AMATH 231 ASSIGNMENT # 5: Greens Theorem Fall 2014

Due Monday, October 20, 2014 at 2pm in box 7, slot 11 (A-M) and 12 (N-Z), located acrossfrom MC4066. Late assignments or assignments submitted to the incorrect dropbox willreceive a grade of zero. Write your solutions clearly and concisely. Marks will be deductedfor poor presentation and incorrect notation.

1. Given the fact that F(x, y, z) = (yz, xz 1, xy) is a conservative vector field, find thepotential function, (x, y, z). [3 marks]

Since F is a conservative vector field, then we know F = . It follows that

x= yz (1a)

y= xz 1 (1b)

z= xy. (1c)

Integrating (1b) with respect to x leads to

(x, y, z) = xyz +K(y, z) (2)

where K is a constant of integration that depends on y and z. The goal is now todetermine K(y, z). Differentiate (2) with respect to y and compare to (1b),

xz +K(y, z)

y= xz 1

which impliesK(y, z)

y= 1

and solving gives K(y, z) = y + K(z) where K(z) is a constant of integration thatdepends on z. It follows that

(x, y, z) = xyz y +K(z)and differentiating this with respect to z and comparing to (1c) leads to

xy +K(z)

z= xy

and henceK(z)

z= 0.

This means K(z) = K0 where K0 is a constant function (that is, K0 no longer dependson x, y, or z.) Therefore, for any K0 a real number, (x, y, z) = xyz y + K0 is apotential function of F. Check that satisfies F = .

2. Decide and justify whether Greens Theorem can be used to evaluate the line integralD

F dx =D

(ln(x+

x2 + y2), 0

) dx

where D is the boundary of [3 marks]

(i) D is the closed disk x2 + y2 1(ii) D is the closed disk (x 2)2 + y2 1

(iii) D is the closed disk (x+ 2)2 + y2 1.Hint: the identity (x+

x2 + y2)(x+x2 + y2) = y2 maybe of use.

Each boundary of D is a simple closed curve; therefore, the only way Greens Theoremcan fail is if F does not satisfy

F is C1 on D D.

So we must determine the domain of F. Because of the ln, x and y must satisfy

x+x2 + y2 > 0. (3)

We need to solve this inequality and determine for which values of x and y are suchthat (3) holds.

Case 1: if x > 0, then clearly x+x2 + y2 > 0 for any y.

Case 2: if x = 0, then x+x2 + y2 =

y2 = |y| > 0 if y 6= 0.

Case 3: if x < 0, then consider the following identity

(x+x2 + y2)(x+

x2 + y2) = y2. (4)

The factor x+x2 + y2 is always positive since x < 0. It follows from this and (4)that the sign of x+

x2 + y2 is the same sign as y2. That is,

Case 3a: if y2 > 0, then x+x2 + y2 > 0 and

Case 3b: if y = 0, then x+x2 + y2 = 0.

Therefore, ln(x+x2 + y2) is not defined when x = 0, y = 0 (case 2) and x < 0, y = 0

(case 3b). Putting this together, ln(x +x2 + y2) is not defined when x 0, y = 0,

which is all of the points on the xaxis for which x 0 (that is, the nonpositive xaxis).The regions described in (i) and (iii) contain portions of the nonpositive xaxis andhence F1 is not C

1 in these two domains; however, the region described in (ii) does notcontain any part of the nonpositive xaxis. It follows that Greens Theorem cannotbe applied to the regions described in (i) and (iii) but can be applied to the regiondescribed in (ii).

3. State whether or not the following sets are (i) connected (ii) simplyconnected.

(a) U1 = R2 {(x, y)| y = 0, x 1}

U1

Figure 1: Consider the square as all of R2. The set U1 = R2 {(x, y)| y = 0, x 1} isdescribed by the gray region.

The set U1 is shown in Figure 1. It is clear form the figure that U1 is connected.Since any simple closed curve in U1 can be drawn around the x-axis and thenshrunken continuously, U1 is simplyconnected. (Example 2.17b from class issimilar to U1.)

(b) U2 = R2 {(x, y)| y = 0, |x| 1}

U2

Figure 2: Consider the square as all of R2. The set U2 = R2 {(x, y)| y = 0, |x| 1} isdescribed by the gray region.

The set U2 is show in Figure 2. The set U2 is connected but since it has a hole(on R2), then it is not simplyconnected. (Example 2.17a from class is similar toU2.)

(c) U3 = {(x, y)|x2 + y2 < 4} {(x, y)|x2 + 4y2 > 4}

The set {(x, y)|x2 + y2 < 4} is an open disk centred at the origin with radius 2.The set {x2 + 4y2 4} is an ellipse centred at the origin with vertical radius 1

U3

Figure 3: The region U3 = {(x, y)|x2 + y2 < 4} {(x, y)|x2 + 4y2 > 4} is shaded in gray.

and horizontal radius 2 since

x2 + 4y2 = 4(x

2

)2+ y2 = 1;

therefore {x2 + 4y2 > 4} is everything outside of the ellipse. The intersection ofboth sets is U3, which is not connected (see Figure 3). Since U3 is not connected,it is not simplyconnected.

4. Let F be a C1 vector field with path independent line integrals in a simply connectedregion U R2. Let f be a C1 scalar field on U . Prove that the vector field H = fFalso has path independent line integrals in U , or find a counterexample to show thatit doesnt. [3 marks]The claim is false. Consider the following counterexample (other examples are of coursepossible): let U = R2, and

F(x, y) = (F1(x, y), F2(x, y)) = (x, y) and f(x, y) = x.

Since U is simply connected we can apply the test for conservative fields; that is,

F2x F1

y= 0

and hence, F is conservative and thus has path independent line integrals on U (bythe second Fundamental Theorem of Calculus for Line Integrals). Now consider

H = fF = (x2, xy)

and applying the test for conservative fields again, we obtain

H2x H1

y= y 0 = y

which means H is not a conservative vector field on U (see Remark 2.20 in the notes).Therefore, by the contrapositive of the first Fundamental Theorem of Calculus for LineIntegrals, H does not have path independent line integrals on U .

5. Determine whether F(x, y) = (3x2y, x3) is conservative or not. If conservative, find thecorresponding potential, (x, y). [0 marks]The test for conservative vector fields can be applied since F is C1 on R2, which is asimplyconnected set, and

F2x F1

y= 3x2 3x2 = 0.

It follows that F is a conservative vector field. To find the potential, set = F whichleads to

x= 3x2y (5a)

y= x3. (5b)

Solving (5a) yields (x, y) = x3y + K(y) where K(y) is a function that depends on yonly and then differentiating this solution with respect to y leads to

y= x3 +

dK

dy.

Comparing with (5b) impliesdK

dy= 0

and solving, leads to K(y) = K0 where K0 is a constant. Therefore, the potential is(x, y) = x3y +K0.

6. Compute~g~F d~x using Greens Theorem:

a) ~F = (2y, x) and ~g = (2 cos t, sin t) for 0 t 2pi. [2 marks]b) ~F = (0, (x2 +1)1) and ~g is the boundary of the rectangle [0, 2] [0, 3], orientated

counterclockwise. [2 marks]

Note: feel free to use clever tricks to avoid integration if you like.Solution:

a) By Greens Theorem we have~g

(2y, x) d~x =

D

(

x(x)

y(2y)

)dxdy = 3

D

dxdy

But the area of the ellipse is 2pi. Therefore, the answer is 6pi.

b) Solution:Use Greens Theorem

~g

(0, (x2 + 1)1) d~x =

D

x(x2 + 1)1 dxdy,

=

30

20

x(x2 + 1)1dx dy,

=

30

[(x2 + 1)1

]20dy,

= 125.

7. Compute~g~F d~x with ~F = (y, x)/(x2 + y2) on the circle described by (x 1)2 +

(y + 1)2 = 1 orientated counter clockwise. Is it easier to evaluate this using Greensfunction or not? How does this compare with the result we found in class?Solution:To compute the line integral requires parameterizing the unit sphere shifted to theright. I personally think thats harder. Either method is fine to evaluate this question.Using Greens theorem we get,

~g

(y, x)x2 + y2

d~x =

D

(

x

(x

x2 + y2

)+

y

(y

x2 + y2

))dxdy,

=

D

0,

= 0.

The calculation above we did in class and therefore not repeated here.