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### Course: Multivariable calculus>Unit 4

Lesson 1: Line integrals for scalar functions

# Line integral example 1

Concrete example using a line integral. Created by Sal Khan.

## Want to join the conversation?

• I'm confused about the dS part. When I think a infinitely small change in arc length, that makes me think of dt, why is dS not dt?
(17 votes)
• dS is a small change in arc length. s is traditionally used in physics for displacement. dt is a small change in the parameter t. most of the time, this is time. t is a parameter of x and y. so dt is a small change in that parameter. dS is a piece of the curve that x and y map out. so dS is a small change along that path.
(52 votes)
• What is an Arc length?
(3 votes)
• In this case Arc length can be thought of as the length of a curve between two chosen points. When Sal mentions 'small arc lengths' around , he means to take a tiny (infinitesimal) segment of the curve which is usually labelled 'ds'. If it helps, then think of 'ds' as the same as 'dx' or 'dy', but instead of being a small change along a coordinate axis, you are now taking a small change along the curve itself. Hope that helps.
(17 votes)
• What if we were to find the volume under the curve ? What would we do? Thanks in Advance.
(3 votes)
• You would probably want to do a double integral of f(x,y) from y = 0 to y = sqrt(1-x^2) and x = 0 to x = 1.
(4 votes)
• Is not it be better use dS=sqrt(1+(dy/dx)^2)dx instead of dS=sqrt((dx/dt)^2+(dy/dt)^2)dt in that way we have not use parametric curves.
(2 votes)
• dS=sqrt(1+(dy/dx)^2)dx would only work if everything was in terms of x, which would complicate matters immensely (since everything is already in terms of t). You would have to find y in terms of x, which for this example is y = sin(arccos(x)) and then find dy/dx, which is dy/dx = -x/sqrt(1-x^2). This is much more difficult, albeit possible, method for finding the arc length of the curve. But I think it's impossible to define f(x,y) is terms of just x, since that would be defining a three-dimensional graph in terms of just one variable. It might be possible (?) but I don't see how.
(4 votes)
• At , is the graph z=xy? I googled the image and it looks different. z should be max at max(x) & max(y) and min(x) & min(y). This is so confusing :( Thanks in advance!
(2 votes)
• Yes, this is the graph z=xy. However, Googling the image will give you different values of min(x), min(y), max(x), and max(y). In Sal's graph, min(x) and min(y) are both zero. In most graphs that you get, you will have a min(x) and a min(y) of something like -5.
(4 votes)
• Is it possible to calculate this if we're not given a parametric equation for the curve? (just a regular y = f(x) equation)
(1 vote)
• Getting a parametric equation out of that is easy: set x = t, y = f(t).
(6 votes)
• Why is the final answer 1? shouldn't you substitute sin ( t ) back into the final equation 1/2(u)^2 - 1/2(u)^2? it should come out to .35 or something, or am I wrong?
(1 vote)
• Sal updated the integration limits, from variable `t` to variable `u`, by doing that there is no need to go back to the definition of `u` after integrating. If you don't update the limits of integration, then you do have to substitute back and use the old integration limits over `t`. Since the result of the integration was:
`1/2 u²`
substituting back the `u` you get:
`1/2 sin²t`
and this you have to evaluate from `0` to `π/2`, so you get:
`1/2 [sin²(π/2) - sin²(0) ]`
And that gives you the same result: `1/2`
(5 votes)
• Does he have the proof for sin(x)^2+cos(x)^2=1?
(0 votes)
• Sal seems to be using words expressions path integral and line integral almost interchangeably. My Vector Calc book has fairly different definitions for them (one dealing with functions, another with vector fields). How crucial is the difference?
(0 votes)
• The difference is not very crucial. If you say path integral when you are integrating over a vector field, most people are going to know that you meant line integral.
(4 votes)
• So, all of this is to find the are of a curtain?
(1 vote)

## Video transcript

The last video was very abstract in general, and I used you know, f of x and g of t, and h of t. What I want to do in this video is do an actual example. Let's say I have f of xy. Let's say that f of xy is equal to xy. And let's say that we have a path in the xy plane, or a curve in the xy plane. And I'm going to define my curve by x being equal to cosine of t, and y being equal to sine of t. And we're going to go from-- you know, we have to define what are our boundaries in our t --and we're going to go from t is equal to 0-- or t is going to be greater than or equal to 0 --and then less than or equal to. We're going to deal in radians, pi over 2. If this was degrees, that would be 90 degrees. So that's our curve. And immediately you might already know what this type of a curve looks like. And I'm going to draw that really fast right here in and we'll try to visualize this. I've actually drafted ahead of time so that we can visualize this. So this curve right here, if I were to just draw it in the standard xy plane-- do that in a different color so we can make the curve green; let's say that is y, and this is right here x --so when t is equal to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1, y is going to be equal to sine of 0, which is 0, so t is equal to 0. We're going to be at x equal to 1, that's cosine of 0, and y is sine of 0, or y is going to be 0, so we're going to be right there. That's what t is equal to; t is equal to 0. When t is equal to pi of 2, what's going to happen? Cosine of pie over 2-- that's the angle; cosine of pi over 2 --is 0. Sine of pi over 2 is 1. We're going to be at the point 0, 1. So this is when we're at t is equal to pi over 2. You might recognize what we're going to draw is actually the first quadrant of the unit circle; when t is equal to pi over 4, or 45 degrees, we're going to be at square root of 2, square root of 2. You can try it out for yourself, but we're just going to have a curve that looks like this. It's going to be the top right of a circle, of the units circle. It's going to have radius 1. And we're going to go in that direction, from t is equal to 0, to t is equal to pi over 2. That's what this curve looks like. But our goal isn't here just to graph a parametric equation. What we want to do is raise a fence out of this kind of base and rise it to this surface. So let's see if we can do that or at least visualize it first, and then we'll use the tools we used in the last video. So right here I've graphed this function, and I've rotated it a little bit so you can see [UNINTELLIGIBLE] case. This right here-- let me get some dark colors out --that right there is the x axis, that in the back is the y axis, and the vertical axis is the z axis. And this is actually 2, this is 1 right here, y equal 1 is right there, so this is graphed that way. So if I were to graphs this contour in the xy plane, it would be under this graph and it would go like something like this--- let me see if I can draw it --it would look something like this. This would be on the xy plane. This is the same exact graph, f of x is equal to xy. This is f of x; f of xy is equal to xy. That's both of these, I just rotated it. In this situation that right there is now the x axis. I rotated to the left, you can kind of imagine. That right there is the x axis, that right there is the y axis-- it was rotated closer to me --that's the z axis. And then this curve, if I were to draw in this rotation, is going to look like this: when t is equal to 0, we're at x is equal to 1, y is equal to 0, and it's going to form a unit circle, or half or quarter of a unit circle like that. And when t is equal to pi over 2, we're going to get there. And what we want to do is find the area of the curtain that's defined. So let's see, let's raise a curtain from this curve up to f of xy. So if we keep raising walls from this up to xy, we're going to have a wall looks something like that. Let me shade it in, color it in so it looks a little bit more substantive. So a wall that looks something like that. If I were to try to do it here this would be under the ceiling, but the wall look something like that right there. We want to find the area of that. We want to find the area of this right here where the base is defined by this curves, and then the ceiling is defined by this surface here, xy, which I graphed and I rotated in two situations. Now in the last video we came up with a, well, you could argue whether it's simple, but the is, well, let's just take small arc lengths-- change in arc lengths, and multiply them by the height at that point. And those small change in arc lengths, we called them ds, and then the height is just f of xy at that point. And we'll take an infinite sum of these, from t is equal to 0 to t will equal pi over 2, and then that should give us the area of this wall. So we said is, well, to figure out the area of that we're just going to take the integral from t is equal to o to t is equal to pi over 2-- it doesn't make a lot of sense when I write it like this --of f of xy times-- or let me even better, instead of writing f of xy, let me just write the actual function. Let's get a little bit more concrete. So f of xy is xy times-- so the particular xy --times the little change in our arc length at that point. I'm going to be very hand-wavy here. This is all a little bit review of the last video. And we figured out in the last video this change in arc length right here, ds, we figured out that we could rewrite that as the square root of the dx versus-- or the derivative of x with respect to t squared --plus the derivative of y with respect to t squared, and then all of that times dt. So I'm just rebuilding the formula that we got in the last video. So this expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy. But you know what? Right from the get go we want everything eventually be in terms of t. So instead of writing x times y, let's substitute the parametric form. So instead of x let's write cosine of t. That is x. x is equal to cosine of t on this curve. That's how we define x, in terms of the parameter t. And then times y, which we're saying is sine of t. That's our y; all I did is rewrote xy in terms of t times ds. ds is this; it's the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared. All of that times dt. And now we just have to find these two derivatives. And it might seem really hard, but it's very easy for us to find the derivative of x with respect to t and the derivative of y with respect to t. I can do it right or down here. Let me lose our graphs for a little bit. We know that the derivative of x with respect to t is just going to be: what's the derivative of cosine of t? Well, it's minus sine of t. And the derivative of y was respect to t? Derivative of a sine of anything is the cosine of that anything. So it's cosine of t. And we can substitute these back into this equation. So remember, we're just trying to find the area of this curtain that has our curve here as kind of its base, and has this function, this surface as it's ceiling. So we go back down here, and let me rewrite this whole thing. So this becomes the integral from t is equal to o to t is equal to pi over 2-- I don't like this color --of cosine of t, sine of t, cosine times sine-- that's just the xy --times ds, which is this expression right here. And now we can write this as-- I'll go switch back to that color I don't like --the derivative of x with respect to t is minus sine of t, and we're going to square it, plus the derivative of y with respect to t, that's cosine of t, and we're going to square it-- let me make my radical a little bit bigger --and then all of that times dt. Now this still might seem like a really hard integral until you realize that this right here, and when you take a negative number and you squared it, this is the same thing. Let me rewrite, do this in the side right here. Minus sine of t squared plus the cosine of t squared, this is equivalent to sine of t squared plus cosine of t squared. You lose the sign information when you square something; it just becomes a positive. So these two things are equivalent. And this is the most basic trig identity. This comes straight out of the unit circle definition: sine squared plus cosine squared, this is just equal to 1. So all this stuff under the radical sign is just equal to 1. And we're taking the square root of 1 which is just 1. So all of this stuff right here will just become 1. And so this whole crazy integral simplifies a good bit and just equals the square root of t equals 0 to t is equal to pi over 2 of-- and I'm going to switch these around just because it will make it a little easier in the next step --of sine of t times cosine of t, dt. All I did, this whole thing equals 1, got rid of it, and I just switched the order of that. It'll make the next up a little bit easier to explain. Now this integral-- You say sine times cosine, what's the antiderivative of that? And the first thing you should recognize is, hey, I have a function or an expression here, and I have its derivative. The derivative of sine is cosine of t. So you might be able to a u substitution in your head; it's a good skill to be able to do in your head. But I'll do it very explicitly here. So if you have something that's derivative, you define that something as u. So you say u is equal to sine of t and then du, dt, the derivative of u with respect to t is equal to cosine of t. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t, dt. And notice right here I have a u. And then cosine of t, dt, this thing right here, that thing is equal to d of u. And then we just have to redefine the boundaries. When t is equal to 0-- I mean so this thing is going to turn into the integral --instead of t is equal to 0, when t is equal to 0 what is u equal to? Sine of 0 is 0, so this goes from u is equal to 0. When t is pie over 2 sine of pi over 2 is 1. So when t is pi over 2, u is equal to 1. So from is equal to 0 to u is equal to 1. Just redid the boundaries in terms of u. And then we have instead of sign of t, I'm going to write u. And instead of cosine of t, dt, I'm just going to write du. And then this is a super-easy integral in terms of u. This is just equal to: the antiderivative of u is u 1/2 times u squared-- we just raised the exponent and then divided by that raised exponent --so 1/2 u squared, and we're going to evaluate it from 0 to 1. And so this is going to be equal to 1/2 times 1 squared minus 1/2 times 0 squared, which is equal to 1/2 times 1 minus 0, which is equal to 1/2. So we did all that work and we got a nice simple answer. The area of this a curtain-- we just performed a line integral --the area of this curtain along this curve right here is-- let me do it in a darker color --on 1/2. You know, if this was in centimeters, it would be 1/2 centimeters squared. So I think that was you know, a pretty neat application of the line integral.