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Current time:0:00Total duration:13:02

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 so let's say I have f of X Y let's say that f of X Y is equal to X Y and let's say that we have a path in the XY plane or curve in the xy-plane I'm going to define my curve say my curve is going to be defined by X being equal to cosine of T and Y being equal to sine of T and we're going to go from it's not you know we have to define what what are our boundaries on our T and we're going to go from T is equal to 0 or 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 and then we'll try to visualize this I've actually grafted 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 let me do that in a different color so we can make the curve green so let's say that is y and that 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 is equal to 1 right 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 at t is equal to t is equal to 0 when T is equal to PI of 2 what's going to happen cosine of PI over 2 right that's the angle cosine of PI over 2 is 0 sine of PI over 2 is 1 so we're going to be at the point 0 1 so this is when we're at T is equal to PI over 2 and 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 your self 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 unit circle it's going to have radius one and we're going to go in that direction from T is equal to zero to T is equal to PI over two that's what this curve looks like but in the last 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 use in the last video so right here I've graphed I've graphed this function and I've done it I've rotated a little bit so you could see the other case this right here let me get some dark colors out that right there is the x-axis that's the x-axis that in the back is the y-axis and the vertical axis is the z axis and this is actually two this is one right here y equal one is right there so this is it graph that way so if I were to graph this contour and 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 let me make that clear this is f of X f of XY is equal to XY that's both of these I just rotated it in this situation this is now that right there is now the x axis I've rotated to the left you can kind of imagine that right there's 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 it in this rotation if I were to draw it in this rotation it's going to look like this when T is equal to 0 where it X is equal to 1 Y is equal to 0 it's going to form a unit circle something like our 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 X Y so if we keep raising walls from this up to X of Y we're going to have a wall that looks something like that gonna have a wall that looks something like that let me shade it in colored 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 the wall would 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 curve and then the ceiling is defined by this surface here XY which I just 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 idea is well let's just take small arc lengths change in arc lengths and multiply them by the height multiply them by the height at that point and those small change in arc lengths we call them D s and then the height is just f of X Y at that point and we'll take an infinite sum of these from T is equal to 0 to t equal to PI over 2 and then that should give us the area of this wall of this wall right there so what we said is well to figure out the area of that we're just going to take the integral from T is equal to 0 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 X Y times or let me be even better instead of writing f of X Y let me just write the actual function let's get a little bit more concrete so f of XY is XY x so the particular XY times the little change in our arc length of that point I'm going to be very hand wavy here this is all a little bit of review of the last video and we figured out in the last video this change in arc length right here this change in arc length D s we figured out that we could rewrite that as the square root of 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 and so I'm just rebuilding the formula that we got in the last video so this expression can be re-written as the integral from t is equal to 0 to T is equal to PI over 2 times X Y but you know what right right from the get-go we want everything of entry to be in terms of T so instead of an at we instead of writing x times y let's substitute the parametric form so instead of X let's write cosine of T so let me write cosine of T right that is X X is equal to cosine of T on this curve right X that's how we define X in terms of the parameter T and then times y which we're saying is sine of T times sine of T that's our Y all I redid all I did is rewrote X Y in terms of T times D s D s is this it's the square root 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 define 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 we'll lose our graphs for a little bit so we have 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 that's minus sine of T and the derivative of Y with respect to T derivative of Y with respect to T derivative of sine of anything it's the cosine of that anything so it's cosine cosine of T and we could substitute these back into this equation so are the remember we're just trying to find the area of this curtain with that has our curve here is kind of its base and has this function the surface as its ceiling so we go back down here and let me rewrite this whole thing so this becomes the integral from T is equal to 0 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 X Y times D s 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 square root of the derivative of X with respect to T is minus sine of t minus sine of T and we're going to square it plus derivative of Y with respect to T that's of T and we're going to square it maybe 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 when you take a negative number and you square it this is the same thing let me rewrite - let me do this in the side right here - sine of T squared plus a cosine of T squared this is equivalent to this is the same thing as sine of T squared plus cosine of T squared right you lose the sine 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 one so all this stuff under the radical sign is just equal to one and we're taking the square root of one which is just one so all of this stuff right here will just become it'll just become one and so this whole crazy integral simplify is a good bit it just equals the square root T equals zero to T is equal to PI over 2 of and I'm going to switch these round just because it'll make it a little easier in the next step of sine of T times cosine of T sine of T times cosine of T DT all I did this whole thing equals one got rid of it and I just switch the order of that it'll make the next step a little bit easier to explain now this integral is hey sine times cosine what's the antiderivative of that and the first thing you should recognize is hey I have I have a 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 do 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 as derivative you define that something is u so you say u is equal to sine of T and then D u 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 D U 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 zero let me so we're going to have this thing is going to turn into the integral instead of T is equal to zero when T is equal to zero what is u equal to sine of zero is zero so this goes from U is equal to zero when T is PI over two T is PI over 2 sine of PI over 2 is 1 so when T is PI over 2 U is equal to 1 so from U is equal to 0 to U is equal to 1 just redid the boundaries in terms of U and then we have instead of sine of T I'm going to write U and instead of cosine of T DT I'm just going to write D U and now 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 right 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 the area of this curtain we just performed a line integral the area of this curtain along along this curve right here is that area let me do it in a darker color is 1/2 you know if this was in centimeters would be 1/2 centimeter squared so I think that was you know a pretty neat application of the line integral