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Current time:0:00Total duration:12:51

Video transcript

now that we have our parameterization right over here let's get down to the business of actually evaluating this surface integral it's a little bit involved but we'll try to do it step by step and so the first thing I'm going to do is figure out what D Sigma is in terms of SNT in terms of our parameters so we can turn this whole thing into a double integral with respect to or a double integral in the s/t plane and remember D Sigma right over here it's just a little chunk of the surface it's a little area of the surface right over there and we saw in in previous videos the ones where we where we learned what a surface integral even is we saw that that D Sigma right over there it is equivalent to the magnitude of the cross product of the partial of our parameterization with respect to one parameter crossed with the parameterization with respect to the other parameter times the differentials of each of the parameters so this is what we're going to use right here it's a pretty simple looking statement but as we'll see taking cross products tend to get a little bit hairy especially cross products of three-dimensional vectors but we'll do it step by step but before we even take the cross product we first have to take the partial of this with respect to s and then the partial of this with respect to T so first let's take the partial with respect to s the partial of R with respect to s so right over here all the stuff with the T in it that's you can just view that as a constant so cosine of T isn't going to change the partial of or the derivative of cosine of s with respect to s is negative sine of s so this is going to be equal to I'll put the negative out front negative cosine of T sine of s I'm going to keep everything that has a T involved purple sine of s and let me make I don't know I'll make the vectors Orange I and then well plus and once again we take the derivative with respect to s cosine of T is just a constant derivative of sine of s with respect to s is cosine of s so it's going to be plus cosine of T cosine of s cosine of T cosine of S cosine of s cosine of s J J and then plus plus the derivative of this with respect to s well this is just a constant the derivative of v with respect to s would just be zero this is the same thing this is a constant with it this does not change with respect to s so it does so our derivative our partial here with respect to s is just 0 so we could write even zero K right 0 I'll just write 0 K right over there and that's nice to see because that'll make our cross product a little bit more straightforward now let's take the partial with respect to T now let's take the partial with respect to T now let's take the partial with respect to T all right and we get so the derivative of this with respect to T now cosine of s is the constant derivative of cosine T with respect to T is negative sine of T so this is going to be negative sine of T cosine of s cosine of s cosine of s I I'll do I'll go to that I'll mail use this blue I and then plus now derivative of this with respect to T derivative of cosine of T is negative sine of T so once again so now we have minus sign like that minus sign minus sine of T sine of s sine of s my hand is already hurting from this this is a painful problem J plus derivative sine of T with respect to T we're taking the partial with respect to T is just cosine of T so plus cosine of T plus cosine of T and now times the K unit vector now we're ready to take the cross product of these two characters right over here and to take the cross products let me write this down so we're going to say we're going to take the cross product of that with that is going to be equal to it I'm going to set this huge matrix or it's really just a three by three matrix so it's going to be huge because going to take up a lot of space to have to write down all this stuff so maybe I'll take up about maybe I'll take up about that what space so that I have space to work in and I'll write my unit vectors up here i JK or at least this is how I like to remember how to take cross products of through of three dimensional vectors take the determinant of this of this 3x3 matrix the first row are just our unit vectors the second row is the first vector that I'm taking the cross product of so this is it's going to be negative I'm just going to rewrite this right over here so it's going to be negative cosine of T sine of s sine of s and then you have cosine of T cosine of s cosine of T cosine of s cosine of s and then you have 0 which will hopefully simplify our calculations and then you have the next vector that's the third row negative sine of T negative sine of T cosine of s and I encourage you to do this on your own if you already know where this is going it's good practice and even if you have to watch this whole thing to see how it's done try to then do it again on your own because this is one of those things you really got to do yourself to really have it sit in negative sine of T sine of s sine of s and then finally cosine of T cosine of T so let's take let's take the determinant now so first we'll think about our I component our I component you would essentially ignore this column the first column and the first row and then take the determinant of this sub matrix right over here so it's going to be I so this is going to be equal to I times something I'll put the something in parenthesis there normally you see there's something in front of the eye but you can swap them there so let me just so it's going to be I times something ignore this column that row this determinant is cosine of T cosine of s times cosine of T which is going to be cosine squared of T right a little neater cosine squared of T cosine of s cosine of s and then from that we need to subtract zero times this but that's just going to be zero so we're just left with that now we're going to do the J component but you probably remember the checkerboard thing when you have to evaluate three by three matrices positive neg positive so you have a negative you write a negative J negative coefficient I guess in front of the J times something and so ignore J's column J zero and so you have negative cosine of T sine of s times cosine of T well that's going to be negative that's going to be negative cosine squared of T negative cosine squared of T times sine of s let me make sure I'm doing that right ignore that and that it's going to be that times that so yep negative close I'm sine of s minus zero times that and so that's just going to be zero so we can ignore it you have a negative times a negative here so they can both become a positive and then finally you have the K component and once again you go back to positive they're positive negative positive on the coefficients that's just evaluating it three by three matrix and then you have plus K times and now this might get a little bit more involved because we won't have the zero to help us out ignore this row ignore this column take the determinant of this sub two-by-two you have negative cosine T sine s times negative sine of T sine s well that's going to be the negatives cancel out so it's going to be cosine of T cosine of T sine of T times sine squared of s times sine squared of s and then from that we're going to subtract the product of these two things but that products going to be negative so you subtract a negative that's the same thing as adding a positive so plus and you're going to have cosine T sine T again plus cosine T let me scroll to the right a little bit we know that plus cosine T sine T again and that's times cosine squared of s times cosine squared of s now this is already looking pretty hairy but it already looks like a simplifications there and that's where the colors are helpful actually now have trouble doing math and anything other than kind of multiple pastel colors because this actually makes it much easier to to see some patterns and so what we can do over here is we can factor out the cosine T sine T and so this is equal to cosine T sine T ty times sine squared s plus cosine squared s and this we know definition of the unit circle this is just going to be equal to one so that was a significant simplification and so now we get our cross product we get our cross product being equal to let me just rewrite it all our cross product R sub s crossed with r sub t is going to be equal to cosine squared cosine of cosine squared T cosine s cosine s times our I unit vector plus cosine squared T plus cosine squared t sine of s sine of s times our J unit vector times our J unit vector plus plus all we have left because this is just one cosine T sine T plus cosine T sine T times our K R K unit vector so that was pretty good but we're still not done we need to figure out the magnitude of this thing remember D Sigma simplifies to the magnitude of this thing times d s DT so let's figure out what the magnitude of that is and this is really the homestretch so I'm really crossing my fingers that I don't make any careless mistakes now so the magnitude of all of this business magnitude of all of this business is going to be equal to the square root and I'm just gonna have to square each of these terms and then add them up the square root of the sum of the squares of each of those terms so the square of this is going to be cosine squared square is cosine to the fourth cosine to the fourth T cosine squared s cosine squared s + cosine to the fourth T sine squared s and I already see a pattern jumping out sine squared s + + cosine squared T sine squared T now the first pattern I see is this if just this first part right over here we can factor out a cosine to the fourth T then we get something like this happening again so let's do that so this these first two terms are equal to cosine to the fourth T times cosine squared s plus sine squared s which once again we know is just one so this whole expression has simplified to cosine to the fourth T plus cosine squared T sine squared T now we can attempt to simplify this again because this term in this term both have a cosine squared T in in them so let's factor those out so this is going to be equal to everything I'm doing is under the radical sign so this is going to be equal to this is going to be equal to a cosine squared T times cosine squared T and when you factor out a cosine squared T here you just have a plus a sine squared T and that's nice because that once again simplified to 1 all of this is under the radical sign maybe I'll keep drawing the radical signs here to make it clear that all of this is still in the radical sign and then this is really really useful for us because the square root of cosine squared of T is just going to be cosine of T so all of that business actually finally simplify to something pretty straightforward so all of this is just going to be equal to cosine cosine of cosine of T so going back to what we wanted before if we want to rewrite what D Sigma is it's just cosine T DS DT so let me write that down so D Sigma and we can use this for the next part D Sigma is equal to cosine of T DS DT and I'll see you in the next part