Query134

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course Mth 277

12/8/2011 1015pm

If your solution to stated problem does not match the given solution, you should self-critique per instructions athttp://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what

you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

At the end of this document, after the qa problems (which provide you with questions and solutions), there is a series of Questions, Problems and Exercises.

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Question: Use Green's Theorem to evaluate Int[ y^3 dx - x^3 dy, C] where C is parameterized by x = cos(theta), y = sin(theta) and 0 <= theta <= 2pi.

Check your answer by computing the line integral without Green's Theorem.

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Your solution:

Int[ y^3 dx - x^3 dy, C]

0 <= theta <= 2pi

M = y^3dy

N = -x^3dx

x = cos(theta)

x = cos(2pi)

= 1

x = cos(0)

= -1

y = sin(theta)

y = sin(2pi)

= 0

y = sin(0)

= 0

0 <= theta <= 2pi

Int[ y^3 dx - x^3 dy, C]

Int[ Int [N_x - M_y], 0,0] -1,1]

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The region is the unit circle.

The limits on your integral would imply that the region is 0 <= y <= 0, -1 <= x <= 1. Those inequalities don't describe the unit circle.

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= Int[ Int[-3x^2 - 3y^2]dy , 0, 0]dx -1,1]

= Int[ -3x^2] dx -1,1

= -x^3]-1,1

= 0

Int[ y^3 dx - x^3 dy, C]

x = cos(theta)

y = sin(theta)

dy = cos(theta)

dx = -sin(theta)

Int[ sin(theta)^3 * -sin(theta) - cos(theta)^3 * cos(theta), 0, 2pi

= Int[(1-cos^2(theta))*sin(theta) - cos(theta)^4(dtheta) 0 , 2pi

= -cos(theta) + cos^3(theta)/3] 0, 2pi - Int[(1/4)(1 + cos(2theta)^2) ]dtheta, 0, 2pi

= [-cos(2pi) + (1/3)cos(2pi)^3 - cos(0) + (1/3)cos(0)^3] - Int[1 + 2cos(2theta) + (1/8){1 + cos(4theta)]dtheta, 0, 2pi

= (1 + 1/3) - (1 + 1/3) - Int[(1/8)(3 + 4cos(2theta) + cos(4theta))]dtheta 0, 2pi

= 0 - (1/8) [3(theta) + 2sin2(theta) + 1/4 sin(4theta)]

= (1/8) [6pi + 2sin(4pi) + 1/4 sin(8pi)] - (1/8) [0 + 2sin(0) + 1/4 sin(0)]

= [(1/8) * 18.8 + 0 + 0] - [0]

= 2.35

confidence rating #$&*:

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Given Solution:

This function is of the form M dx + N dy, with M = y^3 and N = x^3.

This function is integrated over a positively oriented closed curve, so by Green's Theorem the given line integral is equal to the integral of N_x - M_y

over the region enclosed by the curve.

The curve is the unit circle, traversed counterclockwise. The region can be described by -1 <= x <= 1, -sqrt(1 - x^2) <= y <= sqrt(1 - x^2),

so our integral is integral ( integral( N_x - M_y) dy, -sqrt(1 - x^2), sqrt(1 - x^2)), dx, -1, 1)

= integral ( integral( -3 x^2 - 3 y^2 ) dy, -sqrt(1 - x^2), sqrt(1 - x^2)), dx, -1, 1)

An antiderivative of -3 x^2 - 3 y^2 with respect to y is -3 x^2 y - y^3, which between the given limits changes from 3 x^2 sqrt(1 - x^2) + (1 - x^2)^(3/2)

to -3 x^2 sqrt(1 - x^2) - (1 - x^2)^(3/2) , a change of -6 x^2 sqrt(1 - x^2) - 2 (1 - x^2)^(3/2).

We are thus left with

integral ( -6 x^2 sqrt(1 - x^2) - 2 (1 - x^2)^(3/2), x, -1, 1).

Evaluating this integral we get -3 pi / 2.

The line integral around the curve is easily written down. We have x = cos(t), y = sin(t), dx = - sin(t) dt, dy = cos(t) dt so that

Int[ y^3 dx - x^3 dy, C] = int( sin^3(t) * (-sin(t) dt) - cos^3(t) * cos(t) dt , t from 0 to 2 pi)

= int(-sin^4(t) - cos^4(t) dt, 0, 2 pi).

This integral comes out to -3 pi / 2, in agreement with the area integral.

Notes on details of integrations:

x^2 sqrt(1 - x^2) contains the expression x sqrt( 1 - x^2), which can be evaluated with substitution

x^2 sqrt(1 - x^2) can then be written as x ( x sqrt(1 - x^2)), and integrated by parts

sin^4(t) can be written sin^2(t) ( 1 - cos^2(t)) = sin^2(t) - sin^2(t) cos^2(t).

sin^2(t) can be integrated by parts, letting u = sin(t) and dv = sin(t) dt.

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Self-critique (if necessary):

hm.. the second part of the answer was off by (1/2). I knew it wasn't correct because when I solved for greene's theorem, the answer was zero, but the questions in the book that related to

the problem, regularly came out as zero. Which is understandable because it is a closed loop. displacement is zero, therefore the work done is zero. But if you look at it at the total distance traveled, you get a different answer.

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The work done would be zero if the function M `i + N `j was conservative. However there is no guarantee that this is the case.

The main error with your integrals was that they did not correctly describe the region of integration.

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Question: Use Green's Theorem to evaluate the integral Int[2x arctan(y) dx - x^2*y^2 (1 + y^2) dy, C] where C is the square with vertices (0,0), (3,0),

(3,3), (0,3) traversed clockwise.

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Your solution:

Int[2x arctan(y) dx - x^2*y^2 (1 + y^2) dy, C]

M = 2x arctan(y)

N = - x^2*y^2 (1 + y^2)

Int[ Int[ N_x - M_y]dy 0, 3, 3, 0] dx 0, 0, 3, 3

= Int[ Int[ 2x*y^2(1+y^2) - 1/(1+y^2)] dy, dx from (0,0), (3,0),(3,3), (0,3) traversed clockwise

use trig substitution

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At this point you are integrating with respect to area over the rectangle. There is no direction to the integration. If you were integrating around a boundary then the curve would indeed have an orientation.

Your integrand should be 2x*y^2(1+y^2) - 2x/(1+y^2). M = 2 x arctan(y) so M_y = 2 x * ( 1 / (1 + y^2) ).

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= Int [ (2xy^3) / 3 + (2xy^5)/5 - ln(y^2 + y + 1)

= Int [ 54x/3 + 566x/5 - ln(13), dx

= 9x^2 + (283/5)x^2 - ln(13)x, 0 to 3 to 0

= 0 - 81+(2547/5) - 9ln(13) = 567.31

confidence rating #$&*:

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Given Solution:

This function is of the form M dx + N dy, with M = 2x arcTan(y) and N = -x^2 y^2 ( 1 + y^2).

This function is integrated over a positively oriented closed curve, so by Green's Theorem the given line integral is equal to the integral of N_x - M_y over

the region enclosed by the curve.

N_x = 2 x y^2 ( 1 + y^2) and M_y = 2 x / (1 + y^2), so N_x - M_y = -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2).

The region is easily described, so the integrand of our area integral will be -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2). We can choose to integrate first with

respect to x or y. Integrating first with respect to x we get

int(int( -2 x y^2 ( 1 + y^2) - 2 x / (1 + y^2) dx, 0, 3) dy, 0, 3)

= - 2 int (int x dx,0,3) (y^2 ( 1 + y^2) + 1 / (1 + y^2)) dy, 0, 3)

= -2 int ( 9/2 ( y^2 + y^4 + 1 / (1 + y^2) ) ) dy, 0, 3)

= - 9 int ( (y^2 + y^4 + 1 / ( 1 + y^2) dy, 0, 3)

An antiderivative is y^3 / 3 + y^5 / 5 + arcTan(y). This antiderivative changes from 0 at y = 0 to 288 / 5 + arcTan(3) at y = 3, a change of 288 / 5 + arcTan(3).

Our integral is therefore -9 ( 288/5 + arcTan(3) ) = -9 arcTan(3) - 2592 / 5.

The line integral around the square can be broken into four integrals. The four paths can be parameterized as

x = t, y = 0

x = 3, y = t

x = (3 - t), y = 0

x = 0, y = 3 - t

all for 0 <= t <= 4.

On the first and third paths y is constant so the dy term will be zero.

On the second and fourth paths x is constant so the dx term will be zero.

On the fourth path both terms have x as a factor. Since x = 0 on this path, the integrand is therefore zero.

On the third path x = 3 - t so dx = - dt; on the fourth y = 3 - t so dy = - dt.

Our integral around the boundary of the region is therefore

int( 2 t arcTan(0) dt, 0, 3)

+ int ( -(3^2 t^2 ( 1 + t^2) dt, 0, 3)

+ int( 2 ( 3 - t) arcTan(3) * (-dt), 0, 3)

+ int ( 0 (- dt), 0, 3).

arcTan(0) = 0 so the first integral is just zero. This leaves us with

- int( 9 t^2 ( 1 + t^2 ) dt, 0, 3) - 2 arcTan(3) int( (3 - t) dt, 0, 3).

Both integrands are polynomials. You should easily be able to confirm that the result is

-2592 / 5 - 9 arcTan(3).,

in agreement with the previous area integral.

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Self-critique (if necessary):

Shouldve gone from 03 just once.

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Question:

Find the work done when an object moves in the force field F(x,y) = 2y^2i + 3x^2j counterclockwise around the circular path x^2 + y^2 = 4.

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Your solution:

F(x,y) = 2y^2i + 3x^2j

x^2 + y^2 = 4

M = 2y^2

N = 3x^2

M_y = 4y

N_x = 6x

Int[N-MdA]

= Int[ Int[6x - 4y] dy, -sqrt(4-x^2), sqrt(4-x^2) dx, -2, 2

= Int[- 2(4-x^2) + 2(4-x^2) dx, -2, 2

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An antiderivative of 6 x - 4 y with respect to y is 6 x y - 2 y^2; substituting the y limits for y tarms will yield 6 x * ( 2 sqrt(4 - x^2)), or something very similar.

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= [-8x +(2/3)x^3] - [-8x + (2/3)x^3)]

= [16 - 16/3] - [-16 - 16/3]

= 0

confidence rating #$&*:

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The work would be the line integral int ( F dot ds ) around the curve. If ds = dx i + dy j the integral would be

int(2 y^2 dx + 3 x^2 dy, over C),

which is of the form int(M dx + N dy, over C), with M = 2 y^2 and N = 3 x^2.

By Green's Theorem this is the integral of N_x - M_y = 6 x - 4 y the enclosed region.

In terms of x and y the region is -2 <= x <= 2, -sqrt(4 - x^2) <= y <= sqrt(4 - x^2).

The integral would therefore be

int(int ( (6 x - 4 y) dy), -sqrt(4 - x^2), sqrt(4 - x^2))dx, -2, 2)

The antiderivative for the inner integral is 6 x y - 2 y^2. The antiderivative changes by 12 x sqrt( 4 - x^2).

The outer integral is therefore

int( 12 x sqrt(4 - x^2) dx, -2, 2).

An antiderivative is 4 * 2/3 (4 - x^2)^(3/2). The antiderivative has the same value at x = -2 and at x = 2, so the integral is zero.

The path could be parameterized by x = 4 cos(t), y = 4 sin(t). The integral would be

int( 3 * 16 cos^2(t) * (-sin t) dt + 2 * 16 sin^2(t) * cos(t) dt, 0, 2 pi).

The antiderivative is a multiple of cos^3(t) added to a multiple of sin^3(t), and hence will not change between t = 0 and t = 2 pi. The resulting integral is

therefore zero.

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Given Solution: Using Greene's Theorem I got 0 for the answer. Not sure if my algebra is correct, the integration got a sloppy towards the end with the positives and negatives. If its not 0 its 32. But I'm pretty sure its zero, because Work = Force times displacement and the displacement on a circle is zero

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Again you don't know if the force field is conservative so you can't know that the integral is zero.

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Self-critique (if necessary):3

Che

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Check my notes and let me know if you have questions.

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