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Green's theorem

  • Thread starter boneill3
  • Start date
  • #1
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Homework Statement



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

Homework 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]


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
 

Answers and Replies

  • #2
164
2
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.
 
  • #3
127
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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
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,777
911
If the result is supposed to be a constant, then, yes, the limits of the integral have to be numbers, not variables!
 
  • #5
127
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Thanks
 

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