Surface Integral Example Part 2 - Calculating the Surface Differential Taking the cross product to calculate the surface differential in terms of the parameters
Surface Integral Example Part 2 - Calculating the Surface Differential
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- 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
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