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Green's theorem

by boneill3
Tags: green, theorem
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boneill3
#1
May23-09, 11:22 PM
P: 127
1. The problem statement, all variables and given/known data

Use greens theorem to calculate.
[itex]\int_{c}(e^{x}+y^{2})dx+(e^{x}+y^{2})dy[/itex]

Where c is the region between y=x2y=x

2. Relevant equations

Greens Theorem

[itex]\int_{c}f(x.y)dx+g(x,y)dy= \int_{R}\int (\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y})dA[/itex]


3. The attempt at a solution

[itex]\frac{\partial g}{\partial x}= 2x[/itex]
[itex]\frac{\partial g}{\partial x}= 2y[/itex]
Calculate the integral

[itex]\int_{0}^{x}\int_{0}^{\sqrt{y}}2x-2y\text{ }dy dx[/itex]

[itex]=\frac{x^2}{2}-\frac{4x^{5/2}}{5}[/itex]

Does this look right?
regards
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xaos
#2
May24-09, 12:30 AM
P: 161
with f(x,y)=g(x,y)=exp(x)+y*y, dg/dx=exp(x), the second dg/dx is a typo.

if you want the region bounded by y=x^2 and y=x, the inside integral must be from x^2 to x and the outside 0 to 1 with area element dydx, the result needs to be a value rather than a function, just something to get use to with multiple integrals.
boneill3
#3
May24-09, 02:51 PM
P: 127
Thanks

[itex]\int_{0}^{1}\int_{x}^{x^2}2x-2y\text{ }dy dx[/itex]

[itex]=\frac{1}{30}[/itex]

With the outside limits of double integrals eg 0 to 1 do they always have to be constants?
regards

HallsofIvy
#4
May24-09, 08:20 PM
Math
Emeritus
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Thanks
PF Gold
P: 39,682
Green's theorem

If the result is supposed to be a constant, then, yes, the limits of the integral have to be numbers, not variables!
boneill3
#5
May25-09, 03:08 AM
P: 127
Thanks


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