drawing vectors at various points in the plane, showing how they change from point to point.
This information is crucial when dealing with line integrals, which evaluate the field along a specific path. Understanding vector fields helps us compute these integrals and analyze path dependence.
For example, consider the vector field \( \mathbf{F} = (2y^2 x - 3, 2yx^2 + 4) \). Here \( P(x, y) = 2y^2 x - 3 \) is the x-component and \( Q(x, y) = 2yx^2 + 4 \) is the y-component. The behavior of a vector field can be visualized by drawing vectors at various points in the plane, showing how they change from point to point.
This information is crucial when dealing with line integrals, which evaluate the field along a specific path. Understanding vector fields helps us compute these integrals and analyze path dependence.
Conservative Vector Field
A conservative vector field is an important concept because it describes fields where the line integral is path-independent. This means that for conservative fields, the integral of the field between two points is the same regardless of the path taken. The defining characteristic of a conservative vector field is the existence of a potential function \( \phi \) such that \( abla \phi = \mathbf{F} \).
One way to check if a vector field is conservative involves verifying that the mixed partial derivatives of the field's components are equal, i.e., \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). If they are equal, the field is conservative. For example, in the vector field \( \mathbf{F} = (2y^2 x - 3, 2yx^2 + 4) \), we found that \( \frac{\partial P}{\partial y} = 4yx \) and \( \frac{\partial Q}{\partial x} = 4xy \), confirming the field is conservative.
This characteristic is crucial for simplifying problems involving line integrals, as conservative fields allow the use of potential functions to easily compute these integrals.
One way to check if a vector field is conservative involves verifying that the mixed partial derivatives of the field's components are equal, i.e., \( \frac{\partial P}{\partial y} = \frac{\partial Q}{\partial x} \). If they are equal, the field is conservative. For example, in the vector field \( \mathbf{F} = (2y^2 x - 3, 2yx^2 + 4) \), we found that \( \frac{\partial P}{\partial y} = 4yx \) and \( \frac{\partial Q}{\partial x} = 4xy \), confirming the field is conservative.
This characteristic is crucial for simplifying problems involving line integrals, as conservative fields allow the use of potential functions to easily compute these integrals.
Potential Function
A potential function \( \phi(x, y) \) for a conservative vector field is a scalar function such that its gradient \( abla \phi \) equals the vector field \( \mathbf{F} \). Mathematically, this means \( \frac{\partial \phi}{\partial x} = P(x, y) \) and \( \frac{\partial \phi}{\partial y} = Q(x, y) \).
To find the potential function, integrate the x-component \( P(x, y) \) with respect to \( x \) and the y-component \( Q(x, y) \) with respect to \( y \). In our example, integrating \( P(x, y) = 2y^2 x - 3 \) with respect to \( x \), gives us \( \phi(x, y) = y^2 x^2 - 3x + g(y) \), where \( g(y) \) is any function of \( y \).
To determine \( g(y) \), differentiate \( \phi \) with respect to \( y \), set it equal to \( Q(x, y) \), and solve for \( g'(y) \). The result gives a complete expression for \( \phi \) that can be used to evaluate line integrals easily. In this case, \( \phi(x, y) = y^2 x^2 - 3x + 4y \).
To find the potential function, integrate the x-component \( P(x, y) \) with respect to \( x \) and the y-component \( Q(x, y) \) with respect to \( y \). In our example, integrating \( P(x, y) = 2y^2 x - 3 \) with respect to \( x \), gives us \( \phi(x, y) = y^2 x^2 - 3x + g(y) \), where \( g(y) \) is any function of \( y \).
To determine \( g(y) \), differentiate \( \phi \) with respect to \( y \), set it equal to \( Q(x, y) \), and solve for \( g'(y) \). The result gives a complete expression for \( \phi \) that can be used to evaluate line integrals easily. In this case, \( \phi(x, y) = y^2 x^2 - 3x + 4y \).
Theorem 9.9.1
Theorem 9.9.1 is a powerful tool for evaluating line integrals in conservative vector fields. It states that if \( \mathbf{F} \) is conservative with a potential function \( \phi \), then the line integral of \( \mathbf{F} \) over a curve \( C \) between two points is the difference in the values of \( \phi \) at these points.
This theorem provides an efficient way to evaluate integrals without needing to compute them directly along a path. Instead, we can find \( \phi(a, b) \) and \( \phi(c, d) \) for the endpoints \((a, b)\) and \((c, d)\), and compute \( \phi(c, d) - \phi(a, b) \). For our specific problem, we calculated \( \phi(3, 6) = 123 \) and \( \phi(1, 2) = 9 \).
The difference, \( 123 - 9 = 114 \), is the value of the line integral, illustrating the path independence property of conservative vector fields.
This theorem provides an efficient way to evaluate integrals without needing to compute them directly along a path. Instead, we can find \( \phi(a, b) \) and \( \phi(c, d) \) for the endpoints \((a, b)\) and \((c, d)\), and compute \( \phi(c, d) - \phi(a, b) \). For our specific problem, we calculated \( \phi(3, 6) = 123 \) and \( \phi(1, 2) = 9 \).
The difference, \( 123 - 9 = 114 \), is the value of the line integral, illustrating the path independence property of conservative vector fields.
Other exercises in this chapter
Problem 7
Evaluate the given iterated integral. $$ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1-x^{2}-y^{1}} x y e^{x} d z d x d y $$
View solution Problem 7
Find the surface area of the portion of the sphere \(x^{2}+y^{2}+z^{2}=25\) that is above the region in the first quadrant bounded by the graphs of \(x=0, y=0,4
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Evaluate \(\int_{C}(2 x+y) d x+x y d y\) on the given curve \(C\) between \((-1,2)\) and \((2,5)\). $$ y=x+3 $$
View solution Problem 7
In Problems \(7-16\), find the curl and the divergence of the given vector field. $$ \mathbf{F}(x, y, z)=x z \mathbf{i}+y z \mathbf{j}+x y \mathbf{k} $$
View solution