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Current time:0:00Total duration:9:51

Video transcript

now that we've set up the parameter the parameter ization let's try to evaluate the integral and the next thing we'll do is essentially try to express D s right over here in terms of D U and DV and we've seen this before D s is going to be equal to the magnitude of the cross product of the partial of R with respect to u crossed with the partial of R with respect to V D u DV so first let's take the cross product and we'll do that with a 3x3 matrix I'll do that right over right over here we set up and I'll just kind of fill in what R sub U and R sub V is in the actual determinant right over here so first we have our components I J and K and now first let's think about what let's think about what R sub U is the partial of R with respect to u well it's I component is going to be 1 the partial of U with respect to u is just 1 so it's I component is 1 its J component is going to be 0 partial of V with respect to U is 0 V does not change with respect to U so this is going to be 0 partial of this should be parentheses around here partial of this with respect to U is just going to be 1 again this is the partial of V squared with respect to U is just 0 so this is just 1 again and then R sub V the partial of R with respect to V the I component is going to be 0 J component is going to be 1 and the partial of U plus V squared with respect to V is going to be is going to be 2v so it's a pretty straightforward determinant so let's try to evaluate so the I component it's going to be I so we cross out this column this row it's going to be 0 times 2 V minus 1 times 1 so essentially just going to be negative 1 times I so we're going to have negative I so this is equal to negative I and then the J component and we're going to put a negative out front because remember we do that checkerboard pattern so cross out that row column one times two V is two V let me make sure either one times two V is two V minus zero times one so it's going to be two V but since but since it was a the J component which is going to be negative is going to be negative 2 VJ negative 2 VJ let me make sure I did that right if you that column that row 1 times 2 V is 2 V minus 0 is 2 V the checkerboard pattern you have a negative J so you have negative 2 V J and then we have finding the K component cross out that row that column 1 times 1 minus 0 so it's going to be plus plus K so if we want the magnitude of this this whole thing right over here is just going to be is just going to be the square root the square root I'm just taking the magnitude of this part right over here the actual cross product it's going to be of negative 1 squared which is just 1 plus negative 2 V squared which is 4 v squared plus 1 squared which is just 1 so this whole thing is going to evaluate to is going to evaluate to we have 2 + 2 v squared d u DV and if we want to actually let me I almost made a mistake that would have been a disaster 2 + 4 V squared 2 + 4 V squared D u DV and if we want maybe it'll help us a little bit if we factor out a 2 right over here so this is the same thing as can factor out this is the same thing as 2 times 1 plus 2 V squared D u DV if you factor out the 2 you get this is equal to the square root of 2 x times the square root of 1 plus 2 v squared D u DV and I now think we are ready to evaluate the surface integral so let's do it alright so we can let me just write this thing down here so I'm gonna write everything that has to do with V I'm going to write in purple so I'm just going to write the DS part right over here it is the square root of 2 - square root of two times the square root of 1 plus 2 V squared and then we're going to have D U which I'll write in green D U and then DV DV so this is just the D s part this is just the D s now we have the Y right over here Y is just equal to V well I'll write that in purple so Y is just equal to V that is why let me make it very clear and all of this is DS all of this is DS and now I can write the bounds in terms of U and V and so the U part U is the same thing as X it goes between 0 and 1 goes between 0 and 1 and then V V is the same thing as Y and it goes and Y goes between or V goes between 0 and 2 0 and 2 and I now think we're ready to evaluate and so and the U and V variables are not so mixed up so we can actually separate out these two integrals make this a product of two single integrals the first thing if you look at it with respect to D u all this stuff in purple is just a constant with respect to u so we can take it out of the D U integral we can take all of this purple stuff out of this D u integral and so this double integral simplifies to the integral from 0 to 2 of I'll write it as the square root of 2 V times the square root of 1 plus 2 V squared and then so I factored out all of this stuff and then you have times the integral from 0 to 1 D U and then times D and then you have DV D V now this was really complicated I can say okay this is just going to be a function of U it's a constant with respect to V and you could factor this whole thing out and separate the integrals but this is even easier this integral is just going to evaluate to 1 so this whole thing just evaluates to 1 and so we've simplified this into one single integral so this simplifies to and I can even take the square root of two out front the square root of two times the integral from V is equal to zero to V is equal to two of V times the square root of 1 plus 2 V squared DV and so we really are in the homestretch of evaluating the surface integral and here this is basic this is actually a little bit of u substitution that we can do in our head you have a function or this kind of this embedded function right over here 1 plus 2 V squared what's the derivative of 1 + 2 V squared well it would be 4 V we have something almost 4 V here we can make this 4 V by multiplying it by 4 here and then dividing it by 4 out here this doesn't change the value of the integral and so now this part right over here it's pretty straightforward to take the antiderivative the antiderivative of this is going to be we have this this embedded functions derivative right over here so we can kind of just treat it like a like an X or V take the antiderivative with respect to this thing right over here and we get this is essentially 1 + 2 V squared to the one-half power we increment it by 1 so it's 1 plus 2 V squared to the three-halves power and then you divide by 3 halves we multiply by 2/3 so times 2/3 2/3 so this is the antiderivative of that and then of course you still have all of this stuff out front square root of 2 / / 4 square root of 2 / 4 and we are going to evaluate this evaluate this from 0 from 0 to 2 and actually just to simplify it let me factor out the 2/3 so we don't have to a value worried about that at 2 and 0 so I'm going to factor out the 2/3 so times 2 over 3 and actually this will cancel out that becomes a 1 that becomes a 2 and so we are left this stuff over here is 1/6 1/6 times and now if you evaluate this at 2 you have 2 V squared that's going to be 2 times 4 is a 8 plus 1 is 9 9 to the three-halves power so 9 to the 1/2 is 3 3 to the third is 27 so it's going to be equal to 27 27 and then minus this thing evaluated 0 well this is evaluated 0 is just going to be 1 1 of the 3 halves is just 1 so minus 1 and so this gives us oh I almost made a mistake there should be square root of 2 up here square root of 2 square root of 2 over 6 times 27 minus 1 so drum roll this gives us the value of our surface integral is let's see 27 minus 1 is 26 so we get 26 times the square root of 2 over 6 we can simplify a little bit more divide the numerator denominator by 2 and you get you get 13 square roots of 2 over 3 and we are done