Line Integral

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nasi112

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[math]Evaluate \int_C \mathbf{F} \cdot \mathbf{dr}.[/math]
[math]\mathbf{F}(x, y) = <\frac{1}{y} - e^{2x}, 2x - \frac{x}{y^2}>, C \ is \ the \ circle \ (x - 5)^2 + (y + 6)^2 = 16, oriented \ counterclockwise.[/math]
The integral becomes tedious even if I parametrize it. it looks like it is impossible to solve it without a shortcut or a trick.
 
nasi112 said:
[math]Evaluate \int_C \mathbf{F} \cdot \mathbf{dr}.[/math]
[math]\mathbf{F}(x, y) = <\frac{1}{y} - e^{2x}, 2x - \frac{x}{y^2}>, C \ is \ the \ circle \ (x - 5)^2 + (y + 6)^2 = 16, oriented \ counterclockwise.[/math]
The integral becomes tedious even if I parametrize it. it looks like it is impossible to solve it without a shortcut or a trick.
Click to expand...
Remember - circle is a CLOSED path.
 
I know it is closed.

[math]\int_C \mathbf{F} \cdot \mathbf{dr} = \int \frac{1}{y} - e^{2x} \ dx + \int 2x - \frac{x}{y^2} \ dy [/math]
How can I know which one of these integrals will be zero without calculating because the path is closed?
 
The article is focusing on conservative vectors which makes the integral independent of path. But here we have [math]M_y \neq M_x[/math]
In this case, some of the integral will be cancelled which i cannot figure out how, and the remaining will be easy to calculate.
 
[imath]<\frac{1}{y}- e^{2x}, 2x- \frac{x}{y^2}>[/imath] IS a "conservative vector" because
[imath]\frac{\partial \frac{1}{y}- e^{2x}}{\partial y}= -\frac{1}{y^2}[/imath] and
[imath]\frac{\partial 2x- \frac{x}{y^2}}{\partial x}= -\frac{1}{y^2}[/imath]

so the integral around the closed path is 0.
 
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Thanks a lot Professor HallsofIvy for passing by.

i don't agree with your calculations regarding [math]\frac{\partial(2x - \frac{x}{y^2} )}{\partial x}[/math]

[math]\frac{\partial(2x - \frac{x}{y^2} )}{\partial x} \neq \frac{-1}{y^2}[/math]

[math]\frac{\partial(2x - \frac{x}{y^2} )}{\partial x} = 2 - \frac{1}{y^2}[/math]
 
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