Surface Integral Example Part 2 - Calculating the surface differential

Now that we have our parameterization right over here Let's get down to the business of actually evaluating this surface integral And its a little bit involved, but we'll try and do it step by step So the first thing I'm going to do is figure out what D Sigma is In terms of S and T, in terms of our parameters So we can turn this whole thing into a double integral in the S-T plane And remember D Sigma is just a little chunk of the surface And we saw in previous videos, the ones where we learned what a surface integral is We saw that D Sigma right over here is equivalent to the magnitude of the cross-product of the partial of our paramterization with respect to one parameter crossed with the paramterization with respect to the other parameter times the differentials of each of the parameters So this is what we are going to use, right here And it's a pretty simple looking statement, but as we'll see Taking cross products tends 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 this stuff with T in it, you can just view that as a constant So Cosine of T is not going to change, the derivative of Cosine of S with respect to S is negative Sine of S, so this is going to be equal to negative Cosine of T times Sine of S (everything T involved will be purple) (vectors will be orange) Then I, and plus... And we're going to 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 this will be plus Cosine of T times Cosine of S then J, and then plus the derivative of this with respect to S well this is just a constant, the derivative of 5 with respect to S would just be zero This does not change with respect to S So our partial with respect to S is just zero So we will just write here 'zero K' And that's nice to see, because it will make our cross-product a little more straightforward Now let's take the partial with respect to T So the derivative of this with respect to T Now, Cosine of S is a constant, derivative of Cosine of T with respect to T is negative Sine of T So this is going to be negative Sine of T times Cosine of S times I, plus... Now the derivative of this with respect to T, derivative of Cosine of T is negative Sine of T So once again, we have minus Sine of T times Sine of S My hand is already hurting from this, this is a painful problem... Now J, plus the derivative of Sine of T with respect to T is just Cosine of T So 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 To take the cross product we are going to set up this three by three matrix And I'll write my unit vectors up here I, J, K... (this is how I like to remember how to take cross products of 3-dimensional vectors, Take the determinant of this three by three matrix) The first row is just our unit vectors The second row is the first vector I'm taking the cross product of So I'm just going to re-write the top-most vector over here And the last part is zero, which will hopefully simplify our calculations And then you have the next vector, that's the third row I encourage you to do this on your own if you already know where this is going It's good practice Even if you have to watch this whole thing to see how its done try to then do it again on your own This is one of those things that you really have to do by yourself to have it really sit in So let's take the determinate now First we'll think about 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 I, times something (normally you see the something in front of the I, but you can swap it) I'm going to write a little neater... The last bit would be subtracting zero times that, but it would just be zero so we don't write it Now we are going to do the J component, but you probably remember the "checkerboard" thing when you have to evaluate three by three matrices Positive, negative, positive, so you have a negative J times something So you ignore J's column, J's row Let me make sure I'm doing this right... Finally you have the K component, and once again you go back to positive there Positive, negative, positive on the coefficients That's just for evaluating a three by three matrix So you have plus K, times... and this might get a little bit more involved Since we don't have the zero to help us out Ignore this row, ignore this column, take the determinant of this sub two by two Let me scroll to the right a little bit... Now this is already looking pretty hairy, but it looks like a simplification is there That's how the colour is helpful I now have trouble doing math in anything other than kind of multiple pastel colours This makes it much easier to see some patterns What we can do is we can factor out the Cosine of T times Sine of T So this is equal to Cosine T Sine T times Sine squared S plus Cosine squared S And this we know, the definition of the unit circle, this is just equal to 1 So that was a significant simplification Now we get our cross product, we get it being equal to Our cross product R sub S crossed with R sub T is going to be equal to Cosine squared T Cosine S times our I unit vector, plus Cosine squared T Sine of S times our J unit vector Plus (all we have left, because this is just one) Cosine T Sine T Times our 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 simplified to the magnitude of this thing times dsdt So let's figure out what the magnitude of this is This is the home stretch, I'm crossing my fingers that I don't make any careless mistakes now So, the magnitude of all of this business is going to be equal to: The square root of the sum of the squares of each of those terms So the first will be Cosine to the fourth T Cosine squared S Plus, Cosine to the fourth T Sine squared S Plus, Cosine squared T Sine squared T Now, the first pattern I see is this first part, we can factor out a cosine to the fourth T 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 these two terms both have a Cosine squared T in them Let's factor those out (everything I'm doing is under the radical sign) So this is equal to Cosine squared T times Cosine squared T and when you factor out a Cosine squared T here you just have plus a Sine squared T And that's nice because that once again simplified to one (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) This is really really useful for us because the square root of Cosine squared of T is just Cosine of T So ALL of that business actually finally simplified to something pretty straightforward So all of this is going to be equal to Cosine of T Going back to what we wanted before, if we want to re-write what D Sigma is It's just cosine T, dsdt So let me write that down... D sigma is equal to Cosine of T dsdt And I'll see you in the next part