query 134

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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.

Your solution:

x = cos(theta)

y = sin(theta)

0 <= theta <= 2pi

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

Int(Int((dN/dx - dM/dy) dA)

dN/dx = -3x^2, dM/dy = 3y^2

Int(Int((3y^2- (-3x^2)) dA) = Int(Int(3(y^2 + x^2) dA)

Int(Int(3(y^2 + x^2) dA) = Int(Int(3(r^2)*r) dr,0,1)d`theta,0,2`pi)

Int(3r^3 dr) = (3/4)r^4

Int((3/4) d`theta) = (3/4)`theta

((3/4)*2`pi) - ((3/4)*0) = (3/2)`pi

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

= Int(sin(`theta)^3*(-sin(`theta))d`theta,0,2`pi)

=Int( -(((1 - cos(2`theta))/2)^2) dt) = Int( -(1/4)(1 - cos(2`theta))^2 dt)

= Int( -(1/4)(1 - 2cos(2`theta) + cos(2`theta)^2 dt)

= Int( -(1/4)(1 - 2cos(2`theta) + (1+cos(4`theta))/2) dt)

=Int((-3/8) + (1/4)*2cos(2`theta) - (1/8)*2cos(2`theta) d`theta,0,2`pi)

=Int((-3/8) d`theta) + Int((1/4)*2cos(2`theta) d`theta) - Int((1/8)2cos(4`theta) d`theta)

Using u sub for inner fn of cos we have simple cos integrals

= (-3/8)*(`theta) + [(1/4)*sin(2`theta)] - [(1/16)sin(4`theta)], evaluated

=(-3`pi/4)

= Int(cos(`theta))^3*(cos(`theta))d`theta,0,2`pi)

=Int( (((1 + cos(2`theta))/2)^2) dt) = Int( (1/4)(1 + cos(2`theta))^2 dt)

= Int( (1/4)(1+ 2cos(2`theta) + cos(2`theta)^2 dt)

= Int( (1/4)(1 + 2cos(2`theta) + (1+cos(4`theta))/2) dt)

=Int((3/8) + (1/4)*2cos(2`theta) + (1/8)*2cos(2`theta) d`theta,0,2`pi)

=Int((3/8) d`theta) + Int((1/4)*2cos(2`theta) d`theta) + Int((1/8)2cos(4`theta) d`theta)

= (3/8)*(`theta) + [(1/4)*sin(2`theta)] + [(1/16)sin(4`theta)]=((3/8)*2`pi)

Int[ y^3 dx - x^3 dy, C] = (-3`pi/4) - (3`pi/4)

-3`pi/2

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):OK

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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 (0,0), (3,0),(3,3), (0,3)

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

81+(2547/5) - 9ln(13)

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):OK

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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=2x^2

Nx-My=6x-4y

Int((6x - 4y)dy)=6xy-2y^2

6x*`sqrt(4-x^2)-2(4-x^2)

Int(6x*`sqrt(4-x^2)-2(4-x^2)dx)

Int(6x*`sqrt(4-x^2)-2(4-x^2)dx)

Int(6x*`sqrt(4-x^2) dx) - Int(8 dx) + Int(2x^2 dx)

Int(6x*`sqrt(4-x^2) dx)

x^2, u’ = 2x xd Int((1/3)*`sqrt(4-u) du)

(1/3)((2/3)(4-u)^(3/2)

(2/9)(4-x^2)^(3/2)

Int(8 dx)

8x

Int(2x^2 dx)

(2/3)*x^3

Int(6x*`sqrt(4-x^2) - 2(4-x^2)dx) = [(2/9)(4-x^2)^(3/2)] - [8x] + [(2/3)*x^3]

-16 + (15/3) - (16/9)

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:

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Self-critique (if necessary):OK"

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