Multivariable Calculus, Line Integral

  • #1
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Homework Statement:
Calculate a line integral
Relevant Equations:
$$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$
The vector field F wich is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$


C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
 

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  • #2
LCKurtz
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Homework Statement:: Calculate a line integral
Relevant Equations:: $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

The vector field F wich is given by $$\mathbf{F} = \dfrac{(x, y)} {\sqrt {1-x^2-y^2}}$$

And the line integral $$ \int_{C} F \cdot dr $$


C is the path of $$\dfrac{\ (\cos (t), \sin (t))}{ 1+ e^t}$$ , and $$0 ≤ t < \infty $$

How do I calculate this? Anyone got a tip/hint? many thanks
Remember, ##\int_C \vec F \cdot d\vec r = \int_t \vec F(\vec r(t))\cdot \vec r'(t)~dt##. So the obvious hint is plug the formulas in and see what happens. Then come back to show us where you are stuck. There is no substitute for getting your hands dirty.
 
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  • #3
Delta2
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My advice would be to split ##\mathbf{F}## as well the path ##C## into x and y components. For example it will be $$F_x(x,y)=\frac{x}{\sqrt{1-x^2-y^2}}$$ $$F_y(x,y)=\frac{y}{\sqrt{1-x^2-y^2}}$$while for the path C it will be $$x(t)=\frac{\cos t}{1+e^t}$$ $$y(t)=\frac{\sin t}{1+e^t}$$. Then calculate $$x'(t)$$ and $$y'(t)$$ (derivatives of x(t),y(t)with respect to t). Then combine the whole thing to $$\int_0^{\infty} (F_x(x(t),y(t))x'(t)+F_y(x(t),y(t))y'(t))dt$$.
This might look a bit scary but i think if you do all the calculations correctly and the algebra correctly some simplifications will occur (by the use of trigonometric identities).
 

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