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Example of calculating a surface integral part 3

Video transcript

in the last couple of videos we've been slowly moving towards our goal of figuring out the surface area of this torus and we did it by evaluating a surface integral in order to evaluate a surface integral we had to take the parameterization take its partial with respect to s and with respect to T we did that in the first video then we had to take its cross-product we did that in the second video now we're ready to take the magnitude of the cross product we have to take the magnitude of the cross product and then we can evaluate it inside of a double integral and we will have solved or we would have computed an actual surface integral something you see very few times in your education career so this is this is kind of exciting so this was the cross product right here now let's take the magnitude of this thing and you might remember the magnitude of any of any vector it's kind of a Pythagorean theorem in this case it's going to be kind of the distance formula or the Pythagorean theorem in three dimensions so the magnitude this right here is the this is equal to just as a reminder is equal to this right here it's equal to the partial of R cross the carved partial of R with respect to s cross with this partial of R with respect to T let me copy it and paste it that is equal to that right there but an equal sign these two quantities are equal now we want to figure out the magnitude so if we want to take the magnitude the magnitude of this thing so if we are interested in taking the magnitude of that thing that's going to be equal to well this is just a scalar that's multiplying everything so let's just write the scalar out there so B plus a cosine of s times the magnitude of this thing right here and the magnitude of this thing right here is going to be is going to be the the sum of you can imagine it's the square root of this thing dotted with itself or you could say it's the sum of the squares of each of these terms to the one-half power so let me write it like that so it's equal to let me write the the sum of the squares so if you square this you get a squared cosine cosine square of s sine squared of T that's that term plus let me color coded that's that term plus I'll do the magenta Plus that term squared plus a squared cosine squared of s cosine squared of T that's that term and then finally I'll find them do another color this term squared so plus a squared sine squared of s it's going to be all of this business to the one-half power this right here this right here is the same thing as the magnitude of this right here and this is just a scalar that's multiplying by both of these terms so let's see if we can do anything interesting here if this can be simplified in any way we have a squared cosine squared of s we have an a squared cosine squared of s here so let's factor let's factor that out from both of these terms and see what happens so we could I'm just going to rewrite the second part so this is going to be a squared cosine squared of s x times sine squared of t the parentheses sine squared of T plus cosine all I'm going to do in that magenta color not orange plus cosine squared of T sine squared of T plus cosine squared of T and then you're going to have this plus a squared and then you're gonna have this plus a squared sine squared of s and of course all of that is to all of that is to the one-half power now what is this when we have sine squared of T plus cosine squared of T that's nice that's equal to one or the most basic of trig identities so this expression right here simplifies to a squared cosine squared of s plus plus this over here a squared sine squared of s and all of that to the one-half power you might immediately guys you can factor out an a squared this is equal to a squared times cosine squared of s plus sine squared sine squared of s and all of that to the one-half power I'm just focusing on this term right here I'll write this in a second but once again cosine squared plus sine squared of anything is going to be equal to one as long as it's the same anything it's equal to one so this term is a squared to the one-half power or a square the square root of a squared which is just going to be equal to a so all of this all of that crazy business right here just simplifies all of that just simplifies to a so this cross product here simplifies to this times a which is a pretty neat and clean simplification so let me rewrite this so this that simplifies it simplifies to it's equal to a times that and what's that a times B so it's a B a B plus a squared cosine of s plus a squared cosine of s so already we've gotten pretty far and it's nice when you do something so beastly and eventually it gets to something reasonably reasonably simple and just to review what we had to do what our mission was several videos ago is we want to evaluate what this thing is over the region from s over the region over which the surface is defined for so s going from 0 to 2 pi and T going from 0 to 2 PI over this region so we want to integrate this over that region so that region we're going to have we're going to vary s from 0 to 2 pi so D s and then we're going to vary T from 0 to 2 pi DT and this is what we're evaluating we're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization so this is what we can put in there things are getting simple all of a sudden or simpler a B plus a squared cosine of s and what is this equal to so this is going to be equal to well we just take the antiderivative of the inside with respect to s so the antiderivative so let me do the outside of our integral so we're still gonna have to deal with the zero to 2pi and our DT right here but the antiderivative with respect to s right here is going to be a B's just a constant so it's going to be a be s plus what's the antiderivative of cosine of s its sine of s so plus a squared sine of s and we're going to evaluate it from 0 to 2 pi 0 to 2 pi and what is this going to be equal to let's put our boundaries out again or are the T integral that we're going to have to do in a second 0 to 2 pi DT when you put 2 pi here you're going to get a B times 2 pi or 2 pi a B so you're going to have 2 pi a B plus a squared sine of 2 pi sine of 2 pi is 0 so there's not going to be any term there and then minus 0 times 0 times a B which is 0 and they're going to have minus a squared sine of 0 which is also 0 so all of the other terms are all 0 so that's what we're left with it's simplified nicely so now we just have to take the antiderivative of this with respect to T with respect to T and this is a constant in T so this is going to be equal to take the antiderivative with respect to t 2 pi a b t and we need to evaluate that from 0 to 2 pi which is equal to so we put 2 pi in there you have a 2 pi 4 t it'll be a 2 pi times 2 pi a b or we should say 2 pi squared times a b minus 0 times this thing well that's just going to be 0 so we're going to have to write it down so we're done this is the surface area of the torso this is equal to this is exciting it just kind of snuck up on us this is equal to 4 PI squared a B which is kind of a neat formula because it's a very neat and clean you know it has the two pi which is kind of the Dian of a circle we're squaring it which kind of makes sense because we're taking the product of you know you can kind of manage the product of these two circle the the these two circles I'm speaking in very abstract general terms but that kind of feels good and then we're taking this the product of those two radiuses remember let me just copy this thing down here let me actually let me copy this thing because this is our new this is our exciting result let me copy this so copy so all of this work that we did simplified to this which is exciting which is exciting we now know that if you have a Taurus where the radius of the cross-section is a the radius of the cross section is a and the radius of from the center of the Taurus to the middle of the cross sections is B that the surface area of that Taurus is going to be 4 pi squared times a times B which I think is a pretty pretty neat outcome