# Calculus

## Line Integrals # Green’s Theorem

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

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

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Problems 1-2
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Problems 3-9