MA 1505 Tutorial 9 and 10: Line Integral and Green’s Formula

Line integral of a scalar field:

Assume $f: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}$ is a smooth function,

where $U$ is the domain of $f$ .

$\int_{C} f ds =\int_{a}^{b} f(r(t)) \cdot ||r^{'}(t)|| dt$

where $r:[a,b] \rightarrow C$ is a smooth curve.

Line integral of a vector field:

Assume $\mathbf{F}: U \subseteq \mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is a smooth vector function,

$\int_{C} \mathbf{F} \cdot d\mathbf{r} = \int_{a}^{b} \mathbf{F}( \mathbf{r}(t)) \cdot \mathbf{r}^{'}(t) dt$

where $r:[a,b] \rightarrow C$ is a smooth curve.

Green’s Formula:

Assume $\bold{F}=(P,Q)$ is a vector field,

$\oint_{\partial D} \bold{F}\cdot d\bold{s}=\oint_{\partial D} Pdx + Qdy = \iint_{D} (\frac{\partial Q}{\partial x} - \frac{ \partial P}{\partial y}) dxdy,$

where $D$ is the domain and $\partial D$ denotes the boundary of $D.$ The orientation of $\partial D$ satisfies the left hand rule. That means if you walk along the boundary of $D$ , the domain $D$ must be on your left.

$D$ is a simply connected region with boundary consisting four boundaries $C_{1}, C_{2}, C_{3}, C_{4}$ , the orientation is counterclockwise.

In the first graph, $\partial D$ which denotes the boundary of $D$ has only one closed curve $C$ and the orientation of $C$ is counterclockwise. However, in the second graph, $\partial D$ contains two curves, i.e. the blue one and the red one. The orientation on the blue one which is the outer boundary of $D$ is counterclockwise, the orientation on the red one which in the inner boundary of $D$ is clockwise. That means if you walk along the boundary of $D$ , the domain $D$ must be on your left. This is the left hand rule.

Corollary of Green’s Theorem

Assume $D$ is a domain in the plane, $Area(D)$ denotes the area of $D$ , then the area can be calculated from the following formulas:

$Area(D)=\iint_{D} dxdy$

$= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)$

Fundamental Theorem of Line Integral:

$\int_{C}\nabla f \cdot d\bold{r}=f(\text{terminal point})-f(\text{initial point}),$

where $C$ denotes a curve from initial point to terminal point and $f$ is a scalar field.

Conservative Vector Field:

1. $\bold{F}=(P,Q,R)$ is called a conservative vector field, if there exists a scalar field $f$ such that $\nabla f=\bold{F}$ . It is equivalent to these conditions:

$\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x},$ $\frac{\partial Q}{\partial z}=\frac{\partial R}{\partial y},$ $\frac{\partial R}{\partial x}=\frac{\partial P}{\partial z}$ .

2. $\bold{F}=(P,Q)$ is called a conservative vector field, if there exists a scalar field $f$ such that $\nabla f=\bold{F}$ . It is equivalent the condition $\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}.$

Path Independence:

A key property of a conservative vector field is that its integral along a path depends only on the endpoints of that path, not the particular route taken.

For example, if $\bold{F}=(P,Q)$ or $\bold{F}=(P,Q,R)$ is a conservative vector field, then the value of the line integral $I=\int_{C} \bold{F}\cdot d\bold{r}$ depends only on the initial point and terminal point of the curve $C$ . That means if $\bold{F}$ is a conservative vector field, the curves $C_{1}$ and $C_{2}$ have the same initial points and terminal points, then these two line integrals are equal: $\int_{C_{1}} \bold{F}\cdot d\bold{r}=\int_{C_{2}}\bold{F}\cdot d\bold{r}$ . For this reason, a line integral of a conservative vector field is called path independent.

Question 1. For each non-zero constant $a>0$ , let $C_{a}$ denote the curve $y=a \sin x$ , where $0 \leq x \leq \pi.$ Let

$I(a)= \int_{C_{a}} (1+y^{3})dx + (2x+y)dy$

Find the minimum value of $I(a)$ in the domain $a>0.$

Solution.

Method (i). Use the definition of line integration.

Since $y=a \sin x$ , $0 \leq x \leq \pi$ , $dy= a \cos x dx$ ,

$I(a)= \int_{0}^{\pi} (1+a^{3} \sin^{3} x) dx + (2x+ a \sin x) a \cos x dx$

$= \int_{0}^{\pi} ( 1+ a^{3} \sin^{3} x + 2a x \cos x + a^{2} \sin x \cos x) dx.$

Since

$\int_{0}^{\pi} (a^{3} \sin^{3} x) dx = \frac{4}{3} a^{3},$

$\int_{0}^{\pi} (2a x \cos x) dx =-4a,$

$\int_{0}^{\pi} (a^{2} \sin x \cos x) dx =0,$

we get

$I(a)= \pi + \frac{4}{3} a^{3} - 4a$ on $a>0$ .

$I^{'}(a) = 4a^{2}-4$ .

The minimum value is taken at $a=1,$ the $I(1)= \pi -\frac{8}{3}.$

Method (ii). Use Green’s Formula.

Consider the domain $D$ bounded by $y= a\sin x$ and $x=0,$ $P(x,y)=1+y^{3}$ and $Q(x,y)= 2x+y.$

i.e.

$D: 0\leq x \leq \pi, 0\leq y \leq a \sin x.$

From Green’s Formula, pay attention to the orientation,

$\iint_{D} ( 2-3y^{2}) dxdy = -I(a) + \int_{0}^{\pi} dx$

Therefore,

$I(a) = \pi - \iint_{D} ( 2- 3y^{2}) dxdy$

$= \pi - \int_{0}^{\pi} \int_{0}^{a \sin x} (2-3y^{2}) dy dx$

$= \pi - \int_{0}^{\pi} ( 2a \sin x - a^{3} \sin^{3} x) dx$

$= \pi - 4a + \frac{4}{3} a^{3}.$

The derivative of $I(a)$ is $I^{'}(a) = 4 a^{2}-4,$ the minimum value is taken at $a=1,$ and $I(1)= \pi-\frac{8}{3}.$

Question 2. Prove the area of the disc with radius R is $\pi R^{2}$ .

Solution.

Method (i). Definition of Integration.

$Area=4 \int_{0}^{R} \sqrt{R^{2}-x^{2}} dx$

$= 4R^{2} \int_{0}^{\frac{\pi}{2}} cos^{2}\theta d \theta$ $(x=\sin \theta)$

$= 4R^{2} \int_{0}^{\frac{\pi}{2}} \frac{1+ \cos (2\theta)}{2} d\theta$

$= \pi R^{2}.$

Method (ii). Green Formula

$Area(D)=\iint_{D} dxdy$

$= \oint_{\partial D} xdy = \oint_{\partial D} -ydx =\oint_{\partial D} (-\frac{y}{2} dx +\frac{x}{2} dy)$

For $D=\{ x^{2}+y^{2}\leq R^{2} \},$ on $\partial D,$ $x=\cos \theta,$ $y=\sin\theta,$ where $\theta \in [0, 2\pi).$

$Area(D)= \oint_{\partial D} x dy$

$= \int_{0}^{2\pi} R \cos \theta \cdot R \cos \theta d\theta$

$= \pi R^{2}.$

Question 3. Prove the area of the ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} =1$ is $\pi ab$ .

Solution. It is similar to Question 2, and $x=a \cos \theta,$ $y= b \cos \theta.$

Question 4. Calculate

$\int_{C} 2xy dx + (x^{2}+z)dy +y dz$

where $C$ consists two line segments: $C_{1}$ from (0,0,0) to (1,0,2), and $C_{2}$ from (1,0,2) to (3,4,1).

Solution.

Method (i). From the definition of line integral,

$C_{1}: r_{1}(t)=(t,0,2t),$ $C_{2}: r_{2}(t)=(1+2t, 4t, 2-t).$

$\int_{C_{1}} 2xy dx + (x^{2}+z)dy +y dz=0$

$\int_{C_{2}} 2xy dx + (x^{2}+z)dy +y dz$

$=\int_{0}^{1}( 48 t^{2} + 24 t +12) dt$

$=40.$

Method (ii). Check the vector field $\mathbf{F}=(2xy, x^{2}+z, y)$ is a conservative vector field. Since

$\frac{\partial}{\partial y} (2xy)= \frac{\partial}{\partial x} (x^{2}+z)=2x$

$\frac{\partial}{\partial x}(y)=\frac{\partial}{\partial z}(2xy)=0$

$\frac{\partial}{\partial y}(y)=\frac{\partial}{\partial z}(x^{2}+z)=1$

Therefore, $\mathbf{F}$ is a conservative vector field, and we can assume $\nabla f=\mathbf{F}$ , i.e. $f_{x}=2xy, f_{y}=x^{2}+z, f_{z}=y.$ Hence,

$f(x,y,z)=x^{2}y+yz+K$ for some constant $K.$

Therefore, the answer is $f(3,4,1)-f(0,0,0)=40.$

Question 5. Evaluate

$\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,$

where C denotes the boundary with positive orientation of the region between the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=b^{2}$ with $0<a<b.$

Solution.

Method (i). The definition of the line integral.

On circle $x^{2}+y^{2}=b^{2},$ it is counterclockwise, $x= b \cos \theta \text{ and } y= b \sin \theta,$ $\theta$ is from 0 to $2\pi.$

On circle $x^{2}+y^{2}=a^{2},$ it is clockwise, $x=a \cos \theta \text{ and } y= a \sin \theta,$ $\theta$ is from $2\pi$ to 0.

On the circle $C_{b}: x^{2}+y^{2}=b^{2}$ ,

$\oint_{C_{b}} (x^{5}-y^{5})dx + ( x^{5}+y^{5}) dy$

$= b^{6} \int_{0}^{2\pi} ( \cos^{5} \theta - \sin^{5} \theta) \cdot ( - \sin \theta) + ( \cos^{5} \theta + \sin^{5} \theta) \cdot \cos \theta ) d \theta$

$= b^{6} \int_{0}^{2\pi} ( \cos^{6} \theta + \sin^{6} \theta) - \sin \theta \cos \theta ( \cos^{4} \theta - \sin^{4}\theta ) d\theta$

$= b^{6} \int_{0}^{2\pi} ( 1- 3\sin^{2} \theta \cos^{2}\theta - \frac{1}{2} \sin 2\theta \cos 2\theta ) d\theta$

$= b^{6} \int_{0}^{2\pi} ( \frac{5}{8} -\frac{3}{8} \cos 4\theta -\frac{1}{4} \sin 4 \theta ) d\theta$

$= b^{6} \cdot \frac{5}{4} \cdot \pi.$

Pay attention to the orientation, we get the answer is

$\frac{5 \pi}{4} ( b^{6}-a^{6}).$

Method (ii). Green’s Theorem.

$\oint_{C} (x^{5}-y^{5}) dx + (x^{5}+y^{5}) dy,$

$= \iint_{D}( 5 x^{4}+ 5 y^{4}) dxdy$

$= 5 \int_{0}^{2\pi} \int_{a}^{b} r^{5}( \cos^{4} \theta + \sin^{4} \theta) dr d \theta$

$= 5 \int_{a}^{b} r^{5} dr \cdot \int_{0}^{2 \pi} ( \cos^{4} \theta + \sin^{4}\theta) d\theta$

$= \frac{5( b^{6}-a^{6})}{6} \int_{0}^{2\pi} (1-2\sin^{2}\theta \cos^{2} \theta) d\theta$

$= \frac{5(b^{6}-a^{6})}{6} \cdot \frac{3 \pi}{2}$

$= \frac{5}{4} \pi ( b^{6}-a^{6}).$

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ZHANG RONG

MA 1505 Tutorial 9 and 10: Line Integral and Green’s Formula

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