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Green's theorem 
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#1
May2309, 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=x^{2}y=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}}2x2y\text{ }dy dx[/itex] [itex]=\frac{x^2}{2}\frac{4x^{5/2}}{5}[/itex] Does this look right? regards 


#2
May2409, 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. 


#3
May2409, 02:51 PM

P: 127

Thanks
[itex]\int_{0}^{1}\int_{x}^{x^2}2x2y\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
May2409, 08:20 PM

Math
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Thanks
PF Gold
P: 39,363

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!



#5
May2509, 03:08 AM

P: 127

Thanks



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