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# Example of calculating a surface integral part 2

## Video transcript

where we left off in the last video we were finding the surface area of a torus or a doughnut shape and we were doing it by taking a surface integral in order to take a surface integral we had to find the partial of our parameterization with respect to s and the partial with respect to T and now we're ready to take the cross-product and then we can take the magnitudes of the cross product and then we can actually take this double integral and figure out the surface area so let's just do it step by step here we could take the cross-product which is not a non hairy operation this is why you don't see many surface integrals actually get done or many examples done so let's take the cross product of these two fellows so the partial of R with respect to s crossed with in magenta the partial of R with respect to T this will be a little bit of a view of cross products for you you might remember this is going to be equal to the determinant the determinant I'm going to write the unit vectors up here of the first rows I J and K and then the next two rows are going to be let me do that in that yellow color the next two rows the next two rows are going to be the components of these guys so let me copy and paste them you have that right there let me copy and paste put that guy right there then you have this fellow right there copy and paste put them right there and then you got this guy right here this will save us some time copy and paste put them right there and the last row is going to be this guy's components copy and paste put them right here and almost done this guy copy and paste put them right there make sure we know that these are separate terms and finally we could if we don't have to copy and paste it but just since we did it for all of the other terms I'll do it for that zero as well so the cross product of these is literally the determinant of this matrix right here right here it's this determinant it's that determinant and so just sort of a bit of a refresher of taking two permanence this is going to be I times the sub determinant right here if you cross out this column in that row so it's going to be equal to it's going to be equal to I you're not used to seeing the unit vector written first but we can switch the order later times I times the sub matrix right here if you cross out this column in that row so it's going to be this term times 0 which is just 0 minus this term times that term so minus this terms times this term the negative signs are going to cancel out so this will be positive so it's just going to be I times this term times this term without a negative sign right there so I times this term which is a cosine of s it's really that term times that term minus that term times that term but the negatives cancel out that times that is 0 so that's not we can do it so it's a cosine of s times times B plus a cosine of s I'll just all switch to the same color sine of T so we got our I term for the cross product now it's going to be plus actually minus J you remember when you take the determinate you actually have this kind of you have to checkerboard of switching signs so it's now it's going to be minus J minus J times so you cross out that row in that column and it's going to be this term times this term which is just 0 minus this term times this term and once again when you have oh sorry it's quote when you cross out this column and that row so it's going to be that guy it's going to be that guy times that guy - this guy times this guy so it's going to be minus this Tikes times this guy so it's going to be let me do it in yellow so the negative times negative that guy B plus a cosine of s cosine of T times this guy a cosine of s a cosine of s we'll clean it up in a little bit immediately well we'll clean this up and you see this this negative and that negative will cancel out we're just multiplying everything and then finally the K term so plus I'll go to the next line plus K times cross out that row that column it's going to be that times that minus minus that times that so that looks like a kind of a beastly thing but I think if we take step by step it shouldn't be too bad so that times that the negatives are going to cancel out so we have this term right here is going to be a sine of T sine of s and then this term right here is B plus a cosine of s sine of T sine of T so that's that times that and the negatives canceled out that so I didn't put any negatives here minus this times this so if you this times this is going to be a negative number but if you take the negative of it it's going to be a positive value so it's going to be plus plus that a cosine of T sine of s times that times B plus a cosine of s cosine of T now you see why you don't see many examples of surface integrals being done let's see if we can if we can clean this up a little bit especially especially if we can clean up this last term a bit so let's see it what we can do to simplify it so our first term so let's just multiply it out I guess is the easiest way to do it actually the easiest first step would just to be factor out the B plus a cosine of s because that's in every term B plus a cosine of s B plus a cosine of s B plus a cosine of s B plus a cosine of s so let's just factor that out so this whole crazy thing can be written as B plus a cosine of s so we factored it out times I'll put in maybe put some brackets here so you know it multiplies times every component so the I component we need to factor this guy out is going to be a cosine of s sine of T let me write it in green so that's going to be a cosine of s sine of T a cosine of s sine of T times I you're not used to seeing the eye before it so I'm going to write the eye here and then plus plus we're factoring this guy out so you're just going to be left with cosine of T a cosine of s or we can write it as a cosine of s cosine of T that's that right there just putting it in the same order as that times the unit vector J and then when we factored this guy out so we're not going to see we're not going to see that or or that anymore when we factor that out we can multiply this out and what do we get so in green I'll write again so plus so if you multiply sine of T times this thing over here because that's all that we have left after we factor out this thing we get a a sine of s sine of s sine squared of T right we have sine of T times sine of T sine squared of T so that's that over there plus what do we have over here we have a sine of s times cosine squared of T a sine of s times cosine squared of T and all of that times the K unit vector all of that times the K unit vector and so things are looking a little bit more simplified but you might see something jump out at you you have a sine squared and a cosine squared so somehow if I can just make that just sine squared plus cosine squared of T those will simplify to one and we can we can in this term right here we can if we just focus on that term and this is all kind of algebraic manipulation if we just focus on that term if we just focus because this term right here can be rewritten as a sine of s if we factor that out times sine squared of t plus cosine squared of T times our unit vector K right I just factored out an a sine of s from both of these terms this is our most fundamental trig identity from the unit circle this is equal to one so this last term simplifies to a sine of s times K so so far we've gotten pretty far we were able to figure out the cross product of these two I guess partial derivatives of the vector value to our original parameterization there we were able to figure out what this thing this thing right here before we take the magnitude of it it translates to this thing right here let me let me rewrite it well I don't need to rewrite it you know that well I'll rewrite it so that's equal to I'll rewrite it neatly and we'll use this in the next video B plus a cosine of s times open bracket a cosine of s sine of T times I plus switch back to the blue plus a cosine of s cosine of T times J plus switch back to the blue this thing Plus this simplified nicely a sine of s times K times the unit vector K this this right here is this expression right there and I'll finish this video since I'm already over ten minutes and in the next video we're going to take the magnitude of it and then if we have time actually take this double integral and we'll all be done we'll figure out the surface area of this torus