Line Integral Homework: Find Curve & Vector Field for Green's Theorem [Solved]

mottov2
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Homework Statement


Find the parameterized curve gamma and vector field F so that
the \int\gamma F ds = \int\int2xy dx dy by Green's Theorem.
where -2<y<2
1-sqrt(4-y2) < x < 1+sqrt(4-y2)

The Attempt at a Solution

x = 1 + sqrt(4-y2)
(x-1)2=4-y2
(x-1)2+y2=4

so the path is a circle centered at (1,0) with radius of 2.
parametrize this by setting x = 1+2Cos(t) and y = 2Sin(t)

for the vector field I got F = (xy2,2x2y)
but is there more than one possible vector field?
 
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mottov2 said:

Homework Statement


Find the parameterized curve gamma and vector field F so that
the \int\gamma F ds = \int\int2xy dx dy by Green's Theorem.
where -2<y<2
1-sqrt(4-y2) < x < 1+sqrt(4-y2)

The Attempt at a Solution




x = 1 + sqrt(4-y2)
(x-1)2=4-y2
(x-1)2+y2=4

so the path is a circle centered at (1,0) with radius of 2.
parametrize this by setting x = 1+2Cos(t) and y = 2Sin(t)

for the vector field I got F = (xy2,2x2y)
but is there more than one possible vector field?

Yes, there are infinitely many such vector fields. For example, you can modify yours like this:$$
\vec F = \langle xy^2+g(x), 2x^2y+h(y)\rangle$$where h and g can be any differentiable functions.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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