Line Integrals: Full Example

in this video we're going to do a complete example of computing a line integral or sometimes called a path integral in my previous video on line integrals i introduced the big formula the big idea of line integrals and the link to that video and the rest of my vector calculus playlist will be down in the description but in this video we're just going to take the formula for line integrals and see how we can compute it out in well this specific example what i'm claiming is i have some function f x y it's given here and then i'm trying to figure out what is the line integral of that function over a curve of radius 2 centered at the origin what this looks like visually is something like this now there's two things to point out here first is along the bottom we have that curve which is our circle of radius 2 centered at the origin and then the blue curve at the top describes the height of the function f x y above that curve and then what i'm actually computing here the line integral what it represents in this example is the surface area underneath that function going down to the curve and indeed this picture is the same picture i used in the introduction video now we're actually just going to compute it with the specific functions that we have so our formula for a line integral is well this the left hand side tells me the thing i'm trying to compute the line integral over the curve c of my function integrated with respect to arc length integrated dx but the way i compute this is the bottom or the right hand side where if i can parameterize my curve and my function in terms of t then i have this formula to be able to compute it explicitly as just a single variable calculus problem in terms of t so the real challenge for me is to come up with some parameterization that gives me these functions g of t and h of t so let's see how we can do that the first step we're going to have is to find a parameterization for the curve c now so far on the problem all i've been given is a geometric description the cursit is a circle of radius 2 centered at the origin but this is a relatively standard curve and it has the following parametrization namely 2 cosine of t in the i-hat direction and 2 sine of t in the j-hat direction this makes sense that it's a circle because the defining property of a circle is that the radius squared is x squared plus y squared but if you take the x component squared here and the y component squared here because of pythagoras indeed that adds up to two squared in other words a circle of radius two to align with the previous formula the two cosine of t we call g of t and the two sine of t we call h of t so the x and y components get those names g of t and y of t respectively next up i want to now the curve does not just have a parametrization like this it also has to have a domain so i'm going to specify that this is on the domain 0 to 2 pi and 0 to 2 pi means it plots out that circle once if i wanted to say that my path was that i went around the circle twice maybe i go to zero to four pi but implicit here is that i have just the circle and the entirety of the circle covered exactly once so zero to two pi is a good domain next i need to talk about the function so i have that original function f of x y but i want to write it in terms of my parameter i want to replace the x and the y with these functions of t so i'm going to do that i'm going to take my x squared plus y squared over 4 plus xy over 2 the function and everywhere that there was an x i replaced it with the g of t the 2 cosine of t and everywhere there was a y i replaced it with the h of t or the 2 sine of t thankfully this messy formula cleans up really nicely the co squared plus the sine squared collapses by pythagoras and then the 2 sine t cos t and then 2 sine t cos t is by another trig identity just sine of 2t nevertheless i get this much nicer function only of t so those are my two pieces of data that i have my r of t my curve and then my f of t my function and both of these are described now entirely in terms of t so my third step is to take those two pieces of information and plug them into our formula this is the big formula that we had well this is straightforward okay couple different things first let's look at the limits of integration the a and the b well since t was going from 0 to 2 pi well i'll just replace those and i get 0 2 pi for my limits of integration with respect to t next there's an f of g of t and h of t and we've computed what that f is going to be it's just this 1 plus sine of 2 t so i plug that in and then finally underneath the square root i have my g prime which is 2 cosine t and my h prime which is 2 sine of t i plug those in as well and i get this new expression this is substituting in the functions i have again this is a nicely computable example because everything under the square root is just a pythagoras sine squared plus cos squared is 1. and so because of the twos they square up to 4 and the square root takes them back to 2. either way all of that just simplifies to 2. and then this is an easy enough interval if you want to you can test it out it is equal to 4 pi and so my final answer then if i return back to the original problem so having this particular line integral my final answer is that the line integral of this function is just equal to 4 pi this video has been part of my series on vector calculus if you want to check out that playlist in the playlist to any of my other courses check it out down in the description if you have a question please leave a comment on this video and i'll do my best to answer it give the video a like finally for the youtube algorithm and we'll do some more math in the next video

Channel: Dr. Trefor Bazett

Views: 107,476

Keywords: Math, Solution, Example, Line Integrals, Path Integrals, Simplest Example, curve, circle, solve, vector calculus, multivariable calculus, surface area, surface integral, parameterize

Id: fqVEuFldFuA

Length: 5min 39sec (339 seconds)

Published: Mon Sep 28 2020