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Current time:0:00Total duration:10:45

Example of calculating a surface integral part 1

Video transcript

we saw several videos ago that we can parameterize a torus or a doughnut shape as a as a vector position vector-valued function of two parameters and this is the outcome that we had I think I did it over several videos because it was a bit hairy and I'll write I'll write our vector position vector valued function first so we have R as a function of our two parameters s and T and then I'll review a little bit of what all the terms what the s the T and the A's and the B's represent but it's equal to B plus a cosine of s and once again we saw this we saw this several videos ago so you might want to watch the videos on on parameterizing surfaces with two parameters to get to figure out how we got here times the sine of T I'm going to put the S terms and the T terms in different colors times our I unit vector out put the vectors or the unit vectors in this orange color plus during the same yellow plus B plus a cosine a cosine of s times cosine of T times cosine of T times the J unit vector the unit vector in the Y Direction plus plus a sine of s a sine of s times the K unit vector the unit vector in the Z direction and in order to generate the torus or the doughnut shape this is true for our parameters so we don't multi rap multiple times around the torus for s being between this is for s being between zero and 2pi and for T being between zero and 2pi and just as a bit of review where all of this came from and I'm gonna have to do what my plan is for this video over several videos but let's review where all of this came from we draw a doughnut my best effort at a donut right here that looks like a donut or a Taurus and you can imagine a Taurus or this doughnut shape it's kind of the product of two circles you have this circle that's kind of the cross-section of the doughnut at any point you could take it there you could take it over there and then you have the circle that kind of wraps around all of these other circles or these other circles wrap around it and so we when we drove the derived this formula up here this per parameterization a was the radius of these cross sectional circles that's a that's what these a terms were and B was the distance from the center of our torus B was a distance from the center of our tours out to the center of these cross-sections so this was B so you could imagine that B is kind of the radius of the big circle up to the midpoint of the I guess cross-section and a is the radius of the cross-sectional circles and when we parameterised it s the parameter s was essentially telling us how far s was telling us how far we where how are we wrapping around this circle so it's an angle from 0 to 2 pi 2 pi to say where we are in that circle and T tells us how much we've rotated around the larger circle so T was telling us how much we rotated around the larger circle so if you think about it you can specify any point on this doughnut or on this surface or on this torus by telling you an S or a T and so that's why we picked that as the parameterization now the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral and the surface integral we're going to compute will tell us the surface area of this torus so let's this surface right here is Sigma right like that it's being represented by this position vector valued function that is a parameter that is parameterize by these two parameters right there and if we wanted to figure out the surface area if we just kind of set it as a surface integral we saw and I think the last video or at least the last vector calculus video I did that this is a surface integral over the surface here this capital Sigma does not represent some it represents a surface of a bunch of the little D Sigma's a bunch of the little chunks of the surface and just as a review you could imagine a DHD Sigma is a little patch of the surface right there that is a D Sigma and we're taking where it's a double integral here because we want to add up all the D Sigma's in two directions you can imagine one kind of rotating this way around the torus and then the other direction is going in the other direction around the torus so that's why I say double integral and this is just going to give you the surface area which is the whole point of this video and probably the next one or two videos but if you wanted to also multiply these Sigma's times some other values you know there's some scalar field this is in that you cared about you could put that other value right there but here we're just multiplying it by one and we saw in the last video that this is you know it's a way of expressing an idea but you really can't do much computation with this but a way that you can express this in a so that you can actually take the integral you say this is the same thing and we saw this in the last several videos this is the same thing as the double integral over the region over which our parameters are defined so it's this region over here where s and T go from 0 to 2 pi of whatever function this is we just have a 1 here so we don't have to we could just write a 1 if we like it doesn't change much times and this is what we learned times the magnitude of the partial derivative of R with respect to s the magnitude of that crossed with the partial derivative of R with respect to t DS you can take it in either order but D s DT so we saw this in the last video what were going to do now is actually compute this that's the whole point of this video we're going to take the cross product of these two vectors so let's figure out these vectors in the next video we're gonna take the cross-product and in the video after that we'll actually evaluate this double integral you're going to see it's a pretty hairy problem and this is probably the reason that very few people ever see an actual surface integral get computed but let's let's do it anyway so the partial derivative the partial derivative of R with respect to s so this term right here will do the cross product in the next video this term is what we just want to hold T constant and take the partial with respect to just s so this up here that if we distributed the sine of T times B that's just going to be a constant in terms of s so we can ignore that then you have sine of T times this over here so sine of T and a is a constant and you take the derivative of cosine of s that's negative sine of s so it's going to be so the derivative of this with respect to s so the partial with respect to s is going to be minus a I'll write in green the sine of T so you know that's where it came from sine of T and then sine of s right that's the the derivative of this is negative sine of s that's where that negative came from and then I'll write the sine of s right there sine of s times the unit vector I that's the partial of just this X term with respect to s and then we'll do the same thing with the Y term or the J term so plus same logic B times cosine of T with respect to s we take the partial justice becomes 0 so you're left with a a well it's going to be a minus a again right because when you take the derivative of the cosine of s it's going to be negative sine of s so you have a let me do it so you're going to have a minus a this cosine of t minus a cosine of T that's the constant terms sine of s sine of s just taking partial derivatives sine of s J sine of s J and then finally we take the derivative of this with respect to s and that's pretty straightforward just going to be a cosine of s so plus plus a cosine a cosine of s K hopefully you didn't find this confusing the negative signs because the derivative of cosines aren't negative sign so negative sign of s so that's where it's negative sign of s times the constants negative sign of s times the constants the constant cosine of T sine of T so hopefully this makes some sense this is a review of taking a partial derivative now let's say the same thing with respect to T let's do the same thing with respect to T and I'll do that in a different color so we now we want to take the partial of R with respect to T so the partial of R with respect to T is equal to so now this whole term over here is a constant and so it's going to be that whole term times the derivative of this with respect to T which is just cosine of T so it's going to be it's going to be B plus a cosine of s times cosine of T cosine of T I and then plus and it's actually gonna be a minus coz you take the derivative of this with respect to t is going to be minus sine of T so let me put them it's going to be negative and then let me leave some space for this term right here negative sine of T and you're going to have this constant out there right that's a constant in T B plus a cosine of s that's just that term right there derivative of cosine T is negative sine of T times J times J and then the partial of this with respect to T this is just a constant in T so the partial is going to be zero so I'll write plus 0 K plus 0 let me do all my vectors in that same color plus zero times the unit vector K so that gives us our partial derivatives now we have to take the cross product then find the magnitude of the cross product and then evaluate this double integral and I'll do that in the next couple of videos