Let \(R\) be a region in the \(xy\)-plane that is bounded by a closed, piecewise smooth curve \(C,\) and let

\[\mathbf{F} = P\left( {x,y} \right)\mathbf{i} + Q\left( {x,y} \right)\mathbf{j}\]

be a continuous vector function with continuous first partial derivatives \({\large\frac{{\partial P}}{{\partial y}}\normalsize}, {\large\frac{{\partial Q}}{{\partial x}}\normalsize}\) in a some domain containing \(R.\) Then Green’s theorem states that

\[

{\iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right)dxdy} }

= {\oint\limits_C {Pdx + Qdy} ,}

\]

where the symbol \(\oint\limits_C {} \) indicates that the curve (contour) \(C\) is closed and integration is performed counterclockwise around this curve.

If \(Q = x,\) \(P = -y,\) Green’s formula yields:

\[{S = \iint\limits_R {dxdy} }={ \frac{1}{2}\oint\limits_C {xdy – ydx} ,}\]

where \(S\) is the area of the region \(R\) bounded by the contour \(C.\)

We can also write Green’s Theorem in vector form. For this we introduce the so-called curl of a vector field. Let

\[{\mathbf{F} = P\left( {x,y,z} \right)\mathbf{i} }+{ Q\left( {x,y,z} \right)\mathbf{j} }+{ R\left( {x,y,z} \right)\mathbf{k}}\]

be a vector field. Then the curl of the vector field \(\mathbf{F}\) is called the vector denoted by \(\text{rot}\,\mathbf{F}\) or \(\nabla \times \mathbf{F}\), which is equal to

\[ {\text{rot}\,\mathbf{F} = \nabla \times \mathbf{F} } = {\left| {\begin{array}{*{20}{c}} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ P&Q&R \end{array}} \right| } = {{\left( {\frac{{\partial R}}{{\partial y}} – \frac{{\partial Q}}{{\partial z}}} \right)\mathbf{i} } + {\left( {\frac{{\partial P}}{{\partial z}} – \frac{{\partial R}}{{\partial x}}} \right)\mathbf{j} } + {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right)\mathbf{k}.}} \]

In terms of curl, Green’s Theorem can be written as

\[

{\iint\limits_R {\left( \text{rot}\,\mathbf{F} \right) \cdot \mathbf{k}\,dxdy} }

= {\oint\limits_C {\mathbf{F} \cdot d\mathbf{r}} .}

\]

Note that Green’s Theorem is simply Stoke’s Theorem applied to a \(2\)-dimensional plane.

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Using Green’s theorem, evaluate the line integral \(\oint\limits_C {xydx \,+}\) \({\left( {x + y} \right)dy} ,\) where \(C\) is the curve bounding the unit disk \(R.\)### Example 2

Using Green’s formula, evaluate the line integral \(\oint\limits_C {\left( {x – y} \right)dx +}\) \({\left( {x + y} \right)dy},\) where \(C\) is the circle \({x^2} + {y^2} = {a^2}.\)### Example 3

Using Green’s theorem, calculate the integral \(\oint\limits_C {{x^2}ydx – x{y^2}dy}.\) The curve \(C\) is the circle \({x^2} + {y^2} = {a^2}\) (Figure \(1\)), traversed in the counterclockwise direction.### Example 4

Using Green’s formula, evaluate the integral \(\oint\limits_C {\left( {x + y} \right)dx }\) \(-{ \left( {x – y} \right)dy} ,\) where the curve \(C\) is the ellipse \({\large\frac{{{x^2}}}{{{a^2}}}\normalsize} + {\large\frac{{{y^2}}}{{{b^2}}}\normalsize} = 1\) (Figure \(2\)).### Example 5

Using Green’s formula, calculate the line integral \(\oint\limits_C {{y^2}dx \,+}\) \({{{\left( {x + y} \right)}^2}dy}, \) where the contour \(C\) is the triangle \(ABD\) with vertices \(A\left( {a,0} \right),\) \(B\left( {a,a} \right),\) \(D\left( {0,a} \right)\) (Figure \(3\)).### Example 6

Using Green’s theorem, evaluate the line integral \(\oint\limits_C {\left( {y – {x^2}} \right)dx }\) \(-{ \left( {x + {y^2}} \right)dy},\) where the contour \(C\) encloses the sector of the circle with radius \(a\) lying in the first quadrant (Figure \(4\)).### Example 7

Calculate the integral \(\oint\limits_C {\large\frac{{dx – dy}}{{x + y}}\normalsize} \) using Green’s theorem. The contour \(C\) is the boundary of the square with the vertices \(A\left( {1,0} \right),\) \(B\left( {0,1} \right),\) \(D\left( {-1,0} \right),\) \(E\left( {0,-1} \right)\) (Figure \(5\)).### Example 8

Calculate the line integral \(\oint\limits_C {\sqrt {{x^2} + {y^2}} dx }\) \(+{ y\big[ {xy \,+}}\) \({{\ln \left( {x + \sqrt {{x^2} + {y^2}} } \right)} \big]dy} \) using Green’s theorem. The contour of integration \(C\) is the circle \({x^2} + {y^2} = {a^2}\) (Figure \(7\)).### Example 9

Calculate the area of the region \(R\) bounded by the astroid \(x = a\,{\cos ^3}t,\) \(y = a\,{\sin ^3}t,\) \(0 \le t \le 2\pi .\)### Example 1.

Using Green’s theorem, evaluate the line integral \(\oint\limits_C {xydx \,+}\) \({\left( {x + y} \right)dy} ,\) where \(C\) is the curve bounding the unit disk \(R.\)Solution.

The components of the vector field are

\[{P\left( {x,y} \right) = xy,\;\;}\kern-0.3pt{Q\left( {x,y} \right) = x + y.}\]

Using the Green’s formula

\[

{\iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} – \frac{{\partial P}}{{\partial y}}} \right)dxdy} }

= {\oint\limits_C {Pdx + Qdy} }

\]

we transform the line integral into the double integral:

\[

{I = \oint\limits_C {xydx + \left( {x + y} \right)dy} }

= {\iint\limits_R {\left( {\frac{{\partial \left( {x + y} \right)}}{{\partial x}} – \frac{{\partial \left( {xy} \right)}}{{\partial y}}} \right)dxdy} }

= {\iint\limits_R {\left( {1 – x} \right)dxdy} .}

\]

Converting the double integral into polar coordinates, we have

\[

{I = \int\limits_R {\left( {1 – x} \right)dxdy} }

= {\int\limits_0^{2\pi } {\int\limits_0^1 {\left( {1 – r\cos \theta } \right)rdrd\theta } } }

= {\int\limits_0^{2\pi } {\left[ {\int\limits_0^1 {\left( {r – {r^2}\cos \theta } \right)dr} } \right]d\theta } }

= {\int\limits_0^{2\pi } {\left[ {\left. {\left( {\frac{{{r^2}}}{2} – \frac{{{r^3}}}{3}\cos \theta } \right)} \right|_{r = 0}^1} \right]d\theta } }

= {\int\limits_0^{2\pi } {\left( {\frac{1}{2} – \frac{{\cos \theta }}{3}} \right)d\theta } }

= {\left. {\left( {\frac{\theta }{2} – \frac{{\sin \theta }}{3}} \right)} \right|_0^{2\pi } }={ \pi .}

\]

### Example 2.

Using Green’s formula, evaluate the line integral \(\oint\limits_C {\left( {x – y} \right)dx +}\) \({\left( {x + y} \right)dy},\) where \(C\) is the circle \({x^2} + {y^2} = {a^2}.\)Solution.

First we identify the components of the vector field:

\[{P = x – y,\;\;\;}\kern-0.3pt{Q = x + y}\]

and find the partial derivatives:

\[

{\frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( {x + y} \right)}}{{\partial x}} = 1,\;\;\;}\kern-0.3pt

{\frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {x – y} \right)}}{{\partial x}} = – 1.}

\]

Hence, the line integral can be written in the form

\[

{I }={ \oint\limits_C {\left( {x – y} \right)dx + \left( {x + y} \right)dy} }

= {\iint\limits_R {\left( {1 – \left( { – 1} \right)} \right)dxdy} }

= {2\iint\limits_R {dxdy} .}

\]

In the last expression the double integral \(\iint\limits_R {dxdy} \) is equal numerically to the area of the disk \({x^2} + {y^2} = {a^2},\) which is \(\pi {a^2}.\) Then the integral is

\[{I = 2\iint\limits_R {dxdy} }={ 2\pi {a^2}.}\]