# Line Integral with respect to

1. Apr 1, 2016

### SamitC

1. The problem statement, all variables and given/known data
Need to visualize what it means by Line Integral along curve C with respect to x or y axis.
For example suppose the curve is C (I did not find a way to write the C under the integration sign here)

∫ f(x,y) ds is like a fence along C whose height varies as per f(x,y). The line integral with respect to delta arc length ds is the area of the fence within two points on the Curve.

Then what does ∫ f(x,y) dx or ∫ f(x,y) dy or ∫ f(x,y) dx+∫ f(x,y) dy mean ? The curve (and fence) are still the same.
(i have not used the parametrized form here since omitting that will not change the question - i think)

If possible can somebody share a pictorial representation with explanation or share a link with the same/

2. Relevant equations
Does ∫ f(x,y) dx means how much of the fence can be seen from the other side of the x-axis perpendicularly???
If that is so, what is its relevance?

3. The attempt at a solution
Not Applicable

Last edited: Apr 1, 2016
2. Apr 1, 2016

### SamitC

Last edited: Apr 1, 2016
3. Apr 1, 2016

### LCKurtz

Personally, I don't like the fence sitting atop a curve example, because it is so artificial and isn't what these integrals are typically used for. A much better example is working with force fields. Suppose every point in the xy plane lies in a force field. Could be, for example, gravitational or magnetic. So each point $(x,y)$ has an associated force vector $\vec F(x,y) = \langle P(x,y),Q(x,y)\rangle$. Suppose $C$ is a smooth curve in the plane, an object moves along the curve feeling the force, and you want to know how much work it takes to do that. Let's say the curve is parameterized as $\vec R(t) = \langle x(t),y(t)\rangle,~a\le t \le b$. You know that $\vec R'(t)$ is tangent to the curve and if $\hat T(t)$ is a unit vector in the direction of $\vec R'(t)$, then the component of the force in the direction of motion is given by $\vec F\cdot \hat T$. So if the object moves a distance $ds$ along the curve the differential amount of work done by the force would be $dW =\vec F\cdot \hat T~ds$ and the total work done by the force would be$$W =\int_C \vec F\cdot \hat T~ds$$If you want to express this in terms of the $t$ parameterization, note that$$\hat T ds = \frac{\vec R'(t)}{|\vec R'(t)|}\frac{ds}{dt}dt = \vec R'(t)~dt$$so the expression for work done by the force can be written$$W = \int_a^b\vec F\cdot \vec R'~dt$$Texts typically abbreivate $\vec R'(t)~dt$ as $d\vec R$ and you get the compact form$$W = \int_C \vec F\cdot d\vec R = \int_a^b\vec F(x(t),y(t))\cdot \vec R'(t)~dt$$.
Sometimes you will see alternate notation, since $\vec F(x,y) = \langle P(x,y),Q(x,y)\rangle$, you will see $d\vec R$ written as $\langle dx, dy\rangle$ and the line integral abbreviated as $\int_C Pdx + Qdy$. It's all the same and you usually evaluate everything in terms of $t$, manipulating the differentials using the usual formulas. The dx and dy integrals separately calculate the effects on the integral caused by their corresponding variable changing.

Your text may develop it differently but hopefully this example helps give you some intuition about what line integrals are for.

Last edited: Apr 1, 2016