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Now that we have our parameterization right over here

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Let's get down to the business of actually evaluating this surface integral

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And its a little bit involved, but we'll try and do it step by step

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So the first thing I'm going to do is figure out what D Sigma is

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In terms of S and T, in terms of our parameters

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So we can turn this whole thing into a double integral in the S-T plane

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And remember D Sigma is just a little chunk of the surface

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And we saw in previous videos, the ones where we learned what a surface integral is

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We saw that D Sigma right over here is equivalent to the magnitude of the cross-product

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of the partial of our paramterization with respect to one parameter

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crossed with the paramterization with respect to the other parameter

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times the differentials of each of the parameters

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So this is what we are going to use, right here

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And it's a pretty simple looking statement, but as we'll see

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Taking cross products tends to get a little bit hairy

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Especially cross products of three dimensional vectors

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But we'll do it step by step

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But before we even take the cross product we first have to take the partial of this with respect to S

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and then the partial of this with respect to T

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So first let's take the partial with respect to S

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The partial of R with respect to S

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So right over here, all this stuff with T in it, you can just view that as a constant

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So Cosine of T is not going to change, the derivative of Cosine of S with respect to S

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is negative Sine of S, so this is going to be equal to negative Cosine of T times Sine of S

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(everything T involved will be purple)

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(vectors will be orange)

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Then I, and plus...

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And we're going to take the derivative with respect to S, Cosine of T is just a constant

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Derivative of Sine of S with respect to S is Cosine of S

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So this will be plus Cosine of T times Cosine of S

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then J, and then plus the derivative of this with respect to S

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well this is just a constant, the derivative of 5 with respect to S would just be zero

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This does not change with respect to S

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So our partial with respect to S is just zero

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So we will just write here 'zero K'

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And that's nice to see, because it will make our cross-product a little more straightforward

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Now let's take the partial with respect to T

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So the derivative of this with respect to T

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Now, Cosine of S is a constant, derivative of Cosine of T with respect to T is negative Sine of T

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So this is going to be negative Sine of T times Cosine of S

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times I, plus...

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Now the derivative of this with respect to T, derivative of Cosine of T is negative Sine of T

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So once again, we have minus Sine of T times Sine of S

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My hand is already hurting from this, this is a painful problem...

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Now J, plus the derivative of Sine of T with respect to T is just Cosine of T

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So plus Cosine of T

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and now times the K unit vector

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Now we're ready to take the cross product of these two characters right over here

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To take the cross product we are going to set up this three by three matrix

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And I'll write my unit vectors up here

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I, J, K...

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(this is how I like to remember how to take cross products of 3-dimensional vectors,

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Take the determinant of this three by three matrix)

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The first row is just our unit vectors

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The second row is the first vector I'm taking the cross product of

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So I'm just going to re-write the top-most vector over here

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And the last part is zero, which will hopefully simplify our calculations

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And then you have the next vector, that's the third row

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I encourage you to do this on your own if you already know where this is going

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It's good practice

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Even if you have to watch this whole thing to see how its done try to then do it again on your own

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This is one of those things that you really have to do by yourself to have it really sit in

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So let's take the determinate now

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First we'll think about our I component

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You would essentially ignore this column, the first column and the first row

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And then take the determinant of this sub-matrix right over here

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I, times something (normally you see the something in front of the I, but you can swap it)

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I'm going to write a little neater...

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The last bit would be subtracting zero times that, but it would just be zero so we don't write it

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Now we are going to do the J component, but you probably remember the "checkerboard"

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thing when you have to evaluate three by three matrices

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Positive, negative, positive, so you have a negative J times something

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So you ignore J's column, J's row

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Let me make sure I'm doing this right...

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Finally you have the K component, and once again you go back to positive there

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Positive, negative, positive on the coefficients

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That's just for evaluating a three by three matrix

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So you have plus K, times... and this might get a little bit more involved

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Since we don't have the zero to help us out

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Ignore this row, ignore this column, take the determinant of this sub two by two

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Let me scroll to the right a little bit...

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Now this is already looking pretty hairy, but it looks like a simplification is there

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That's how the colour is helpful

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I now have trouble doing math in anything other than kind of multiple pastel colours

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This makes it much easier to see some patterns

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What we can do is we can factor out the Cosine of T times Sine of T

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So this is equal to Cosine T Sine T times Sine squared S plus Cosine squared S

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And this we know, the definition of the unit circle, this is just equal to 1

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So that was a significant simplification

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Now we get our cross product, we get it being equal to

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Our cross product R sub S crossed with R sub T is going to be equal to

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Cosine squared T Cosine S times our I unit vector, plus Cosine squared T Sine of S times our J unit vector

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Plus (all we have left, because this is just one) Cosine T Sine T

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Times our K unit vector

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So that was pretty good, but we're still not done

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We need to figure out the magnitude of this thing

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Remember: D Sigma simplified to the magnitude of this thing times dsdt

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So let's figure out what the magnitude of this is

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This is the home stretch, I'm crossing my fingers that I don't make any careless mistakes now

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So, the magnitude of all of this business is going to be equal to:

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The square root of the sum of the squares of each of those terms

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So the first will be Cosine to the fourth T Cosine squared S

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Plus, Cosine to the fourth T Sine squared S

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Plus, Cosine squared T Sine squared T

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Now, the first pattern I see is this first part, we can factor out a cosine to the fourth T

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These first two terms are equal to Cosine to the fourth T times Cosine squared S plus Sine squared S

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Which once again we know is just one

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So this whole expression has simplified to Cosine to the fourth T

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plus Cosine squared T Sine squared T

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Now we can attempt to simplify this again, because these two terms both have a Cosine squared T in them

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Let's factor those out

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(everything I'm doing is under the radical sign)

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So this is equal to Cosine squared T times Cosine squared T

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and when you factor out a Cosine squared T here you just have plus a Sine squared T

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And that's nice because that once again simplified to one

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(All of this is under the radical sign, I'll keep drawing it here to keep it clear that this is still under the radical)

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This is really really useful for us because the square root of Cosine squared of T is just Cosine of T

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So ALL of that business actually finally simplified to something pretty straightforward

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So all of this is going to be equal to Cosine of T

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Going back to what we wanted before, if we want to re-write what D Sigma is

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It's just cosine T, dsdt

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So let me write that down...

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D sigma is equal to Cosine of T dsdt

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And I'll see you in the next part