Green’s Theorem
Solved Problems
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.
Solution.
Figure 1. We write the components of the vector fields and their partial derivatives:
\[P\left( {x,y} \right) = {x^2}y,\;\; Q\left( {x,y} \right) = - x{y^2},\;\;\ \frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( { - x{y^2}} \right)}}{{\partial x}} = - {y^2},\;\; \frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {{x^2}y} \right)}}{{\partial y}} = {x^2}.\]
Then
\[I = \oint\limits_C {{x^2}ydx - x{y^2}dy} = \iint\limits_R {\left( { - {y^2} - {x^2}} \right)dxdy} = - \iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy} ,\]
where \(R\) is the circle with radius \(a\) centered at the origin. Transforming to polar coordinates, we obtain
\[I = - \iint\limits_R {\left( {{x^2} + {y^2}} \right)dxdy} = - \int\limits_0^{2\pi } {d\theta } \int\limits_0^a {\left( {{r^2}{{\cos }^2}\theta + {r^2}{\sin^2}\theta } \right)rdr} = - \int\limits_0^{2\pi } {d\theta } \int\limits_0^a {{r^3}dr} = - 2\pi \cdot \left[ {\left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^a} \right] = - \frac{{\pi {a^4}}}{2}.\]
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 \[\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\] (Figure \(2\)).
Solution.
Figure 2. We use Green's formula:
\[\oint\limits_C {Pdx + Qdy} = \iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy} .\]
Here
\[P = x + y,\;\; Q = - \left( {x - y} \right),\;\; \frac{{\partial Q}}{{\partial x}} = - 1,\;\; \frac{{\partial P}}{{\partial y}} = 1.\]
Hence,
\[I = \oint\limits_C {\left( {x + y} \right)dx - \left( {x - y} \right)dy} = \iint\limits_R {\left( { - 1 - 1} \right)dxdy} = - 2\iint\limits_R {dxdy} .\]
As the double integral \(\iint\limits_R {dxdy} \) is equal to the area of the ellipse \(\pi ab,\) the integral is
\[I = - 2\iint\limits_R {dxdy} = - 2\pi ab.\]
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 \(3\)).
Solution.
Figure 3. In the given line integral \(P = {y^2},\) \(Q = {\left( {x + y} \right)^2},\) so that
\[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( {{{\left( {x + y} \right)}^2}} \right)}}{{\partial x}} = 2\left( {x + y} \right),\;\; \frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {{y^2}} \right)}}{{\partial y}} = 2y.\]
Then, by Green's formula,
\[I = \oint\limits_C {{y^2}dx + {{\left( {x + y} \right)}^2}dy} = \iint\limits_R {\left[ {2\left( {x + y} \right) - 2y} \right]dxdy} = 2\iint\limits_R {xdxdy} .\]
It is easy to see that the equation of the side \(AD\) is \(y = -x + a.\) Hence, the latter double integral becomes
\[I = 2\iint\limits_R {xdxdy} = 2\int\limits_0^a {\left[ {\int\limits_{ - x + a}^a {dy} } \right]xdx} = 2\int\limits_0^a {\left[ {\left. y \right|_{ - x + a}^a} \right]xdx} = 2\int\limits_0^a {\left[ {a - \left( { - x + a} \right)} \right]xdx} = 2\int\limits_0^a {{x^2}dx} = 2\left[ {\left. {\left( {\frac{{{x^3}}}{3}} \right)} \right|_0^a} \right] = \frac{{2{a^3}}}{3}.\]
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 \(4\)).
Solution.
Figure 4. Applying Green's formula
\[\oint\limits_C {Pdx + Qdy} = \iint\limits_R {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)dxdy}, \]
we identify:
\[P = y - {x^2},\;\; Q = - \left( {x + {y^2}} \right),\;\; \frac{{\partial Q}}{{\partial x}} = - \frac{{\partial \left( {x + {y^2}} \right)}}{{\partial x}} = - 1,\;\; \frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {y - {x^2}} \right)}}{{\partial y}} = 1.\]
Hence,
\[I = \oint\limits_C {\left( {y - {x^2}} \right)dx - \left( {x + {y^2}} \right)dy} = \iint\limits_R {\left( { - 1 - 1} \right)dxdy} = - 2\iint\limits_R {dxdy} .\]
Transforming this integral to polar coordinates, we get
\[I = - 2\iint\limits_R {dxdy} = - 2\int\limits_0^{\frac{\pi }{2}} {d\theta } \int\limits_0^a {rdr} = - 2 \cdot \frac{\pi }{2} \cdot \left[ {\left. {\left( {\frac{{{r^2}}}{2}} \right)} \right|_0^a} \right] = - \frac{{\pi {a^2}}}{2}.\]
Example 7.
Calculate the integral \[\oint\limits_C {\frac{{dx - dy}}{{x + y}}} \] 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 \(5\)).
Solution.
Figure 5. According to Green's theorem, we can write
\[P = \frac{1}{{x + y}},\;\; Q = - \frac{1}{{x + y}},\;\; \frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left( { - \frac{1}{{x + y}}} \right)}}{{\partial x}} = \frac{1}{{{{\left( {x + y} \right)}^2}}},\;\; \frac{{\partial P}}{{\partial y}} = \frac{{\partial \left( {\frac{1}{{x + y}}} \right)}}{{\partial y}} = - \frac{1}{{{{\left( {x + y} \right)}^2}}}.\]
Consequently,
\[I = \oint\limits_C {\frac{{dx - dy}}{{x + y}}} = \iint\limits_R {\left( {\frac{1}{{{{\left( {x + y} \right)}^2}}} + \frac{1}{{{{\left( {x + y} \right)}^2}}}} \right)dxdy} = 2\iint\limits_R {\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}} .\]
Find the equations of the sides of the square:
\[AB:y = - x + 1,\]
\[BD:y = x + 1,\]
\[DE:y = - x - 1,\]
\[EA:y = x - 1.\]
It is convenient to change variables. Let \(u = y + x,\) \(v = y - x.\) In terms of \(u\) and \(v\), we have
\[y = - x + 1,\;\; \Rightarrow y + x = 1,\;\; \Rightarrow u = 1,\]
\[y = x + 1,\;\; \Rightarrow y - x = 1,\;\; \Rightarrow v = 1,\]
\[y = - x - 1,\;\; \Rightarrow y + x = - 1,\;\; \Rightarrow u = - 1,\]
\[y = x - 1,\;\; \Rightarrow y - x = - 1,\;\; \Rightarrow v = - 1.\]
As it can be seen, the pullback \(S\) of the initial region \(R\) is the square, shown in Figure \(6.\)
Figure 6. Calculate the Jacobian of the inverse transformation.
\[\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial u}}{{\partial x}}}&{\frac{{\partial u}}{{\partial y}}}\\
{\frac{{\partial v}}{{\partial x}}}&{\frac{{\partial v}}{{\partial y}}}
\end{array}} \right|
= \left| {\begin{array}{*{20}{c}}
{\frac{{\partial \left( {y + x} \right)}}{{\partial x}}}&{\frac{{\partial \left( {y + x} \right)}}{{\partial y}}}\\
{\frac{{\partial \left( {y - x} \right)}}{{\partial x}}}&{\frac{{\partial \left( {y - x} \right)}}{{\partial y}}}
\end{array}} \right|
= \left| {\begin{array}{*{20}{c}}
1&1\\
{ - 1}&1
\end{array}} \right| = 2.\]
Then the absolute value of the Jacobian of the original transformation is given by
\[\left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right| = \left| {{{\left( {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right)}^{ - 1}}} \right| = \frac{1}{2}.\]
Hence,
\[dxdy = \left| {\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}} \right|dudv = \frac{1}{2}dudv,\]
so that the integral is
\[I = 2\iint\limits_R {\frac{{dxdy}}{{{{\left( {x + y} \right)}^2}}}}
= 2\iint\limits_S {\frac{{\frac{1}{2}dudv}}{{{u^2}}}}
= \iint\limits_S {\frac{{dudv}}{{{u^2}}}}
= \int\limits_{ - 1}^1 {dv} \int\limits_{ - 1}^1 {\frac{{du}}{{{u^2}}}}
= \left[ {\left. v \right|_{ - 1}^1} \right] \cdot \left[ {\left. {\left( { - \frac{1}{u}} \right)} \right|_{ - 1}^1} \right]
= \left[ {1 - \left( { - 1} \right)} \right] \cdot \left[ { - 1 + \left( { - 1} \right)} \right] = - 4.\]
Example 8.
Calculate the line integral
\[\oint\limits_C {\sqrt {{x^2} + {y^2}} dx + y \left[ {xy \, + \ln \left( {x + \sqrt {{x^2} + {y^2}} } \right)} \right]dy} \]
using Green's theorem. The contour of integration \(C\) is the circle \[{x^2} + {y^2} = {a^2}\] (Figure \(7\)).
Solution.
Figure 7. Identify the components of the vector field and find the partial derivatives:
\[P = \sqrt {{x^2} + {y^2}} ,\;\;Q = y\left[ {xy + \ln \left( {x + \sqrt {{x^2} + {y^2}} } \right)} \right],\]
\[\frac{{\partial P}}{{\partial y}} = \frac{{\partial \sqrt {{x^2} + {y^2}} }}{{\partial y}} = \frac{y}{{\sqrt {{x^2} + {y^2}} }},\]
\[\frac{{\partial Q}}{{\partial x}} = \frac{{\partial \left\{ {y\left[ {xy + \ln \left( {x + \sqrt {{x^2} + {y^2}} } \right)} \right]} \right\}}}{{\partial x}}
= y\left( {y + \frac{{1 + \frac{{2x}}{{2\sqrt {{x^2} + {y^2}} }}}}{{x + \sqrt {{x^2} + {y^2}} }}} \right)
= y\left( {y + \frac{{\frac{{x + \sqrt {{x^2} + {y^2}} }}{{\sqrt {{x^2} + {y^2}} }}}}{{x + \sqrt {{x^2} + {y^2}} }}} \right)
= y\left( {y + \frac{1}{{\sqrt {{x^2} + {y^2}} }}} \right)
= {y^2} + \frac{y}{{\sqrt {{x^2} + {y^2}} }}.\]
Then by Green's formula,
\[I = \oint\limits_C {\sqrt {{x^2} + {y^2}} dx + y\left[ {xy + \ln \left( {x + \sqrt {{x^2} + {y^2}} } \right)} \right]dy} = \iint\limits_R {\left( {{y^2} + \frac{y}{{\sqrt {{x^2} + {y^2}} }} - \frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)dxdy} = \iint\limits_R {{y^2}dxdy} .\]
It is convenient to transform the double integral to polar coordinates.
\[I = \iint\limits_R {{y^2}dxdy} = \int\limits_0^{2\pi } {\left[ {\int\limits_0^a {{r^2}{{\sin }^2}\theta \cdot rdr} } \right]d\theta } = \int\limits_0^{2\pi } {{{\sin }^2}\theta d\theta } \int\limits_0^a {{r^3}dr} .\]
Here
\[\int\limits_0^{2\pi } {{{\sin }^2}\theta d\theta } = \int\limits_0^{2\pi } {\frac{{1 - \cos 2\theta }}{2}d\theta } = \frac{1}{2}\left[ {\left. {\left( {\theta - \frac{{\sin 2\theta }}{2}} \right)} \right|_0^{2\pi }} \right] = \frac{1}{2} \cdot 2\pi = \pi ,\]
\[\int\limits_0^a {{r^3}dr} = \left. {\left( {\frac{{{r^4}}}{4}} \right)} \right|_0^a = \frac{{{a^4}}}{4}.\]
Thus, the integral is
\[I = \frac{{\pi {a^4}}}{4}.\]
Example 9.
Calculate the area of the region \(R\) bounded by the astroid
\[x = a\,{\cos ^3}t, y = a\,{\sin ^3}t, 0 \le t \le 2\pi .\]
Solution.
We will use the line integral to define the area of the region, using the formula
\[S = \frac{1}{2} \oint\limits_C {xdy - ydx} .\]
In parametric form this formula is represented as
\[S = \frac{1}{2}\int\limits_\alpha ^\beta {\left( {x\left( t \right)\frac{{dy}}{{dt}} - y\left( t \right)\frac{{dx}}{{dt}}} \right)dt} .\]
Substituting the equations of the astroid, we obtain
\[S = \frac{1}{2}\int\limits_\alpha ^\beta {\left[ {a\,{{\cos }^3}t \cdot \frac{{d\left( {a\,{{\sin }^3}t} \right)}}{{dt}} - a\,{{\sin }^3}t \cdot \frac{{d\left( {a\,{{\cos }^3}t} \right)}}{{dt}}} \right]dt} = \frac{1}{2}\int\limits_0^{2\pi } {\left[ {a\,{{\cos }^3}t \cdot 3a\,{{\sin }^2}t\cos t - a\,{{\sin }^3}t \cdot 3a\,{{\cos }^2}t\left( { - \sin t} \right)} \right]dt} = \frac{{3{a^2}}}{2}\int\limits_0^{2\pi } {\left[ {{{\cos }^4}t\,{{\sin }^2}t + {{\sin }^4}t\,{{\cos }^2}t} \right]dt} = \frac{{3{a^2}}}{8}\int\limits_0^{2\pi } {\left[ {4\,{\sin^2}t\,{{\cos }^2}t \left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \right]dt} = \frac{{3{a^2}}}{8}\int\limits_0^{2\pi } {{{\left( {\sin 2t} \right)}^2}dt} = \frac{{3{a^2}}}{8}\int\limits_0^{2\pi } {\frac{{1 - \cos 4t}}{2}dt} = \frac{{3{a^2}}}{{16}}\int\limits_0^{2\pi } {\left( {1 - \cos 4t} \right)dt} = \frac{{3{a^2}}}{{16}}\left[ {\left. {\left( {t - \frac{{\sin 4t}}{4}} \right)} \right|_0^{2\pi }} \right] = \frac{{3{a^2}}}{{16}} \cdot 2\pi = \frac{{3{a^2}\pi }}{8}.\]
