Line Integral

nasi112

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EvaluateCFdr.Evaluate \int_C \mathbf{F} \cdot \mathbf{dr}.
F(x,y)=<1ye2x,2xxy2>,C is the circle (x5)2+(y+6)2=16,oriented counterclockwise.\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.
The integral becomes tedious even if I parametrize it. it looks like it is impossible to solve it without a shortcut or a trick.
 
EvaluateCFdr.Evaluate \int_C \mathbf{F} \cdot \mathbf{dr}.
F(x,y)=<1ye2x,2xxy2>,C is the circle (x5)2+(y+6)2=16,oriented counterclockwise.\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.
The integral becomes tedious even if I parametrize it. it looks like it is impossible to solve it without a shortcut or a trick.
Remember - circle is a CLOSED path.
 
I know it is closed.

CFdr=1ye2x dx+2xxy2 dy\int_C \mathbf{F} \cdot \mathbf{dr} = \int \frac{1}{y} - e^{2x} \ dx + \int 2x - \frac{x}{y^2} \ dy
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 MyMxM_y \neq M_x
In this case, some of the integral will be cancelled which i cannot figure out how, and the remaining will be easy to calculate.
 
Why?

Cd(xy)=0\int_C d\left(\frac{x}{y}\right) = 0
 
<1ye2x,2xxy2><\frac{1}{y}- e^{2x}, 2x- \frac{x}{y^2}> IS a "conservative vector" because
1ye2xy=1y2\frac{\partial \frac{1}{y}- e^{2x}}{\partial y}= -\frac{1}{y^2} and
2xxy2x=1y2\frac{\partial 2x- \frac{x}{y^2}}{\partial x}= -\frac{1}{y^2}

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 (2xxy2)x\frac{\partial(2x - \frac{x}{y^2} )}{\partial x}

(2xxy2)x1y2\frac{\partial(2x - \frac{x}{y^2} )}{\partial x} \neq \frac{-1}{y^2}

(2xxy2)x=21y2\frac{\partial(2x - \frac{x}{y^2} )}{\partial x} = 2 - \frac{1}{y^2}
 
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