If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://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.
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:
Confidence rating:
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.
Self-critique (if necessary):
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.
Your solution:
Confidence rating:
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.
Self-critique (if necessary):
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.
Your solution:
Confidence rating:
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.
Given Solution:
Self-critique (if necessary):