Lemma 1

Let \(D\) be a \(y\)-simple region and let \(C\) be its boundary. Suppose \(P \colon\, D \to {\mathbb R}\) is of class \(C^1\). Then \[ \int_{C^+} P {\,d} x = - \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! P}{\partial y} {\it {\,d} x}\, {\it dy}. \] (The left-hand side denotes the line integral \(\int_{C^+} P {\it {\,d} x} + Q\, {\it dy}\), where \(Q = 0\).)

proof

Suppose \(D\) is described by \[ a \le x \le b \qquad \phi_1 (x) \le y \le \phi_2 (x). \]

We decompose \(C^+\) by writing \(C^{+} = C^{+}_1 \,+\, B_2^{+} \,+\, C_2^{-} \,+\, B_1^{-}\) (see Figure 18.2). By Fubini’s theorem, we can evaluate the double integral as an iterated integral and then use the fundamental theorem of calculus: \begin{eqnarray*} \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! P}{\partial y} (x,y) {\it {\,d} x}\, {\it dy} & = & \int_a^b \int_{\phi_1(x)}^{\phi_2 (x)} \frac{\partial\! P}{\partial y} (x,y)\, {\it dy}\, {\it {\,d} x}\\[8pt] & = & \int_a^b [P (x, \phi_2(x)) -P (x, \phi_1 (x)) ] {\it {\,d} x}. \end{eqnarray*}

However, because \(C_1^{+}\) can be parametrized by \(x \mapsto (x, \phi_1(x)), a\le x \le b\), and \(C_2^{+}\) can be parametrized by \(x \mapsto (x, \phi_2 (x)), a\le x \le b\), we have \[ \int_a^b P (x, \phi_1 (x)) {\it {\,d} x} = \int_{C_1^{+}} P (x,y) {\it {\,d} x} \] and \[ \int_a^bP(x,\phi_2(x))\,{\it {\,d} x}=\int_{C_{2}^{+}}P(x,y)\,{\it {\,d} x}. \]

Thus, by reversing orientations, \[ - \int_a^b P (x, \phi_2 (x)) {\it {\,d} x} = \int_{C_2^{-}} P (x,y) {\it {\,d} x}. \]

Hence, \[ \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! P}{\partial y} {\it {\,d} x}\, {\it dy} = - \int_{C_1^{+}} P {\it {\,d} x} - \int_{C_2^{-}} P {\it {\,d} x}. \]

Because \(x\) is constant on \(B_2^{+}\) and \(B_1^{-}\), we have \[ \int_{B_2^{+}} P {\it {\,d} x}=0 = \int_{B_1^{-}} P {\it {\,d} x}, \] so \[ \int_{C^{+}} P {\it {\,d} x} = \int_{C^{+}_1} P {\it {\,d} x}+ \int_{B^{+}_2} P {\it {\,d} x} + \int_{C_2^{-}} P {\it {\,d} x} + \int_{B_1^{-}} P {\it {\,d} x} = \int_{C_1^{+}} P {\it {\,d} x} + \int_{C_2^{-}} P {\it {\,d} x}. \]

Thus, \[ \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! P}{\partial y} {\it {\,d} x}\, {\it dy} = - \int_{C_1^{+}} P {\it {\,d} x} - \int_{C_2^{-}} P {\it {\,d} x} = - \int_{C^{+}} P {\it {\,d} x}. \]

Lemma 2

Let \(D\) be an \(x\)-simple region with boundary \(C\). Then if \(Q\colon\, D \to {\mathbb R}\) is \(C^1\), \[ \int_{C^{+}} Q \,{\it dy} = \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! Q}{\partial x} {\it {\,d} x}\, {\it dy}. \]

proof

Suppose \(D\) is given by \[ \psi_1 (y) \le x \le \psi_2 (y), c \le y \le d. \]

Using the notation of Figure 18.3, and noting that \(y\) is constant on \(B^+_1\) and \(B^-_2\), we have \[ \int_{C^{+}} Q \,{\it dy} = \int_{C_1^{-} +B_1^{+} + C_2^{+} + B_2^{-}} Q \,{\it dy} = \int_{C_2^{+}} Q \,{\it dy} + \int_{C_1^-} Q \,{\it dy}, \] where \(C_2^{+}\) is the curve parametrized by \(y \mapsto (\psi_2 (y), y), c \le y \le d\), and \(C_1^{+}\) is the curve \(y \mapsto ( \psi_1 (y), y), c \le y \le d\). Applying Fubini’s theorem and the fundamental theorem of calculus, we obtain \begin{eqnarray*} \intop\!\!\!\intop\nolimits_{D} \frac{\partial\! Q}{\partial x} {\it {\,d} x}\, {\it dy} &=& \int_c^d \int_{\psi_1 (y)}^{\psi_2 (y)} \frac{\partial\! Q}{\partial x} {\it {\,d} x}\,{\it dy} = \int_c^d [ Q (\psi_2 (y),y) - Q (\psi_1 (y), y)]\, {\it dy}\\[6pt] & = & \int_{C_2^{+}} Q \,{\it dy} - \int_{C_1^{+}} Q \,{\it dy} = \int_{C^{+}_2} Q \,{\it dy} + \int_{C^{-}_1} Q \,{\it dy} = \int_{C^{+}} Q \,{\it dy}.\\[-25pt] \end{eqnarray*}

Theorem 1 Green’s Theorem

Let \(D\) be a simple region and let \(C\) be its boundary. Suppose \(P \colon\, D \to {\mathbb R}\) and \(Q\colon\, D \to {\mathbb R}\) are of class \(C^1\). Then \[ \int_{C^{+}} P {\it {\,d} x} + Q \,{\it dy} = \intop\!\!\!\intop\nolimits_{D} \left( \frac{\partial\! Q}{\partial x} - \frac{\partial\! P}{\partial y} \right) {\it {\,d} x}\, {\it dy}. \]

Historical Note

George Green

George Green was born in Nottingham England in 1793 and died in 1841. No picture of him is known to exist. Little is known about his early life before the age of 30 other than that he was a self-taught man, who developed a passion for mathematics at an early age. In 1833, at age 40, Green enrolled as a student in Cambridge. Amazingly, five years earlier, he had published (at his own expense) his first and most famous work “An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism.” Here he proved a theorem similar to the Green’s Theorem of this section and also introduced other very important mathematical concepts such as Green’s functions, now so ubiquitous in mathematical analysis. He was the first person to create a theory of electricity and magnetism, a theory that later formed the basis of the work of Maxwell and others.

example 1

Verify Green’s theorem for \(P (x,y) =x\) and \(Q(x,y)=xy\), where \(D\) is the unit disc \(x^2 + y^2 \le 1\).

solution We do this by evaluating both sides in Green’s theorem directly. The boundary of \(D\) is the unit circle parametrized by \(x =\cos t, y = \sin t, 0 \le t \le 2 \pi\), and so \begin{eqnarray*} \int_{\partial\! D} P {\it {\,d} x} + Q\, {\it dy} & = & \int_0^{2 \pi} [( \cos t) ({-}\!\sin t) + \cos t \sin t \cos t] {\it {\,d} t} \\[6pt] & = & \left[\frac{\cos^2 t}{2} \right]_0^{2 \pi}+ \left[ - \frac{\cos^3 t}{3} \right]_0^{2 \pi} =0. \end{eqnarray*}

On the other hand, \[ \intop\!\!\!\intop\nolimits_{D}\left( \frac{\partial\! Q}{\partial x} - \frac{\partial\! P}{\partial y}\right) {\it dx}\, {\it dy} = \intop\!\!\!\intop\nolimits_{D} y {\it {\,d} x}\, {\it dy}, \] which is also zero by symmetry. Thus, Green’s theorem is verified in this case.

Theorem 2 Area of a Region

If \(C\) is a simple closed curve that bounds a region to which Green’s theorem applies, then the area of the region \(D\) bounded by \(C = \partial\! D\) is \[ A = \frac{1}{2} \int_{\partial\! D} x {\it dy} - y {\it {\,d} x}. \]

proof

Let \(P (x,y) = - y, Q (x,y) =x\); then by Green’s theorem we have \begin{eqnarray*} \frac{1}{2} \int_{\partial\! D} x {\it dy} \,-\, y {\it {\,d} x} &=& \frac{1}{2} \intop\!\!\!\intop\nolimits_{D} \left[ \frac{\partial x}{\partial x} - \frac{\partial (-y)}{\partial y} \right] {\it dx}\, {\it dy} \\ [6pt] &=& \frac{1}{2} \intop\!\!\!\intop\nolimits_{D} [1\,+\,1] {\it {\,d} x}\, {\it dy} = \intop\!\!\!\intop\nolimits_{D} {\it {\,d} x}\, {\it dy} =A. \\[-35pt] \end{eqnarray*}

example 2

Let \(a > 0\). Compute the area (see Figure 18.6) of the region enclosed by the hypocycloid defined by \(x^{2/3} + y^{2/3} = a^{2/3}\) using the parametrization \[ x =a \cos^3 \theta , y =a \sin^3 \theta , 0 \le \theta \le 2 \pi. \]

Figure 18.6: The hypocycloid \(x =a\, \cos^{3} \theta, y = a\, \sin^{3} \theta, 0 \le \theta \le 2 \pi\).

solution From the preceding box, and using the trigonometric identities \(\cos^2\theta+\sin^2\theta=1, \sin 2\theta=2\sin\theta\cos\theta\), and \(\sin^2\phi=(1-\cos2\phi)/2\), we get \begin{eqnarray*} A & = & \frac{1}{2} \int_{\partial\! D} x {\it dy} - y {\it {\,d} x} \\[8pt] & = & \frac{1}{2} \int_0^{2 \pi} [(a \cos^3 \theta) ( 3 a \sin^2 \theta \cos \theta) - ( a \sin^3 \theta)(-3 a \cos^2 \theta\sin \theta) ] {\,d} \theta \\[8pt] & = & \frac{3}{2} a^2 \int_0^{2 \pi} ( \sin^2 \theta \cos^4 \theta + \cos^2 \theta \sin^4 \theta) {\,d} \theta = \frac{3}{2} a^2 \int_0^{2 \pi} \sin^2 \theta \cos^2 \theta {\,d} \theta \\[8pt] & = & \frac{3}{8} a^2 \int_0^{2 \pi} \sin^2 2 \theta {\,d} \theta = \frac{3}{8} a^2 \int_0^{2 \pi} \left( \frac{1 - \cos\, 4 \theta}{2} \right) d \theta \\[8pt] & = & \frac{3}{16} a^2 \int_0^{2 \pi} {\,d} \theta - \frac{3}{16} a^2 \int_0^{2 \pi} \cos 4 \theta {\,d} \theta = \frac{3}{8} \pi a^2.\\[-28pt] \end{eqnarray*}

Theorem 3 Vector Form of Green’s Theorem

Let \(D \subset {\mathbb R}^2\) be a region to which Green’s theorem applies, let \(\partial\! D\) be its (positively oriented) boundary, and let \({\bf F} = P {\bf i} + Q {\bf j}\) be a \(C^1\) vector field on \(D\). Then \[ \int_{\partial\! D} {\bf F} \,{\cdot}\, d {\bf s}= \intop\!\!\!\intop\nolimits_{D} ({\rm curl}\, {\bf F}) \,{\cdot}\, {\bf k}\, {\it {\,d} A} = \intop\!\!\!\intop\nolimits_{D} ( {\nabla} \times {\bf F}) \,{\cdot}\, {\bf k}\, {\it {\,d} A} \] (see Figure 18.7).

example 3

Let \({\bf F} = ( xy^2 , y+ x)\). Integrate \(({\nabla} \times {\bf F}) \,{\cdot}\, {\bf k}\) over the region in the first quadrant bounded by the curves \(y = x^2\) and \(y=x\).

solution Method 1. We first compute the curl \[ {\nabla} \times {\bf F} = \left(0, 0, \frac{\partial\! F_2}{\partial x} -\frac{\partial\! F_1}{\partial y} \right) = (1- 2 xy) {\bf k}. \]

Thus, \(({ \nabla} \times {\bf F} ) \,{\cdot}\, {\bf k} = 1 - 2 xy\). This can be integrated over the given region \(D\) (see Figure 18.8) using an iterated integral as follows: \begin{eqnarray*} \intop\!\!\!\intop\nolimits_{D} ({\nabla} \times {\bf F}) \,{\cdot}\,{\bf k} {\it {\,d} x}\, {\it dy} &=& \int_0^1 \int_{x^2}^x (1-2xy)\, {\it dy}\, {\it {\,d} x} = \int_0^1 [y-xy^2]|_{x^2}^x {\it {\,d} x} \\[6pt] & = & \int_0^1 [x - x^3 - x^2 + x^5] {\it {\,d} x} = \frac{1}{2} - \frac{1}{4} - \frac{1}{3} + \frac{1}{6} = \frac{1}{12}. \end{eqnarray*}

Figure 18.8: The region bounded by the curves \(y=x^{2}\) and \(y=x\).

Method 2. Here we use Theorem 3 to obtain \[ \intop\!\!\!\intop\nolimits_{D} ( {\nabla} \times {\bf F} ) \,{\cdot}\, {\bf k} {\it {\,d} x}\, {\it dy} = \int_{\partial\! D} {\bf F} \,{\cdot}\,d {\bf s}. \]

The line integral of \({\bf F}\) along the curve \(y=x\) from left to right is \[ \int_0^1 F_1 {\it {\,d} x} + F_2\, {\it dy} = \int_0^1 ( x^3+ 2x)\, {\it {\,d} x} = \frac{1}{4}+1 = \frac{5}{4}. \]

Along the curve \(y = x^2\) we get \[ \int_0^1 F_1 {\it {\,d} x} + F_2\, {\it dy} = \int_0^1 x^5 {\it {\,d} x} + ( x+ x^2) (2 x {\it {\,d} x}) = \frac{1}{6} + \frac{2}{3} + \frac{1}{2} = \frac{4}{3}. \]

Thus, remembering that the integral along \(y=x\) is to be taken from right to left, as in Figure 18.8, \[ \int_{\partial\! D} {\bf F} \,{\cdot}\, d {\bf s} = \frac{4}{3} - \frac{5}{4} = \frac{1}{12}. \]

Theorem 4 Divergence Theorem in the Plane

Let \(D \subset {\mathbb R}^2\) be a region to which Green’s theorem applies and let \(\partial\! D\) be its boundary. Let \({\bf n}\) denote the outward unit normal to \(\partial\! D\). If \({\bf c} \colon\, [a,b] \to {\mathbb R}^2 , t \mapsto {\bf c} (t) = (x(t) , y(t))\) is a positively oriented parametrization of \(\partial\! D, {\bf n}\) is given by \[ {\bf n} = \frac{(y' (t), - x' (t))}{\sqrt{[x' (t) ]^2 + [y' (t) ]^2}} \] (see Figure 18.9). Let \({\bf F} = P {\bf i}+ Q {\bf j}\) be a \(C^1\) vector field on \(D\). Then \[ \int_{\partial\! D} {\bf F} \,{\cdot}\, {\bf n} {\,d} s = \intop\!\!\!\intop\nolimits_{D} {\rm div\ } {\bf F}\, {\,d}\! A. \]

proof

Recall that \({\bf c}' (t) = ( x' (t), y' (t))\) is tangent to \(\partial\! D\), and note that \({\bf n} \,{\cdot}\, {\bf c}' =0\). Thus, \({\bf n}\) is normal to the boundary. The sign of \({\bf n}\) is chosen to make it correspond to the outward (rather than the inward) direction. By the definition of the line integral (see Section 7.2), \begin{eqnarray*} \int_{\partial\! D} {\bf F} \,{\cdot} {\bf n}\, ds & = & \int_a^b \frac{P (x(t), y(t)) y' (t) - Q (x(t), y(t)) x'(t)}{\sqrt{[x' (t)]^2+ [y' (t) ]^2}} {\textstyle \sqrt{[x'(t)]^2 + [y' (t) ]^2}}\, {\it {\,d} t} \\[6pt] & = & \int_a^b [P (x (t), y(t)) y' (t) - Q(x(t), y(t)) x' (t)]\, {\it {\,d} t} \\[6pt] & =& \int_{\partial\! D} P {\it dy} - Q {\it {\,d} x}. \end{eqnarray*}

By Green’s theorem, this equals \[ \intop\!\!\!\intop\nolimits_{D} \left( \frac{\partial\! P}{\partial x}+ \frac{\partial\! Q}{\partial y} \right) {\it dx}\, {\it dy} = \intop\!\!\!\intop\nolimits_{D} \hbox{div } {\bf F} {\it {\,d} A}. \]

example 4

Let \({\bf F} = y^3 {\bf i} + x^5 {\bf j}\). Compute the integral of the normal component of \({\bf F}\) around the unit square.

solution This can be done using the divergence theorem. Indeed \[ \int_{\partial\! D} {\bf F} \,{\cdot}\, {\bf n}\, ds = \intop\!\!\!\intop\nolimits_{D} \hbox{div } {\bf F} {\it {\,d} A}. \]

But div \({\bf F}=0\), and so the integral is zero.