Proving a special case of Green’s theorem

Question

Let $A$ be a region in the plane such that $$a\le x\le b \text{ and }g_1(x)\le y\le g_2(x).$$ We can parametrize the two functions as follows $$\gamma_1(t)=(t, g_1(t)) \text{ and } \gamma_2(t)=(t,g_2(t)),$$ where $a\le t\le b$, now consider the vector field defined by $$F(x,y)=(P(x,y),Q(x,y)).$$

The author said that if we want to prove (one half) of green’s theorem for this region $A$, we have to show that $$-\iint_A\frac{\partial P}{\partial y}\,dy\,dx=\int_C P\,dx.$$ Since $$\iint_A\frac{\partial P}{\partial y}\,dy\,dx= \int_a^b\int_{g_1(x)}^{g_2(x)} \frac{\partial P}{\partial y} \,dy\,dx= \int_a^b P(x,g_2(x))-P(x,g_1(x))\,dx \\ =\int_{\gamma_2}P\,dx-\int_{\gamma_1}P\,dx$$

First thing I don’t understand what does he means by "a half", the second thing is the last step, I can’t figure out how he jumps to it.

Answer 1

Green's theorem states that $$\int_A \frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y} \, dA = \oint_C P\, dx + Q \, dy.$$ In particular if $P = 0$ or $Q = 0$ it states that $$\int_A \frac{\partial Q}{\partial x} \, dA = \oint_C Q \, dy \quad \text{and} \quad - \int_A \frac{\partial P}{\partial y} \, dA = \oint_C P\, dx. $$ These are the "halves" of Green's theorem. If you can prove one of them the other proof will be the same (mutatis mutandis) and you can add them together to obtain the full Green's theorem.

As for the second question, the boundary consists of four parts: bottom, right, top, and left. The parameterizations of each of these can be given by

1. Bottom ($\gamma_{\rm bottom}$): $x=t$, $y = g_1(t)$, $a \le t \le b$
2. Right ($\gamma_{\rm right}$): $x=b$, $y = t$, $g_1(b) \le t \le g_2(b)$
3. Top ($\gamma_{\rm top}$): $x = t$, $y = g_2(t)$, $a \le t \le b$
4. Left ($\gamma_{\rm left}$): $x = a$, $y = t$, $g_1(a) \le t \le g_2(b)$

Since $C$ has a counterclockwise parameterization, you have $$\oint_C P \, dx = \int_{\gamma_{\rm bottom}} P \, dx + \int_{\gamma_{\rm right}} P \, dx - \int_{\gamma_{\rm top}} P \, dx - \int_{\gamma_{\rm left}} P \, dx.$$

When you evaluate these integrals you have $$\int_{\gamma_{\rm bottom}} P \, dx = \int_a^b P(t,g_1(t)) \, dt$$ $$\int_{\gamma_{\rm right}} P \, dx = 0$$ and so on.