If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

# Using a line integral to find work

Using a line integral to find the work done by a vector field example. Created by Sal Khan.

## Want to join the conversation?

• If a particle return back to same initial point then displacement is 0, thus workdone is 0(work done =force*displacement)...

for any close path displacement is 0 thus the work done is also 0... but in the above vedio we got -2pi. how it is possible?? can u pls explain me.

• What work really depends on is the field. If you have a conservative field, then you're right, any movement results in 0 net work done if you return to the original spot. With most vector valued functions however, fields are non-conservative. In a non-conservative field, you will always have done work if you move from a rest point.

In the video, the field is non-conservative, because as Sal showed, moving around a circle nets you some work.
• why did the i and j (unit vectors) disappeared after the dot product at ?
• yes the result is a scalar, and the way you calculate the dot product two vectors is to multiply each element from each vector and add them like this V1 = a1i+b1j and V2= a2i+b2j then V1 dot V2 = a1*a2+b1*b2 which is a scalar without the i and J components.
• So when we say the work done was -2pi, that is the work done by the field. Would the work done by us to keep the particle moving in that path then be 2pi?
• Like you stated, if the field does -2pi units of work, then the force pushing against the field must do +2pi units of work. Pavle, the logic you use is slightly skewed. Recall, work is force times distance (and as we have learned now, it is actually the tangential component of force in the direction we are moving times the distance moved), so if the particle doesn't move, then there is no work done by anything, the field or us.
If the field did -2pi units of work, then the particle definitely moved. the amount of work the field "absorbed" (negative) is the same amount that we "put in" (positive).

Pavle, you would be correct in saying that the particle does not move if we applied the same Force as the vector field but in the negative direction (in this case -yi+xj), but remember force is different than work.

Another way to think about it is through the law of conservation of energy, energy can not be created or destroyed, just change forms. Therefore, if the field does negative work to us (take energy from us) then we must do positive work to the field (supply energy to it). So yes, if the field does -2pi units of work, then we do +2pi units of work. If the field does positive work, then we do the same amount of negative work, etc.
(assuming that we are the ones moving the particle in this instance)
• Is it possible to evaluate the line integral without using a parameter?
• Convert the parametric function into single variable function and find the integral of the real-valued function.
(1 vote)
• At , how is he graphing the vectors? at , how does he know whether it's clockwise or counterclockwise? thanks
• f(x,y)=yi+xy ---> f(1,2)=2i+1j ---> means plot the vector with the components of 2i+1j at the point (1,2). Make sense now?
Also he graphed ccw because when you plug in t values, the x and y values correspond to a ccw motion.
• Why does Sal not write the integral sign with the circle at ? It is a closed loop.
• You could write the closed-loop integral symbol here, but it is not necessary. It is normally used for mathematic definitions, when you want to say "this is only true for closed loops".
• Doesn't r(t) = cos(t) i + sin(t) j point outward, rather than tangentially along the circular path?
• Yes, r(t) will point outward from the origin to that circle. However, r'(t) points tangentially along the circle.
• How are you sure that it's a counter clockwise curve?
• If we take the 1st quadrant (where x and y are positive), we see that as t gets larger the cos will get smaller, so x starts at 1 when t=0 and moves to 0 when t=pi/2. We also see that as t gets larger sin gets larger, thus y=0 when t=0 and y=1 when t=pi/2. If we graph these as points (x,y) marking a new point for each value of t we see that a quarter of a circle is drawn in the counter clockwise direction. Extending this to the other 3 quadrants we see it's a counter clockwise curve.