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## Multivariable calculus

### 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?
• 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.
• What is an Arc length?
• 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.
• What if we were to find the volume under the curve ? What would we do? Thanks in Advance.
• 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.
• 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?
• 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`
• Does he have the proof for sin(x)^2+cos(x)^2=1?
• 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.
• 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.
• 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).
• 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!
• 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.
• 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?
(1 vote)
• 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.
• Yes, and it is. But we did a change of variables, and now `x = cos(t)` and `y = sin(t)` so `xy = cos(t)·sin(t)`.
Nowhere in the video does a `cos(x)·sin(y)` appear.