Proof for a region of both types. Suppose \(D = \{(x,y) \mid a\leq x\leq b,\ g_1(x)\leq y\leq g_2(x)\}\) is a type I region. It is enough to show \(-\iint _D P_y\,dA = \oint _C P\,dx\); the companion identity \(\iint _D Q_x\,dA = \oint _C Q\,dy\) follows in the same way from a description of \(D\) as a type II region. By Theorem 3.3.1 and the fundamental theorem of calculus, \[\iint \limits _D \frac {\partial P}{\partial y}\,dA = \int _a^b\int _{g_1(x)}^ {g_2(x)}\frac {\partial P}{\partial y}\,dy\,dx = \int _a^b\big [P(x,g_2(x)) - P(x,g_1(x))\big ]\,dx .\] On the other hand, \(C\) consists of the lower curve \(C_1\): \(y = g_1(x)\) from left to right, the upper curve \(C_3\): \(y = g_2(x)\) from right to left, and possibly vertical segments \(C_2\) and \(C_4\), on which \(dx = 0\). So \[\oint _C P\,dx = \int _a^b P(x, g_1(x))\,dx - \int _a^b P(x, g_2(x))\,dx ,\] which is \(-\iint _D P_y\,dA\). Adding the two identities gives the theorem. A general region is cut into pieces of both types; the line integrals along the cuts are traversed once in each direction and cancel. □

Solution. With \(P = x^4\) and \(Q = xy\), \(\frac {\partial Q}{\partial x} - \frac {\partial P}{\partial y} = y - 0 = y\), and \(C\) is positively oriented around \(D = \{(x,y) \mid 0\leq x\leq 1,\ 0\leq y \leq 1 -x\}\). By Green’s theorem, \[\oint _C x^4\,dx + xy\,dy = \iint \limits _D y\,dA = \int ^1_0 \int ^{1-x}_0 y\,dy\,dx = \int _0^1\frac {(1-x)^2}{2}\,dx = \frac {1}{6} .\] Computing the three line integrals separately gives the same result with more work.

Solution. The region \(D\) bounded by \(C\) is the disc \(x^2 + y^2 \leq 9\), and \(Q_x - P_y = 7 - 3 = 4\). By Green’s theorem, \[\oint _C \left ( 3y - e^{\sin x}\right ) dx + \left ( 7x + \sqrt {y^4 + 1}\right ) dy = \iint \limits _D 4\,dA = 4\cdot 9\pi = 36\pi .\] The awkward terms \(e^{\sin x}\) and \(\sqrt {y^4 + 1}\) contribute nothing, since each depends only on the variable it is integrated against.

Solution. Parametrise the ellipse by \(x = a \cos t\), \(y = b\sin t\), \(0\leq t \leq 2\pi \), which traverses it counter-clockwise. By Theorem 4.8.4, \[A = \frac {1}{2}\oint _C x\,dy - y\,dx = \frac {1}{2}\int ^{2\pi }_0 \left [(a\cos t) (b\cos t) - (b\sin t ) (-a\sin t)\right ] dt = \frac {ab}{2}\int ^{2\pi }_0 dt = \pi ab .\]

Solution. Green’s theorem extends to regions with holes when the boundary is oriented so that the region is always on the left: counter-clockwise on the outer circle and clockwise on the inner one. The cuts used to reduce the region to simply connected pieces cancel. Here \(Q_x - P_y = -3x^2 - 3y^2 = -3r^2\), so \[\oint _C y^3\,dx - x^3\,dy = \int _0^{2\pi }\int _1^2 (-3r^2)\,r\,dr\,d\theta = -3\cdot 2\pi \cdot \frac {16 - 1}{4} = -\frac {45\pi }{2} .\]

Solution. Differentiating, \(P_y = \frac {-(x^2+y^2) + 2y^2}{(x^2+y^2)^2} = \frac {y^2 - x^2}{(x^2+y^2)^2}\), and \(Q_x = \frac {(x^2 + y^2) - 2x^2}{(x^2 + y^2)^2}\) is the same. On the unit circle \(\mathbf r(t) = \langle \cos t,\sin t\rangle \), \(\mathbf F\cdot \mathbf r' = \sin ^2t + \cos ^2t = 1\), so the circulation is \(2\pi \). Green’s theorem requires \(P\) and \(Q\) to have continuous derivatives on the whole disc, and \(\mathbf F\) is not defined at the origin. The punctured plane is not simply connected, and on it \(\mathbf F\) is not conservative, although \(P_y = Q_x\): the hypothesis on the domain in Theorem 4.2.2 cannot be dropped.