Verify The Conclusion Of Green's Theorem By Evaluating Both Sides Of The Equation For The Field F= -2yi+2xj.
Green's theorem establishes a fundamental relationship between a line integral around a simple closed curve in the plane and a double integral over the region it encloses. Specifically, it states that for a continuously differentiable vector field \(\mathbf{F} = P\mathbf{i} + Q\mathbf{j}\), the circulation around the boundary is equal to the sum of the curl over the region:
\[
\oint{C} P\,dx + Q\,dy = \iint{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA
\]
In this article, we aim to verify this theorem explicitly for the vector field \(\mathbf{F} = -2 y \mathbf{i} + 2 x \mathbf{j}\), by calculating both the line integral around the boundary and the double integral over the region \(D\). This process not only confirms the validity of Green's theorem for this specific field but also demonstrates the method of applying the theorem in practice.
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Understanding the Vector Field and Its Components
Components of the Field \(\mathbf{F}\)
The given vector field is:
\[
\mathbf{F} = -2 y \mathbf{i} + 2 x \mathbf{j}
\]
which implies:
- \(P(x, y) = -2 y\)
- \(Q(x, y) = 2 x\)
Understanding these components is crucial for both the line integral and the double integral calculations.
Region of Integration \(D\)
To evaluate the integrals, we must specify the region \(D\). A common choice that simplifies calculations and illustrates Green's theorem well is the unit square:
\[
D: \{ (x, y) \mid 0 \leq x \leq 1, \quad 0 \leq y \leq 1 \}
\]
This choice is convenient and illustrative, but the process can be generalized to other regions as long as the boundary is a simple, closed, positively oriented curve.
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Calculating the Line Integral \(\oint_{C} P\,dx + Q\,dy\)
Parameterizing the Boundary Curve \(C\)
The boundary \(C\) of the region \(D\) (the unit square) consists of four line segments:
- Bottom edge: from \((0,0)\) to \((1,0)\)
- Right edge: from \((1,0)\) to \((1,1)\)
- Top edge: from \((1,1)\) to \((0,1)\)
- Left edge: from \((0,1)\) to \((0,0)\)
Each segment can be parameterized individually.
Evaluating the Line Integral on Each Segment
The line integral is:
\[
\oint{C} P\, dx + Q\, dy = \sum{i=1}^{4} \int{Ci} P\, dx + Q\, dy
\]
Let's evaluate each:
Segment 1: Bottom edge \((0,0) \to (1,0)\)
- Parameterization: \(x = t, y = 0\), with \(t \in [0,1]\)
- Differential: \(dx = dt, dy = 0\)
- \(P = -2 y = 0\)
- \(Q = 2 x = 2 t\)
Integral:
\[
\int{0}^{1} P\, dx + Q\, dy = \int{0}^{1} 0 \cdot dt + 2 t \cdot 0 = 0
\]
Segment 2: Right edge \((1,0) \to (1,1)\)
- Parameterization: \(x=1, y= t\), with \(t \in [0,1]\)
- Differential: \(dx=0, dy=dt\)
- \(P = -2 y = -2 t\)
- \(Q= 2 x= 2\)
Integral:
\[
\int{0}^{1} P\, dx + Q\, dy = \int{0}^{1} (-2 t) \cdot 0 + 2 \cdot dt = \int_{0}^{1} 2\, dt = 2
\]
Segment 3: Top edge \((1,1) \to (0,1)\)
- Parameterization: \(x= t, y= 1\), \(t \in [1,0]\) (or reverse)
- To maintain positive orientation (counterclockwise), we set \(t\) from 1 to 0:
\(x= t, y=1\), with \(t \in [1,0]\)
- Differential: \(dx= dt, dy=0\)
- \(P= -2 y= -2 \cdot 1= -2\)
- \(Q= 2 x= 2 t\)
Integral:
\[
\int{t=1}^{0} P\, dx + Q\, dy = \int{1}^{0} (-2) \cdot dt + 2 t \cdot 0 = \int_{1}^{0} -2\, dt = -2 (0 - 1) = 2
\]
(Note: Reversing limits introduces a sign change, so the integral is positive 2.)
Segment 4: Left edge \((0,1) \to (0,0)\)
- Parameterization: \(x=0, y= t\), \(t \in [1,0]\)
- Differential: \(dx=0, dy= dt\)
- \(P= -2 y= -2 t\)
- \(Q= 2 x= 0\)
Integral:
\[
\int{1}^{0} P\, dx + Q\, dy = \int{1}^{0} (-2 t)\cdot 0 + 0 \cdot dt= 0
\]
Total line integral:
Adding all segments:
\[
0 + 2 + 2 + 0 = 4
\]
Thus,
\[
\oint_{C} P\, dx + Q\, dy = \boxed{4}
\]
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Calculating the Double Integral \(\iint_{D} \left( \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \right) dA\)
Computing Partial Derivatives
Recall:
- \(P = -2 y\)
- \(Q= 2 x\)
Calculate derivatives:
\[
\frac{\partial Q}{\partial x} = \frac{\partial}{\partial x} (2 x) = 2
\]
\[
\frac{\partial P}{\partial y} = \frac{\partial}{\partial y} (-2 y) = -2
\]
Substitute into the integrand:
\[
\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} = 2 - (-2) = 4
\]
Note: The integrand is constant over \(D\).
Evaluating the Double Integral over \(D\)
Since the integrand is constant (4) over the unit square:
\[
\iint_{D} 4\, dA = 4 \times \text{Area of } D
\]
The area of the unit square:
\[
\text{Area} = 1 \times 1 = 1
\]
Thus,
\[
\iint_{D} \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA = 4 \times 1 = 4
\]
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Verification of Green's Theorem
Having computed both sides, we compare:
- Line integral (circulation): \(4\)
- Double integral (curl over region): \(4\)
Since both are equal, the verification confirms the validity of Green's theorem for the field \(\mathbf{F} = -2 y \mathbf{i} + 2 x \mathbf{j}\) over the unit square.
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Discussion and Generalization
Implications of the Result
This explicit calculation demonstrates how Green's theorem provides a powerful method to evaluate line integrals by converting them into double integrals, often simplifying calculations. The consistent result validates the theoretical foundation and confirms the theorem's correctness in this context.
Extension to Other Regions and Fields
While this example used the unit square for simplicity, the procedure extends to more complex regions, provided the boundary is positively oriented and simple, and the vector field is sufficiently smooth. The key steps involve:
- Parameterizing each boundary segment
- Calculating the line integral for each
- Computing the double integral of the curl over the region
This method emphasizes the importance of understanding both the boundary's geometry and the behavior of
