Example of calculating a surface integral part 1 Example of calculating a surface integral part 1
Example of calculating a surface integral part 1
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- We saw several videos ago that we can parameterize a torus or
- a doughnut shape as a position vector-valued function
- of two parameters.
- And this is the outcome that we had.
- I think I did it over several videos because
- it was a bit hairy.
- And I'll write our position vector-valued function first.
- So we have r as a function of our two parameters s and t.
- And then I'll review a little bit of what all the terms--
- what the s, the t, and the a's and the b's represent.
- But it's equal to b plus a cosine of s.
- And once again, we saw this several videos ago.
- So you might want to watch the videos on parameterizing
- surfaces with two parameters to figure out how we got here.
- Times the sine of t.
- I'm going to put the s terms and the t terms
- in different colors.
- Times our i unit vector.
- I'll put the vectors, or the unit vectors
- in this orange color.
- Plus-- I'll do it in the same yellow.
- Plus b plus a cosine of s times cosine of t times the j unit
- vector-- the unit factor in the y direction.
- Plus a sine of s times the k unit vector or the unit
- vector in the z direction.
- And in order to generate the torus or the doughnut shape,
- this is true for our parameters-- so we don't wrap
- multiple times around the torus-- for s being between 0
- and 2 pi and for t being between 0 and 2 pi.
- And just as a bit of review, we're all of this came from--
- and I'm going to have to do what my plan is for this
- video over several videos.
- But let's review where all of this came from.
- Let me draw a doughnut.
- My best effort at a doughnut right here.
- That looks like a doughnut or a torus.
- And you can imagine a torus, or this doughnut shape is kind of
- the product of two circles.
- You have the circle that's kind of the cross section of
- the doughnut at any point.
- You could take it there.
- You could take it over there.
- And then you have the circle that kind of wraps around all
- of these other circles or these other circles wrap around it.
- And so, when we derived this formula up here or this
- parameterization, a was the radius of these cross
- sectional circles.
- That's a.
- That's what these a terms were.
- And b was the distance from the center of our torus out to the
- center of these cross sections.
- So this was b.
- So you can imagine that b is kind of the radius of the big
- circle up to the midpoint of the, I guess, cross section.
- And a is the radius of the cross sectional circles.
- And when we parameterized it, the parameter s was essentially
- telling us how far-- s was telling us how far or where are
- we wrapping around this circle.
- So it's an angle from 0 to 2 pi to say where
- we are on that circle.
- And t tells us how much we've rotated around
- the larger circle.
- So if you think about it, you can specify any point on this
- doughnut or on this surface or on this torus by telling
- you an s or a t.
- And so that's why we picked that as the parameterization.
- Now, the whole reason why I'm even revisiting this stuff that
- we saw several videos ago is we're going to actually use it
- to compute an actual surface integral.
- And the surface integral we're going to compute will tell us
- the surface area of this torus.
- So this surface right here is sigma, like that and it's being
- represented by this position vector-valued function.
- That is parameterized by these two parameters right there.
- And if we wanted to figure out the surface area, if we just
- kind of set it as the surface integral we saw in, I think,
- the last video at least the last vector calculus video I
- did that this is a surface integral over the surface.
- Here this capital Sigma does not represent a sum, it
- represents a surface of a bunch of the little d sigmas--
- a bunch of the little chunks of the surface.
- And just as a review, you can imagine each d sigma
- is a little patch of the surface right there.
- That is a d sigma.
- It's a double integral here because we want to add up
- all of the d sigmas in 2 directions.
- You can imagine one kind of rotating this way around the
- torus and then the other direction is going in the other
- direction around the torus.
- So that's why it's a double integral.
- And this is just going to give you the surface area, which is
- the whole point of this video and probably the next
- one or two videos.
- But if you wanted to also multiply these sigmas times
- some other value-- there's some scalar field that this is in
- that you cared about-- you could put that other
- value right there.
- But here we're just multiplying it by 1.
- And we saw in the last video that it's a way of expressing
- an idea, but you really can't do much computation with this.
- But a way that you can express this so that you can actually
- take the integral, you say this is the same thing-- and we saw
- this in the last several videos.
- This is the same thing as the double integral over the
- region over which our parameters are defined.
- So it's this region over here where s and t go from 0 to 2 pi
- of whatever function this is.
- We just have a 1 here, so we could just write a 1 if we
- like; it doesn't change much.
- Times-- and this is what we learned.
- Times the magnitude of the partial derivative of
- r with respect to s.
- The magnitude of that crossed with the partial derivative
- of r with respect to t ds.
- You could take it in either order, but ds dt.
- So we saw this in the last video.
- What we're going to do now is actually compute this.
- That's the whole point of this video.
- We're going to take the cross product of these two vectors.
- So let's figure out these vectors.
- Then in the next video we're going to take
- the cross product.
- And then the video after that we'll actually evaluate
- this double integral.
- And you're going to see it's a pretty hairy problem and this
- is probably the reason that very few people ever see an
- actual surface integral get computed.
- But let's do it anyway.
- So the partial derivative of r with respect to s-- so
- this term right here.
- We'll do the cross product in the next video.
- This term is what?
- We just want to hold t constant and took the partial
- with respect to just s.
- So this up here, if you distributed the sine of t times
- b-- that's just going to be a constant in terms of s,
- so we can ignore that.
- Then you have sine of t times this over here.
- So sine of t and a is a constant.
- And you take the derivative of cosine of s.
- That's negative sine of s.
- So the derivative of this with respect to s or the partial
- with respect to s is going to be minus a-- I'll write in
- green the sine of t, so you know that's where it came from.
- Sine of t and then sine of s.
- The derivative of this is negative sine of s.
- That's where that negative came from.
- And then I'll write the sine of s right there.
- Times the unit vector i.
- That's the partial of just this x term with respect to s.
- And then we'll do the same thing with the y
- term or the j term.
- So plus-- same logic-- b times cosine of t with respect to s.
- When you take the partial [INAUDIBLE]
- becomes 0 so you're left with a-- well, it's going to be a
- minus a again because when you take the derivative of the
- cosine of s it's going to be negative sine of s.
- Let me do it.
- You're going to have a minus a.
- This cosine of t.
- Minus a cosine of t.
- That's the constant terms.
- Sine of s.
- Just taking partial derivatives.
- Sine of s j.
- And then finally, we take the derivative of this
- with respect to s.
- And that's pretty straightforward.
- It's just going to be a cosine of s.
- So plus a cosine of s k.
- Now hopefully you didn't find this confusing.
- The negative sines because the derivative of cosines
- are negative sines.
- So negative sine of s.
- That's why it's negative sine of s times the constant.
- Negative sine of s times the constant-- the constant
- cosine of t sine of t.
- So hopefully this makes some sense just as a review of
- taking a partial derivative.
- Now let's do the same thing with respect to t.
- And I'll do that in a different color.
- So we're now going to take the partial of r with respect to t.
- So the partial of r with respect to t is equal to-- so
- now this whole term over here is a constant, and so it's
- going to be that whole term times the derivative of this
- with respect to t, which is just cosine of t.
- So it's going to be b plus a cosine of s
- times cosine of t i.
- And then, plus-- and it's actually going to be a minus
- because when you take the derivative of this with
- respect to t it's going to be minus sine of t.
- So it's going to be negative and then let me leave some
- space for this term right here.
- Negative sine of t.
- And you're going to have this constant out there.
- That's a constant in t.
- b plus a cosine of s.
- That's just that term right there.
- Derivative of cosine t is negative sine of t times j.
- And then the partial of this with respect to t-- this
- is just a constant in t.
- So the partial's going to be 0.
- So I'll write plus 0k.
- Let me do all my vectors in that same color.
- Plus 0 times the unit vector k.
- So that gives us our partial derivatives.
- Now we have to take their cross product, then find the
- magnitude of the cross product, and then evaluate this
- double integral.
- And I'll do that in the next couple of videos.
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