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Current time:0:00Total duration:9:58

Video transcript

in the last video we set out to figure out the surface area of the walls of this weird-looking building where the ceiling of the walls was defined by the function f of X Y is equal to X plus y squared and then the base of this building or the the contour of its walls was defined by the path where we have a circle of radius 2 along here then we go down along the y axis and then we take another left and we go along the x axis and that was our building and in the last video we figured out this first walls surface here in fact you can think of it our original problem is we wanted to figure out the line integral along the closed path so it was a closed line integral along the closed path C of f of X Y of f of X Y and where every oh and we're always multiplying f of XY times a little bit a little small distance of our path D s we're writing this in the most abstract way possible and what we saw in the last video is the easiest way to do this is to break this up into multiple paths or into multiple problems so you can imagine this whole contour this whole path we call C but we could call this part that we figured out in the last video c1 this part we can call let me make a pointer c2 and then this point right here is c3 so we could redefine or we can break up this line integral this closed line integral into three non closed line angles this will be equal to the line integral along the path c1 of f of XY DS plus the line integral along c2 of F of XY f of XY d s plus the line integral and you might have guessed it along c3 of F of XY D s and in the in the last video we got as far as figuring out this first part this first curvy wall right here it's the surface area we figured out was 4 plus 2 pi 4 plus 2 pi now we got to figure out the other two parts so let's do C 2 let's do this line integral next and in order to do it we need to do another parametrizations and why it's going to be different than what we did for this part we're no longer along this circle we're just along the y-axis so as long as we're there X is definitely going to be equal to zero so that's my parametrizations X is equal to zero for along the y-axis X is definitely equal to zero and then Y we could say it starts off at Y is equal to two Y is equal to two maybe we'll say Y is equal to two minus T for T is between zero T is greater than or equal to zero less than or equal to two and that should work when T is equal to zero we're at this point right there and then as T increases towards two we move down the y-axis until eventually when T is equal to two we're at that point right there so that's our parametrizations and so let's evaluate this line in actually we could do our derivatives two if we like what's the derivative let's limit right here what's DX DT pretty straightforward derivative 0 0 and dy DT is equal to is equal to the derivative of this is just minus 1 right 2 minus T derivative of minus T is just minus 1 and so let's just break it up so we have this thing right here so we have the integral along c2 but let's instead of writing c2 well I'll leave c2 there but we'll say we're going from T is equal to 0 to 2 of f of XY f of XY so f of XY is this thing right here is X plus y squared and then times D s times D s and we know from the last several videos D s can be re-written as the square root of DX DT squared so 0 squared plus dy DT squared so minus 1 squared is 1 all of that times DT and obviously this is pretty nice it clean gets very nice and clean this is 0 plus 1 square root this just becomes 1 and then what is X X if we write it in terms of our parameterization is always going to be equal to 0 and then Y squared is going to be 2 minus T squared so this is going to be 2 minus T squared so this whole crazy thing five - we're going to go from T is equal to 0 to t is equal to 2 the X disappears in our parametrizations X just stays zero regardless of what T is of what and then you have Y squared but Y is the same thing as 2 minus T so 2 minus T squared and then you have your DT sitting out there this is pretty straightforward I always find it easier I always find it easier to find when you're finding an antiderivative of this although you can do this in your head I like to just actually multiply out this binomial so this is going to be equal to the antiderivative from T is equal to 0 to T is equal to 2 of 4 minus 2 minus 4t plus T squared plus T squared just like that DT and then this is pretty straightforward this is going to be the antiderivative of this is for t minus 2t squared right when you take the derivative two times minus 2 is minus 4t and then you have plus plus 1/3 T to the third right these are just simple anti derivatives and we need to evaluate it from 0 to 2 and so let's evaluate it at 2 4 times 2 is 8 let me just pick a new color 4 times 2 is 8 minus 2 times 2 squared so 2 times 4 so minus 8 plus 1/3 times 2 to the third power so 1/3 times 8 so these cancel out we have 8 minus 8 and we just have 8/3 so this just becomes 8/3 and then we have to put a 0 in minus zero evaluated here but it's just going to be 0 we have 4 times 0 2 times 0 all of these going to be 0 so minus 0 so just like that we found our surface area of our second wall of our second wall this turned out being this right here is 8/3 and now we have our last wall and then we could just add them up so we have our last wall we're looking so just I'll do another parameterization I want to have the graph there well maybe I can paste it again edit so there's the graph again and now we're going to do our last wall so our last wall is this one right here which is we could write it you know this was c3 let me switch colors here so this is C we're going to go along contour c3 of f of X Y D s which is the same thing as let's do a parametrizations along this curve if we just say X is equal to T very straightforward for T is greater than or equal to 0 less than or equal to 2 and then this whole time that we're along the x axis Y is going to be equal to 0 that's pretty straightforward parametrizations so this is going to be equal to we're going to go from T is equal to 0 to T is equal to 2 T is equal to 2 of f of X Y which is I'll write in terms of X right now X&Y X plus y squared times D s now what is DX well let me write D's right here times DX that's what we're dealing with now we know what D s is d s is equal to the square root of DX DT squared plus dy DT squared times DT we proved that in the first video or we didn't rigorously prove it but we got the sense of why this is true and what's the derivative of X with respect to T well that's just 1 so this is just going to be a 1 1 squared same thing and derivative of Y with respect to T is 0 so this is 0 1 plus 0 is 1 square root of 1 is 1 so this thing just becomes this thing just becomes DT DT D s is going to be equal to DT in this case so this just becomes a DT and then our X is going to be equal to a T that's part our definition of our parametrizations and y is 0 so we can ignore it so this was a super simple integral so this simplified down to we're going to go from 0 to 2 of T DT which is equal to the antiderivative T is 1/2 T squared and we're going to go zero to two which is equal to 1/2 times 2 squared 2 squared is 4 times 1/2 is 2 and then minus 1/2 times 0 squared minus 0 so the this third walls area right there is just two pretty pretty straightforward so that right there the area there is just 2 and so to answer our question what was this line integral evaluated over this closed path of f of X Y well we just add up these numbers we have 4 plus 2 pi plus 8/3 plus 2 well what is this 8/3 is the same thing as 2 and 2/3 so we have 4 plus 2 and 2/3 is 6 and 2/3 plus another 2 is 8 and 2/3 so this whole thing becomes 8 and 2/3 if you write it as a mixed numbers plus 2 pi and we're done and we're done now we can start trying to do line integrals with vector valued functions