If \( R \) is a closed region in the \( xy \)-plane bounded by a simple closed curve \( C \), and if \( M(x,y) \) and \( N(x,y) \) are continuous functions of \( x \) and \( y \) having continuous derivatives in \( R \), then:
\[ \oint\limits_{C} M\,dx + N\,dy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx\,dy \]where \( C \) is traversed in the positive direction (counterclockwise).
Visual Interpretation
Green's Theorem establishes a relationship between:
- Line integral around a simple closed curve \( C \)
- Double integral over the plane region \( R \) bounded by \( C \)
It connects the circulation around the boundary with the sum of circulations inside the region.
Key Points
- The theorem applies to simply-connected regions (no holes)
- The curve \( C \) must be piecewise smooth
- Positive direction means counterclockwise traversal
- Special cases:
- When \( M = -y \) and \( N = x \), gives area: \( A = \frac{1}{2}\oint_C x\,dy - y\,dx \)
- When \( \frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \), the integral is zero (conservative field)
Physical Interpretation
Green's Theorem has important physical meanings in:
- Fluid dynamics: Relates the macroscopic circulation around a boundary to the microscopic rotation inside
- Electromagnetism: Connects line integrals of fields to flux through surfaces
- Area calculations: Provides a method to compute areas using boundary integrals
Example 1: Verify Green's theorem for \(\oint\limits_{C} (y-\sin x)dx + \cos x dy\) where \(C\) is the triangle bounded by \(y = 0\), \(x = \frac{π}{2}\), and \(y = \frac{2x}{π}\).
Solution:From Green's Theorem:
\[ \oint\limits_{C} M dx + N dy = \iint\limits_{R} \left( \frac{∂N}{∂x} - \frac{∂M}{∂y} \right) dx dy \]Where:
\[ M = y - \sin x \quad \text{and} \quad N = \cos x \] \[ \frac{∂M}{∂y} = 1 \quad \text{and} \quad \frac{∂N}{∂x} = -\sin x \]The region \(R\) has limits:
\[ x \in [0, π/2], \quad y \in [0, 2x/π] \]Compute the double integral:
\[ \iint\limits_{R} (-\sin x - 1) dx dy = -\int_{0}^{π/2} \int_{0}^{2x/π} (1 + \sin x) dy dx \] \[ = -\frac{2}{π} \int_{0}^{π/2} x(1 + \sin x) dx \] \[ = -\frac{2}{π} \left[ \frac{x^2}{2} + (-x \cos x + \sin x) \right]_0^{π/2} \] \[ = -\frac{2}{π} \left( \frac{π^2}{8} + 1 \right) = -\left( \frac{π}{4} + \frac{2}{π} \right) \]We'll evaluate the line integral directly along the three segments:
Segment 1: Along y=0 from x=0 to x=π/2
\[ y = 0, dy = 0 \] \[ \int_{0}^{π/2} (0 - \sin x)dx + \cos x(0) = -\int_{0}^{π/2} \sin x dx = -1 \]Segment 2: Along x=π/2 from y=0 to y=1
\[ x = π/2, dx = 0 \] \[ \int_{0}^{1} (y - 1)(0) + 0 \cdot dy = 0 \]Segment 3: Along y=2x/π from (π/2,1) to (0,0)
\[ dy = \frac{2}{π}dx \] \[ \int_{π/2}^{0} \left(\frac{2x}{π} - \sin x\right)dx + \cos x \left(\frac{2}{π}dx\right) \] \[ = \int_{π/2}^{0} \left(\frac{2x}{π} - \sin x + \frac{2}{π}\cos x\right)dx = \frac{π}{4} - 1 - \frac{2}{π} \]Summing all segments:
\[ -1 + 0 + \left(\frac{π}{4} - 1 - \frac{2}{π}\right) = -\left(\frac{π}{4} + \frac{2}{π}\right) \]Verification Complete
\[ \text{Both methods yield: } \boxed{ -\left( \frac{π}{4} + \frac{2}{π} \right) } \]Green's theorem is verified for this case.
Geometric Interpretation
The negative result indicates the net clockwise circulation around the triangle when viewing the vector field \(\vec{F} = (y-\sin x)\hat{i} + \cos x\hat{j}\).
The theorem successfully relates this boundary circulation to the sum of microscopic rotations (curl) inside the region.
Example 2: Verify Green's theorem for \(\oint\limits_{C} (xy + y^2)dx + x^2 dy\) where \(C\) is bounded by \(y = x\) and \(y = x^2\).
Solution: \[ M = xy + y^2, \quad N = x^2 \]The curves intersect when \(x = x^2\), i.e., at \(x = 0\) and \(x = 1\).
\[ \oint\limits_{C} M dx + N dy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx dy \] \[ \frac{\partial N}{\partial x} = 2x, \quad \frac{\partial M}{\partial y} = x + 2y \] \[ \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} = 2x - (x + 2y) = x - 2y \]The region \(R\) has limits:
\[ x \in [0, 1], \quad y \in [x^2, x] \] \[ \iint\limits_{R} (x - 2y) dx dy = \int_{0}^{1} \int_{x^2}^{x} (x - 2y) dy dx \] \[ = \int_{0}^{1} \left[ xy - y^2 \right]_{y=x^2}^{y=x} dx \] \[ = \int_{0}^{1} \left( (x^2 - x^2) - (x^3 - x^4) \right) dx \] \[ = \int_{0}^{1} (x^4 - x^3) dx = \left[ \frac{x^5}{5} - \frac{x^4}{4} \right]_0^1 = \frac{1}{5} - \frac{1}{4} = -\frac{1}{20} \]We'll evaluate the line integral directly along two segments:
Segment 1: Along y=x² from (0,0) to (1,1)
\[ y = x^2, \quad dy = 2x dx \] \[ \int_{0}^{1} \left( x(x^2) + (x^2)^2 \right) dx + x^2 (2x dx) \] \[ = \int_{0}^{1} (x^3 + x^4 + 2x^3) dx = \int_{0}^{1} (3x^3 + x^4) dx \] \[ = \left[ \frac{3x^4}{4} + \frac{x^5}{5} \right]_0^1 = \frac{3}{4} + \frac{1}{5} = \frac{19}{20} \]Segment 2: Along y=x from (1,1) to (0,0)
\[ y = x, \quad dy = dx \] \[ \int_{1}^{0} \left( x(x) + x^2 \right) dx + x^2 dx = \int_{1}^{0} (x^2 + x^2 + x^2) dx \] \[ = \int_{1}^{0} 3x^2 dx = \left[ x^3 \right]_1^0 = -1 \]Summing both segments:
\[ \frac{19}{20} - 1 = -\frac{1}{20} \]Verification Complete
\[ \text{Both methods yield: } \boxed{ -\frac{1}{20} } \]Green's theorem is verified for this case.
Physical Interpretation
The negative value indicates net clockwise circulation around the boundary. The theorem successfully relates this boundary circulation to the sum of microscopic rotations (curl) inside the region between the parabola and the line.
Example 3: Verify Green's theorem for \(\oint\limits_{C} (x^2 - 2xy)dx + (x^2 y + 3)dy\) where \(C\) is the boundary of the region defined by \(y^2 = 8x\) and \(x = 2\).
Solution:Given:
\[ M = x^2 - 2xy, \quad N = x^2 y + 3 \] \[ \frac{\partial M}{\partial y} = -2x, \quad \frac{\partial N}{\partial x} = 2xy \]Apply Green's Theorem:
\[ \oint\limits_{C} M dx + N dy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dx dy = \iint\limits_{R} (2xy + 2x) dx dy \]The region \(R\) is bounded by \(y^2 = 8x\) and \(x = 2\):
\[ x \in [0, 2], \quad y \in [-\sqrt{8x}, \sqrt{8x}] \]Compute the double integral:
\[ \int_{0}^{2} \int_{-\sqrt{8x}}^{\sqrt{8x}} (2xy + 2x) dy dx \] \[ = \int_{0}^{2} \left[ xy^2 + 2xy \right]_{-\sqrt{8x}}^{\sqrt{8x}} dx \] \[ = \int_{0}^{2} \left( (8x^2 + 2x\sqrt{8x}) - (8x^2 - 2x\sqrt{8x}) \right) dx \] \[ = \int_{0}^{2} 4x\sqrt{8x} dx = 8\sqrt{2} \int_{0}^{2} x^{3/2} dx \] \[ = 8\sqrt{2} \left[ \frac{2}{5} x^{5/2} \right]_0^2 = 8\sqrt{2} \cdot \frac{2}{5} \cdot 2^{5/2} = \frac{128}{5} \]The boundary \(C\) consists of two parts:
Part 1: Along parabola \(y^2 = 8x\) from (2,4) to (2,-4)
\[ x = 2, \quad dx = 0 \] \[ \int_{4}^{-4} (4 - 4y)(0) + (4y + 3) dy = \int_{4}^{-4} (4y + 3) dy \] \[ = \left[ 2y^2 + 3y \right]_4^{-4} = (32 - 12) - (32 + 12) = -24 \]Part 2: Along line \(x=2\) from (2,-4) to (2,4)
\[ \text{(This would be the reverse of Part 1, giving } +24\text{)} \]However, we must parameterize the parabola properly:
\[ \text{Let } y = \sqrt{8x} \text{ (upper half)} \quad \text{and} \quad y = -\sqrt{8x} \text{ (lower half)} \]Proper evaluation gives the same result as Green's theorem: \(\frac{128}{5}\)
\[ \text{Green's theorem yields: } \boxed{ \dfrac{128}{5} } \]The direct line integral evaluation requires proper parameterization of both halves of the parabola to match this result.
Note on Evaluation
The original solution contained some sign errors in the direct evaluation. The correct approach using Green's theorem gives \(\frac{128}{5}\) as shown above. The direct line integral method requires careful parameterization of both the upper and lower halves of the parabola \(y^2 = 8x\) from \(x=0\) to \(x=2\), plus the vertical line segment at \(x=2\).
Example 4: Evaluate \(\oint\limits_{C} (3x^2 - 8y^2)dx + (4y - 6xy)dy\) where \(C\) encloses the region bounded by \(y = \sqrt{x}\) and \(y = x^2\).
Solution: \[ M = 3x^2 - 8y^2 \\ N = 4y - 6xy \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(4y - 6xy) = -6y \\ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2 - 8y^2) = -16y \] \[ \oint\limits_{C} Mdx + Ndy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dxdy \\ = \iint\limits_{R} (-6y - (-16y)) dxdy \\ = \iint\limits_{R} 10y \, dxdy \]The curves \(y = \sqrt{x}\) and \(y = x^2\) intersect when:
\[ \sqrt{x} = x^2 \Rightarrow x = 0 \text{ or } x = 1 \]For \(x \in [0,1]\), \(y\) ranges from \(x^2\) to \(\sqrt{x}\)
\[ \int_{0}^{1} \int_{x^2}^{\sqrt{x}} 10y \, dydx \\ = 10 \int_{0}^{1} \left[ \frac{y^2}{2} \right]_{x^2}^{\sqrt{x}} dx \\ = 5 \int_{0}^{1} (x - x^4) dx \\ = 5 \left[ \frac{x^2}{2} - \frac{x^5}{5} \right]_0^1 \\ = 5 \left( \frac{1}{2} - \frac{1}{5} \right) \\ = 5 \left( \frac{5}{10} - \frac{2}{10} \right) \\ = 5 \times \frac{3}{10} = \frac{3}{2} \]Example 5: Evaluate \(\oint\limits_{C} (2x^2 - y^2)dx + (x^2 + y^2)dy\) where \(C\) is the boundary of the area enclosed by the x-axis and upper half of the circle \(x^2 + y^2 = a^2\).
Solution: \[ M = 2x^2 - y^2 \\ N = x^2 + y^2 \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(x^2 + y^2) = 2x \\ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(2x^2 - y^2) = -2y \] \[ \oint\limits_{C} Mdx + Ndy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dxdy \\ = \iint\limits_{R} (2x - (-2y)) dxdy \\ = \iint\limits_{R} (2x + 2y) dxdy \\ = 2\iint\limits_{R} (x + y) dxdy \]The region \(R\) is the upper semicircle:
\[ x \in [-a, a], \quad y \in [0, \sqrt{a^2 - x^2}] \]We can evaluate the integral in polar coordinates for simplicity:
\[ x = r\cos\theta, \quad y = r\sin\theta \\ dxdy = r dr d\theta \\ \text{Limits: } r \in [0, a], \theta \in [0, \pi] \] \[ 2\int_{0}^{\pi} \int_{0}^{a} (r\cos\theta + r\sin\theta) r dr d\theta \\ = 2\int_{0}^{\pi} \int_{0}^{a} r^2 (\cos\theta + \sin\theta) dr d\theta \\ = 2\int_{0}^{\pi} (\cos\theta + \sin\theta) d\theta \int_{0}^{a} r^2 dr \\ = 2\left[\sin\theta - \cos\theta\right]_0^{\pi} \cdot \left[\frac{r^3}{3}\right]_0^a \\ = 2\left[(0 - (-1)) - (0 - 1)\right] \cdot \frac{a^3}{3} \\ = 2(1 + 1) \cdot \frac{a^3}{3} = \frac{4a^3}{3} \]Example 6: Evaluate \(\oint\limits_{C} (x^2 - \cosh y)dx + (y + \sin x)dy\) where \(C\) is the boundary of the rectangle \(0 \le x \le \pi\), \(0 \le y \le 1\).
Solution: \[ M = x^2 - \cosh y \\ N = y + \sin x \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(y + \sin x) = \cos x \\ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(x^2 - \cosh y) = -\sinh y \] \[ \oint\limits_{C} Mdx + Ndy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dxdy \\ = \iint\limits_{R} (\cos x - (-\sinh y)) dxdy \\ = \iint\limits_{R} (\cos x + \sinh y) dxdy \]The region \(R\) is the rectangle:
\[ x \in [0, \pi], \quad y \in [0, 1] \] \[ \int_{0}^{\pi} \int_{0}^{1} (\cos x + \sinh y) dy dx \\ = \int_{0}^{\pi} \left[ y\cos x + \cosh y \right]_{0}^{1} dx \\ = \int_{0}^{\pi} \left( \cos x + \cosh 1 - (0 + \cosh 0) \right) dx \\ = \int_{0}^{\pi} (\cos x + \cosh 1 - 1) dx \\ = \left[ \sin x + x(\cosh 1 - 1) \right]_0^{\pi} \\ = (0 + \pi(\cosh 1 - 1)) - (0 + 0) \\ = \pi(\cosh 1 - 1) \]Example 7: Evaluate \(\oint\limits_{C} (x^2 - \cosh y)dx + (y + \sin x)dy\) where \(C\) is the boundary of the rectangle \(0 \le x \le \pi\), \(0 \le y \le 1\).
Solution: \[ M = 3x^2 - 8y^2 \\ N = 4y - 6xy \] \[ \frac{\partial N}{\partial x} = \frac{\partial}{\partial x}(4y - 6xy) = -6y \\ \frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(3x^2 - 8y^2) = -16y \] \[ \oint\limits_{C} Mdx + Ndy = \iint\limits_{R} \left( \frac{\partial N}{\partial x} - \frac{\partial M}{\partial y} \right) dxdy \\ = \iint\limits_{R} (-6y - (-16y)) dxdy \\ = \iint\limits_{R} 10y \, dxdy \]The triangular region \(R\) has boundaries:
\[ x \in [0, 1], \quad y \in [0, 1-x] \] \[ \int_{0}^{1} \int_{0}^{1-x} 10y \, dy dx \\ = 10 \int_{0}^{1} \left[ \frac{y^2}{2} \right]_{0}^{1-x} dx \\ = 5 \int_{0}^{1} (1 - x)^2 dx \\ = 5 \left[ -\frac{(1-x)^3}{3} \right]_0^1 \\ = 5 \left( 0 - (-\frac{1}{3}) \right) \\ = \frac{5}{3} \]