Chapter 6: Problem 428
True or False? Justify your answer with a proof or a counter example. For vector field \(\mathbf{F}(x, y)=P(x, y) \mathbf{i}+Q(x, y) \mathbf{j}, \quad\) if \(P_{y}(x, y)=Q_{x}(x, y)\) in open region \(D\) then \(\int_{\partial D} P d x+Q d y=0\).
Short Answer
Expert verified
True; by Green's Theorem, the integral is zero if \(P_y = Q_x\).
Step by step solution
01
Understand the Given Condition
The statement provides a condition for the vector field \(\mathbf{F}(x, y)=P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}\). This condition is \(P_y(x, y) = Q_x(x, y)\). This equality suggests that the vector field has some specific property, namely it satisfies the criteria for being a conservative vector field if \(\mathbf{F}\) is defined on a simply-connected region.
02
Recognize the Implication of the Condition
The condition \(P_y = Q_x\) is specifically the criteria for a vector field \(\mathbf{F}\) to be conservative in a simply-connected domain. This means \(\mathbf{F}\) can be expressed as the gradient of a scalar potential function. If \(\mathbf{F}\) is conservative, the line integral of \(\mathbf{F}\) over any closed curve in the domain is zero.
03
Apply Green's Theorem
Green's Theorem states that \(\int_{\partial D} P \, dx + Q \, dy = \int\int_D (Q_x - P_y) \, dA\). In the given problem, \(P_y = Q_x\), so the double integral becomes zero: \(\int\int_D (Q_x - P_y) \, dA = 0\).
04
Conclude with the Theorem's Result
Since the area integral equals zero, by Green's Theorem, the closed line integral around the boundary \(\partial D\) must also equal zero. Therefore, \(\int_{\partial D} P \, dx + Q \, dy = 0\).
05
Verify Conclusion Against Original Question
The problem statement claims indeed that for the given condition \(P_y = Q_x\), the line integral over the boundary \(\partial D\) is zero. This holds true based on the application of Green's Theorem.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Conservative Vector Field
A vector field \(\mathbf{F}(x, y) = P(x, y) \mathbf{i} + Q(x, y) \mathbf{j}\)is termed conservative if it can be expressed as the gradient of a scalar potential function. This property indicates that the work done in moving along a path in this field only depends on the endpoints of the path, not the specific path taken. For a vector field to be conservative in a simply-connected domain:
- The partial derivatives of its components must satisfy the condition \(P_y = Q_x\).
- The field can be integrated over any closed path, resulting in a value of zero.
Line Integral
Line integrals extend the concept of an integral to functions along a curve, rather than straightforwardly across an interval. They are particularly useful in physics and engineering:
- They compute work performed by a force field in moving an object along a path.
- Their evaluation depends on the parametrization of the curve.
Scalar Potential Function
A scalar potential function, \(\phi(x, y)\), is a scalar field related to a conservative vector field, which provides a deep insight into the nature of the field:
- It can be seen as a function \(\mathbf{F}\) where \(abla \phi = \mathbf{F}\), effectively describing how the field varies in space.
- For a conservative vector field, the potential function is found by solving \(abla \phi = P\mathbf{i} + Q\mathbf{j}\), ensuring \(\phi_x = P\) and \(\phi_y = Q\).
- It simplifies the evaluation of integrals, turning a complex path integral into a straightforward difference of potential values: \(\phi(B) - \phi(A)\) over endpoints \(A\) and \(B\).