Example 18.4.2 We illustrate the theorem by computing both sides of $$\int_{\partial D} x^4\,dx + xy\,dy=\dint{D} y-0\,dA,$$ where $D$ is the triangular region with corners $(0,0)$, $(1,0)$, $(0,1)$.

Starting with the double integral: $$\dint{D} y-0\,dA=\int_0^1\int_0^{1-x} y\,dy\,dx= \int_0^1 {(1-x)^2\over2}\,dx=\left.-{(1-x)^3\over6}\right|_0^1={1\over6}.$$

There is no single formula to describe the boundary of $D$, so to compute the left side directly we need to compute three separate integrals corresponding to the three sides of the triangle, and each of these integrals we break into two integrals, the "$dx$'' part and the "$dy$'' part. The three sides are described by $y=0$, $y=1-x$, and $x=0$. The integrals are then $$\eqalign{ \int_{\partial D}\!\!\! x^4\,dx + xy\,dy&= \int_0^1 x^4\,dx+\int_0^0 0\,dy+\int_1^0 x^4\,dx+\int_0^1 (1-y)y\,dy+ \int_0^0 0\,dx+\int_1^0 0\,dy\cr &={1\over5}+0-{1\over5}+{1\over6}+0+0={1\over6}.\cr}$$

Alternately, we could describe the three sides in vector form as $\langle t,0\rangle$, $\langle 1-t,t\rangle$, and $\langle 0,1-t\rangle$. Note that in each case, as $t$ ranges from 0 to 1, we follow the corresponding side in the correct direction. Now $$\eqalign{ \int_{\partial D} x^4\,dx + xy\,dy&= \int_0^1 t^4 + t\cdot 0\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt +\int_0^1 0 + 0\,dt\cr &=\int_0^1 t^4\,dt + \int_0^1 -(1-t)^4 + (1-t)t\,dt ={1\over6}.\cr }$$ $\square$

Example 18.4.3 An ellipse centered at the origin, with its two principal axes aligned with the $x$ and $y$ axes, is given by $${x^2\over a^2}+{y^2\over b^2}=1.$$ We find the area of the interior of the ellipse via Green's theorem. To do this we need a vector equation for the boundary; one such equation is $\langle a\cos t,b\sin t\rangle$, as $t$ ranges from 0 to $2\pi$. We can easily verify this by substitution: $${x^2\over a^2}+{y^2\over b^2}= {a^2\cos^2 t\over a^2}+{b^2\sin^2t\over b^2}= \cos^2t+\sin^2t=1.$$ Let's consider the three possibilities for $P$ and $Q$ above: Using 0 and $x$ gives $$\oint_C 0\,dx+x\,dy=\int_0^{2\pi} a\cos(t)b\cos(t)\,dt= \int_0^{2\pi} ab\cos^2(t)\,dt.$$ Using $-y$ and 0 gives $$\oint_C -y\,dx+0\,dy=\int_0^{2\pi} -b\sin(t)(-a\sin(t))\,dt= \int_0^{2\pi} ab\sin^2(t)\,dt.$$ Finally, using $-y/2$ and $x/2$ gives $$\eqalign{ \oint_C -{y\over2}\,dx+{x\over2}\,dy&= \int_0^{2\pi} -{b\sin(t)\over2}(-a\sin(t))\,dt +{a\cos(t)\over2}(b\cos(t))\,dt\cr &=\int_0^{2\pi} {ab\sin^2t\over2}+{ab\cos^2t\over2}\,dt= \int_0^{2\pi} {ab\over2}\,dt=\pi ab.\cr}$$ The first two integrals are not particularly difficult, but the third is very easy, though the choice of $P$ and $Q$ seems more complicated. $\square$