# Line integral confusion

Hi , so I was solving this example . I have no problem in calculation . But at the end of it , when they asked about the closed line integral , I wondered how did the line integral on both paths be less than that on one of them . Starting from the meaning of a line integral to be the work done by a force on a certain path , how could the work acting on two paths decreases , I know by calculation we are moving in an opposite direction on the 2nd path so that's why it has decreased , but when I come to my intuition , I stop being convinced ... ( by the way it's the figure to right not left )

• can you type the example? the image is barely legible. Making your question easier to understand will increase the likelihood of a response. – anon01 Jul 6 '16 at 0:53
• You can easily show that your vector field is non-conservative by showing that curl $\vec v$ is not zero. Please note the fact that curl $\vec v =0$ does not necessarily mean that the field is conservative. – Farcher Jul 6 '16 at 8:01

The notion that a line integral is the "work done be a force" only applies under very certain circumstances! Specifically, if the vector field you're integrating over is conservative. Consider for instance, that I'm in one of those "lazy river" (a big circular flowing tube of water) things, and I'm swimming. If I go from the 12 o'clock position to the 6 o'clock position swimming with the water (say, clockwise) then it's very easy: the vector of my motion and the water's motion always aligned. If I go the other way (counterclockwise) then it's a lot of work, because we're always misaligned. Even though I ended up at the same place, it was a different amount of work.}

As another example of a nonconservative field, suppose I have a tub of water, and a windmill I'm swishing around through the water to charge a battery. If the windmill is pointing the right way (forward) it charges the battery; if it points backwards it drains the battery. The vector field says which direction I'm pointing the windmill as I move it around, where my orientation is only a function of my position. Obviously the optimal thing is to always keep it aligned with my motion, so I get net power. But if I don't do this, then I'll get power proportional to the dot product of my orientation ($s$) and the motion ($dl$). This won't necessarily be conservative either -- otherwise I couldn't charge it by swinging it around in a circle!

If the question is about why the total path integral is smaller by absolute value than path integrals on each of the two parts, then the answer is the following.

Intuitively and physically, in the case of a simple conservative force, like gravity, the work for a closed path, as well as the corresponding closed loop integral, should be zero, and if the loop is split into two parts, then the work on each of the parts should be opposite of the work on another part, and they should cancel each other. So, it shouldn't be suprising that contribution from one path can cancel out at least some contribution from another part, if what often happens is that contributions cancel each other completely.

For a simplest example of that, consider the simplest definition for work $W=\mathbf{F}\cdot\mathbf{S}$. If a particle moves from a point A to a point B along the direction of the force $\mathbf{F}$, then work on that part is $W_{1}=F\cdot S$ and positive. If it moves back, from B to A, completing the loop, then $W_{2}=-F\cdot S$. So the total work would be $W=W_{1}+W_{2}=F\cdot S+(-F\cdot S)=0$.

For another example, imagine a body moving straight up from a point A (0, 0) to a point B (0, 1) and backwards, straight down to A, closing the loop. Let the force be gravity $\mathbf{F}=-m\mathbf{g}=-mg\hat{\mathbf{y}}$, pointing down. Work when moving upwards is $W_{1}=\int_{0}^{1}(-mg\hat{\mathbf{y}})\cdot(dx\hat{\mathbf{x}}+dy\hat{\mathbf{y}}+dz\hat{\mathbf{z}})=-\int_{0}^{1}mg\hat{\mathbf{y}}\cdot dy\hat{\mathbf{y}}=-mg\int_{0}^{1}dy=-mg$, and when moving backwards is $W_{2}=\int_{1}^{0}(-mg\hat{\mathbf{y}})\cdot(dx\hat{\mathbf{x}}+dy\hat{\mathbf{y}}+dz\hat{\mathbf{z}})=-W_{1}=mg$. $W=W_{1}+W_{2}=-mg+mg=0$.

Any non-zero contribution to work disappeared.

In the given textbook example $\mathbf{v}$ is not an expression for a conservative force, so the total path integral is not equal to zero.

Formally, the reason why e. g. in the textbook example contributions from (i) and (ii) added up to 11 in (1) was that both integrals, 1 and 10, were positive. But in going backwards over (2) the contribution is negative: $\int_{b}^{a}\mathbf{v}\cdot d\mathbf{l}=-\int_{a}^{b}\mathbf{v}\cdot d\mathbf{l}=-10$. So $\oint\mathbf{v}\cdot d\mathbf{l}=\underset{(1)}{\int_{a}^{b}}\mathbf{v}\cdot d\mathbf{l}+\underset{(2)}{\int_{b}^{a}}\mathbf{v}\cdot d\mathbf{l}=11+(-10)=1$.

• I see.. Thank you , I understand now where my confusion lies . It's because I think of work as being an effort , but obviously it's not the case . Otherwise , as we increase the amount of displacement, effort must increases whether moving in a positive or negative sense , sense won't matter in the first case . But I got it wrong in my head .. How would you recommend me to think about work ? – user65035 Jul 6 '16 at 22:13
• I would say that work of a given force is also about whether the force contributes to the displacement that happened. When a particles moves, there may be multiple forces acting on it. These forces are directed differently. If you lift a weight from the ground, then the force of your arms does positive work but the force of gravity does negative work. In this example, you could say that motion happened because of your force but despite of or against gravity. – ejojwjrek Jul 7 '16 at 13:06
• If you move something from one point on the ground to another point on the ground (in the horizontal plane), then you do positive work and gravity does zero work since gravity acts downwards, being perpendicular to the movement (and scalar product $\mathbf{F}\cdot\mathbf{S}$ of perpendicular vectors is zero as well). In the last example gravity didn't contribute to the motion at all. Work is also related to energy and is measured in the same units, joules. – ejojwjrek Jul 7 '16 at 13:06