4.1 Line integral and Green's theorem

1. To find a complex integral, it is common to parametrize the curve $C$ as $z(t) = x(t) + iy(t)$.

This curve is a straight line connecting a point $(0,1)$ and a point $(1,0)$. Then we can parametrize this line such as $z(t) = (0,1) + (1,-1)t, 0 \leq t \leq 1$. Thus, $x(t) = t, y(t) = 1 - t, 0 \leq t \leq 1$. Therefore,

$$\int_{c}y dx = \int_{0}^{1}(1-t)dt = t - \frac{t^2}{2} \mid_{0}^{1} = \frac{1}{2}$$

Alternate solution Since $y = 1-x$. we can integrate directly.

$$\int_{c}y dx = \int_{0}^{1}(1-x)dx = x - \frac{x^2}{2} \mid_{0}^{1} = \frac{1}{2}$$

This curve is a straight line connecting a point $(0,1)$ and a point $(1,0)$. Then we can parametrize this line such as $z(t) = (0,1) + (1,-1)t, 0 \leq t \leq 1$. Thus, $x(t) = t, y(t) = 1 - t, 0 \leq t \leq 1$, $dy = -dt$. Therefore,

$$\int_{c}x^2 dy = \int_{0}^{1}t^2 (-dt) = -\frac{t^3}{3} \mid_{0}^{1} = -\frac{1}{3}$$

This is a curve $y = x^2$ connecting $(-1,1)$ and $(1,1)$. Then we can parametrize this curve by $x(t) = t, y(t) = t^2, -1 \leq t \leq 1$, $dy = 2tdt$. Thus,

$$\int_{c}(xy dx - y^2 dy) = \int_{-1}^{1}(t^3 - t^4(2t)) dt = 0 (t^3, t^5$$   are odd function$$)$$

This curve is a circle of the radius 1 with the center at the origin. Then we can parametrize by $x(t) = \cos{t}, y(t) = \sin{t}, 0 \leq t \leq 2\pi$. Now $dx = -\sin{t}dt, dy = \cos{t}dt$. Then

$$\int_{c}(xy dx - x^3 dy) = \int_{0}^{2\pi}(-\sin^{2}{t}\cos{t} - \cos^{4}{t} )dt$$

Note that if we let $u = \sin{t}$, then $du = \cos{t}dt$, $$\begin{array}{ll} t :& 0 \to 2\pi\\ u :& 0 \to 0 \end{array}$$. Then

$$\int_{0}^{2\pi}(-\sin^{2}{t}\cos{t} dt = \int_{0}^{0} u^2 du = 0$$

Next

$$\int_{0}^{2\pi}\cos^{4}{t} dt = \int_{0}^{2\pi}(\cos^{2}{t})^{2} dt = \int_{0}^{2\pi}(\frac{1 + \cos{2t}}{2})^2 dt$$

$$= \frac{1}{4}\int_{0}^{2\pi}(1 + 2\cos{2t} + \cos^{2}{2t}) dt$$

$$= \frac{1}{4}\int_{0}^{2\pi}(1 = 2\cos{2t} + \frac{1 + \cos{4t}}{2}) dt$$

$$= \frac{1}{4}[\frac{3t}{2} + \cos{2t} + \frac{\sin{4t}}{8}\mid_{0}^{2\pi}]$$

$$= \frac{3\pi}{4}$$

2.

(a) This curve is already parametrized. Thus,

$$\int_{c}(x^2 + y) dt = \int_{0}^{1}(t + 1 - t^2) dt = \frac{t^2}{2} + t - \frac{t^3}{3}\mid_{0}^{1}$$

$$= \frac{1}{2} + 1 - \frac{1}{3} = \frac{7}{6}$$

(b) This curve is already parametrized. Thus,

$$\int_{c}xy^2 dt = \int_{0}^{\frac{\pi}{2}}\sin{t}\sin^{2}{t} dt$$

$$= \int_{0}^{\frac{\pi}{2}}\sin^{3}{t} dt = \frac{2!!}{3!!} = \frac{2}{3}$$

Note $\int_{0}^{\frac{\pi}{2}}\sin^{n}{t} dt = \left\{\begin{array}{l} \frac{(n-1)!!... ...d})\\ \frac{(n-1)!!}{n!!} \frac{\pi}{2} (n \mbox{even}) \end{array}\right.$

$n!! = n(n-2)(n-4) \cdots$.

3. Green's theorem is the line integral of $P(x,y), Q(x,y)$ such that the partial derivatives are continuous on the simply connected domain $R$ surrounded by a single closed curve $C$ and it can be expressed as a double integral in the simply connected domain $R$. That is

$$\int_{c}Pdx + Qdy = \int\int_{R}(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y})dxdy$$

(a) Using Green's theorem, we have

$$\int_{c}(xy^2 dx - xy^2 dy) = \int\int_{R}(\frac{\partial(-xy^2)}{\partial{x}} - \frac{\partial(x^2 y)}{\partial{y}})dx dy$$

$$\int\int_{R}(-y^2 - x^2)dxdy = -\int\int_{R}(x^2 + y^2)dx dy$$

$$= - \int_{0}^{2\pi}\int_{r =0}^{1} r^2 \vert J\vert dr d\theta$$

Now we find the Jacobian $J$. Then $J = \frac{\partial(x,y)}{\partial(r,\theta)} = \left\vert\begin{array}{cc} \fr... ...y}{\partial r} & \frac{\partial y}{\partial \theta} \end{array}\right\vert = r$.

Thus,

$$\int_{c}(xy^2 dx - xy^2 dy) = - \int_{0}^{2\pi}\int_{r =0}^{1} r^2 r dr d\theta$$

$$= - \int_{0}^{2\pi} d\theta \int_{r =0}^{1} r^3 dr = - 2\pi \frac{1}{4} = -\frac{\pi}{2}$$

(b) Using Green's theorem, we have

$$\int_{c}(y dx + 2x dy) = \int\int_{R}(\frac{\partial(2x)}{\partial{x}} - \frac{\partial(y)}{\partial{y}})dx dy$$

$$\int\int_{R}(2 - 1)dxdy = \int\int_{R}dx dy$$

$$= = \frac{\pi}{4}$$