Query 134

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course Mth 277

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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.YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

Worked on paper. Took a while to realize why -1<=x<=1 and y its parameters. but understood.

for my partials for greens theorum. I did partial y derive first, then to get the integral -3x^2sqrt(1-x^2) - (-sqrt(1-x^2)^3)

then for that integral with respect to x, I've got some chain rules with u substitution.

int( -3x^2(-x/(sqrt(1-x^2)) + (-6x(sqrt(1-x^2)) + (-3x(sqrt(1-x^2)) from -1 to 1. for these parameters I got a value of 0.

I""m not sure why I got 0. or if I was supposed to fill in for x or y with cos(theta) or sin(theta)

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I don't see enough detail to tell exactly what you did here.

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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):

I'm honestly not seeing where I messed up here.

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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:

split it up into the M dx and N dy form. and take the partial deriv of x for the N function, and y for the M function.

Then integrate (this is all on paper). with respect to x from 0 to 3, then y from 3 to 0.

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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):

It seems my integral for the y part is wrong, and I had a sign error. I think i see where i messed up. but what is this about the part where you broke it into 4 integrals and evalutated it there. are you just showing that you can do it

different ways?

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The boundary of the region consists of segments of four different straight lines. There is no way to parameterize two different lines using the same parameterization, so you need to break the line integral up into four separate integrals.

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Of course it's easier to do the area integral, which doesn't need to be split up.

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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:

just getting out of the way, solve for x and y to find the parameters.

-2<=x<=2

-sqrt(4-x^2)<=y<=sqrt(4-x^2)

getting the equation into the M and N form. it should be int(2y^2dx + 3x^2dy) over C, now take the partial deravation using Greens Theorum to get into the equation N_x - M_y = 6x - 4y

so the int(6x-4y)dx, -2,2 = 6 - 4xy from x= -2 to 2 = 6

actually do the y integration first, because you have an actual value for x, and not y.

should get the answer to be zero. because it went from -2 to 2

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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, My derivative was hard to get for me because of the u substitution. I set u = 4-x^2 and du = -2x dx. It took me a while to realize that the x was taken care of after you took out the constant 12 out of the

integral.

12 int(x*sqrt(4-x^2) ) was the integrand, I kept ending up with 4x(4-x^2)^3/2 instead of just 4 in front with no x. "

12 int(x*sqrt(4-x^2) ) was the integrand, I kept ending up with 4x(4-x^2)^3/2 instead of just 4 in front with no x. "

&#Good responses. See my notes and let me know if you have questions. &#