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I am a little confused on the computation of a Line Integral of a Vector Field.

Here is what I have so far:

$$ \int_C \mathbf F \cdot d \vec r$$ (F is a vector field of n dimensions ($$ n \ge 2- dimensions$$)

I know that you need a parametrization of the Curve C (c=[{x(t),y(t)} $\in$ a $\le$ t $\le$ b])

and that ||d$\vec r$||= $\sqrt{(dx/dt)^2+(dy/dt)^2} dt$

But other then that I am a tad bit lost :P, So any help would be appreciated.

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Think of the integral as: $$ \int \boldsymbol{F} \cdot d\vec{r} = \int_a^b\left[F_x(t) \hat{\boldsymbol{x}} + F_y(t)\hat{\boldsymbol{y}}\right]\cdot \left(dx(t)\hat{\boldsymbol{x}} + dy(t)\hat{\boldsymbol{y}}\right) \\ = \int_a^b F_x(t)dx(t) + F_y(t)dy(t) $$ If you have the parametrizations of $x$, $y$ and $\boldsymbol{F}$ as functions of $t$ then this is the easier path (no pun intended) than evaluating $\lvert d\vec{r} \rvert$.

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  • $\begingroup$ Thanks a lot, that was really helpful. $\endgroup$ – user129570 Feb 18 '14 at 0:45
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Let C be a smooth curve in $R^2$ and $\vec F=\vec F(x,y,z)$ is a vector field, the Line Integral of $\vec F$ along C is: $$\int_C \vec F\cdot d\vec r=\int_{t_A}^{t_B} \vec F(t)\cdot \frac{d\vec r}{dt}dt$$ where $\vec r(t)=(x(t), y(t))$ is the parametric representation of the curve C. Note that the result of this is always a scalar.

Line Integrals of vector fields can also be given by: $$\int_C \vec F \times d\vec r=\int_{t_A}^{t_B} \vec F(t) \times \frac{d\vec r}{dt}dt$$ The result of this will be a vector. The $\times$'s is the cross product.

You can also watch this video. It should be helpful. And here is an example.

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