QA134Mth

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course Mth 277

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.

In the context of a function F(x, y) = F_1 (x, y) i + F_2 (x, y) j we find that the line integral of F dot `ds around a rectangle with sides of length `dx and `dy parallel to the coordinate axes, in the vicinity of the point (x, y), is close to the value of the area integral of (-F_1_y + F_2_x), taken over the entire rectangle.

In the notation of the text, F_1(x, y) is the function M(x, y) and F_2 (x, y) is the function N(x, y), so that F(x, y) = M i + N j. The line integral of F dot `ds around a closed curve C is just the integral

integral( M dx + N dy, integrated around the curve C).

If the functions M and N are reasonably well-behaved and the curve C isn't too pathological, then the result for the original small rectangle can be extended to the entire curve and the integral is equal to the area integral (which will be expressed as a double integral):

integral ( (N_x - M_y) dA, integrated over the region inside the curve).

This is a fairly remarkable result, indicating that the behavior of a vector function inside a region can be predicted by its behavior just on the boundary of the region (and vice versa). This result and its extensions (Stokes' Theorem and the Divergence Theorem) are of wide applicability in the theory of electromagnetism, thermal energy flows and hydrodynamics, to mention only a few of the many applications.

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Question:

`q001. Identify the functions M and N in the following:

integral( 3 x^2 y dx + 5 x e^sqrt(x^2 + y^2) dy, integrated along the arc of a circle of radius 1 centered at the origin).

integral( sqrt( x y) dy - x y / (x^2 + y^2) dx, integrated around boundary of the rectangle 1 <= x <= 3, 1 <= y <= 2)

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Your solution:

int(3x^2ydx+5xe^sqrt(x^2+y^2)dy cent at origin,r=1

M(x,y)=3x^2y

N(x,y)=5xe^sqrt(x^2+y^2)

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

M(x, y)=-xy/(x^2+y^2)

N(x, y)=sqrt(x)

confidence rating #$&*:

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Given Solution:

M is the coefficient of dx, N the coefficient of dy.

So for the first integral, M(x, y) = 3 x^2 y and N(x, y) = 5 x e^sqrt(x^2 + y^2)

For the second integral, N(x, y) = sqrt( x y) and M(x, y) = - x y / (x^2 + y^2).

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Question:

`q002. What is the integrand for the area integral corresponding to each of the integrals in the preceding? Those integrals are

integral( 3 x^2 y dx + 5 x e^sqrt(x^2 + y^2) dy, integrated along the arc of a circle of radius 1 centered at the origin).

and

integral( sqrt( x y) dy - x y / (x^2 + y^2) dx, integrated around boundary of the rectangle 1 <= x <= 3, 1 <= y <= 2).

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3x^2ydx+5xe^sqrt(x^2+y^2)dy cent at origin,r=1

M(x,y)=3x^2y

M_y = 3 x^2

M_x=6xy

N(x,y)=5xe^sqrt(x^2+y^2)

N_x=5e^sqrt(x^2+y^2)+10xye^sqrt(x^2+y^2)

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

M(x, y)=-xy/(x^2+y^2)

N(x, y)=sqrt(x)

confidence rating #$&*:

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Given Solution:

The area integral has integrand N_x - M_y .

For the first integral , M(x, y) = 3 x^2 y and N(x, y) = 5 x e^sqrt(x^2 + y^2). So M_y = 3 x^2, and N_x = 5 e^sqrt(x^2 + y^2) + 10 x y e^sqrt(x^2 + y^2).

The integrand is therefore 3 x^2 - (5 e^sqrt(x^2 + y^2) + 10 x y e^sqrt(x^2 + y^2)).

For the second integral, N(x, y) = sqrt( x y) and M(x, y) = - x y / (x^2 + y^2). The integrand is N_x - M_y = y / (2 sqrt( x y ) ) - ( -x / (x^2 + y^2) + 2 x y^2 / (x^2 + y^2)^2).

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question:

`q003. Parameterize the curves and set up the integrals given in the first question.

The integrals are

integral( 3 x^2 y dx + 5 x e^sqrt(x^2 + y^2) dy, integrated along the arc of a circle of radius 1 centered at the origin).

integral( sqrt( x y) dy - x y / (x^2 + y^2) dx, integrated around boundary of the rectangle 1 <= x <= 3, 1 <= y <= 2)

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

int(3x^2ydx+5xe^sqrt(x^2+y^2)dy r=1 C(0,0)

int(-3cos^2(t)sin^2(t)+5cos^2(t) 0 to 2 pi

int(sqrt(xy)dy-xy/(x^2+y^2)dx1<=x<=3,1<=y<=2)

The rect is giving me some trouble

confidence rating #$&*:

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Given Solution:

The circle of the first integral can be parameterized by x(t) = cos(t), y(t) = sin(t), 0 <= t M= 2 pi. Note that this parameterization takes use in the positive (i.e., counterclockwise) direction around the circle. With this parameterization out integral is

integral ( 3 cos^2(t) sin(t) * (-sin(t) dt) + 5 cos(t) e^sqrt(cos^2(t) + sin^2(t)) * cos(t) dt, t from 0 to 2 pi)

= integral ( -3 cos^2(t) sin^2(t) + 5 cos^2(t), t from 0 to 2 pi).

This integral is perfectly feasible using standard methods.

The rectangle of the second integral can be parameterized on each of its four sides.

The first side runs from x = 1 to x = 3, with y = 1, so that dy = 0, yielding integral( -x * 1 / (x^2 + 1^2) dx, x from 1 to 3) . The integrand simplifies to -x / (1 + x^2).

The second side runs from y = 1 to y = 2, with x = 3, so that dx = 0, yielding integral (sqrt(3 y) dy, y from 2 to 3).

The third side runs from x = 3 to x = 1 with y = 2 and dy = 0, yielding integral( -2 x / (x^2 + 4) dx, x from 3 to 1).

The fourth side yields integral ( sqrt(y) dy, y from 2 to 1).

All these integrals are straightforward (the first and third use substitution u = 1 + x^2 and u = 4 + x^2, respectively).

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