Daily Archives: October 23, 2013

MA 1505 Mathematics I

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

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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.

429px-Green's-theorem-simple-region.svg

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

macroscopic_microscopic_circulationmacroscopic_microscopic_circulation_hole

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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