Surface integrals
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Partial Derivatives of Vector-Valued Functions
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Introduction to the Surface Integral
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Example of calculating a surface integral part 1
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Example of calculating a surface integral part 2
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Example of calculating a surface integral part 3
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Surface Integral Example Part 1 - Parameterizing the Unit Sphere
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Surface Integral Example Part 2 - Calculating the Surface Differential
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Surface Integral Example Part 3 - The Home Stretch
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Surface Integral Ex2 part 1 - Parameterizing the Surface
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Surface Integral Ex2 part 2 - Evaluating Integral
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Surface Integral Ex3 part 1 - Parameterizing the Outside Surface
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Surface Integral Ex3 part 2 - Evaluating the Outside Surface
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Surface Integral Ex3 part 3 - Top surface
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Surface Integral Ex3 part 4 - Home Stretch
Example of calculating a surface integral part 2 Example of calculating a surface integral part 2
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- Where we left off in the last video, we were finding the
- surface area of a torus, or a doughnut shape.
- And we were doing it by taking a surface integral.
- And in order to take a surface integral, we had to find the
- partial of our parameterization with respect to s, and the
- partial with respect to t, and now we're ready to take
- the cross product.
- And then we can take the magnitude of the cross product.
- And then we can actually take this double integral and
- figure out the surface area.
- So let's just do it step by step.
- Here we could take the cross product, which is not
- a non-hairy operation.
- This is why you don't see many surface integrals actually get
- done, or many examples done.
- Let's take the cross product of these two fellows.
- So the partial of r with respect to s, crossed with--
- in magenta-- the partial of r with respect to t.
- This will be a little bit of review of cross
- products for you.
- You might remember this is going to be equal
- to the determinant.
- I'm going to write the unit vectors up here.
- The first row is i, j, and k.
- And then the next 2 rows are going to be-- let me do that in
- that yellow color-- the next 2 rows are going to be the
- components of these guys.
- So let me copy and paste them.
- You have that right there.
- Copy and paste.
- Put that guy right there.
- Then you have this fellow right there.
- Copy and paste.
- Put him right there.
- And then you got this guy right here.
- This'll save us some time.
- Copy and paste.
- Put him right there.
- Then the last row is going to be this guy's components.
- Copy and paste.
- Put him right here.
- Almost done.
- This guy-- copy and paste.
- Put him right there.
- Make sure we know that these are separate terms.
- And finally, we don't have to copy and paste it, but just
- since we did for all of the other terms, I'll do it
- for that 0, as well.
- So the cross product of these is literally the determinant
- of this matrix right here.
- And so, just as a bit of a refresher of taking
- determinants, this is going to be i times the subdeterminant
- right here, if you cross out this column and that row.
- So it's going to be equal to i-- you're not used to seeing
- the unit vector written first, but we can switch the order
- later-- times i times the submatrix right here.
- If you cross out this column and that row.
- So it's going to be this term times 0-- which is just
- 0-- minus this term times that term.
- So minus this term times this term- the negative signs are
- going to cancel out, so this'll be positive.
- So it's just going to be i times this term times this
- term, without a negative sign right there.
- So i times this term, which is a cosine of s.
- It's really that term times that term, minus that term
- times that term, but the negatives cancel out.
- That times that is 0.
- So that's how we can do this.
- It's a cosine of s times b plus a cosine of s-- I'll just all
- switch to the same color-- sine of t.
- So we've got our i term for the cross product.
- Now it's going to be minus j-- remember when you take the
- determinant, you actually have this, kind of, you have to
- checker board of switching sines.
- So now it's going to be minus j times-- and you cross out that
- row and that column-- and it's going to be this term times
- this term-- which is just 0-- minus this term
- times this term.
- And once again, when you have-- oh, sorry.
- When you cross out this column and that row.
- So it's going to be that guy times that guy, minus
- this guy times this guy.
- So it's going to be minus this guy times this guy-- so it's
- going to be-- let me do it in yellow.
- So the negative times negative that guy, b plus a cosine of s
- cosine of t times this guy, a cosine of s.
- We'll clean it up in a little bit.
- Well, we'll clean this up, and you see this negative and that
- negative will cancel out.
- We're just multiplying everything.
- And then finally, the k term.
- So plus-- I'll go to the next line-- plus k times-- cross out
- that row, that column-- it's going to be that times that,
- minus that times that.
- So that looks like a kind of a beastly thing.
- But I think if we take it step by step, it
- shouldn't be too bad.
- So that times that.
- The negatives are going to cancel out.
- So this term right here is going to be a sine
- of t, sine of s.
- And then this term right here is b plus a
- cosine of s sine of t.
- So that's that times that-- and the negatives canceled out,
- that's why I didn't put any negatives here-- minus
- this times this.
- So this times this is going to be a negative number.
- But if you take the negative of it, it's going to
- be a positive value.
- So it's going to be plus that a cosine of t
- sine of s times that.
- Times b plus a cosine of s cosine of t.
- Now you see why you don't see many examples of surface
- integrals being done.
- Let's see if we can clean this up a little bit, especially if
- we can clean up this last term a bit.
- So let's see what we can do to simplify it.
- So our first term.
- So let's just multiply it out, I guess is the
- easiest way to do it.
- Actually, the easiest first step would just be factor out
- the b plus a cosine of s.
- Because that's in every term. b plus a cosine of s.
- b plus a cosine of s.
- b plus a cosine of s. b plus a cosine of s.
- So let's just factor that out.
- So this whole crazy thing can be written as b plus a cosine
- of s-- so we factored it out-- times--.
- I'll put in some brackets here, so you don't multiply
- times every component.
- So the i component, when you factor this guy out, is going
- to be a cosine of s sine of t.
- Let me write it in green.
- So it's going to be a cosine of s sine of t times i-- you're
- not used to seeing the i before, so I'm going to write
- the i here-- and then plus--.
- We're factoring this guy out, so you're just going to be
- left with cosine of t, a cosine of s.
- Or we can write it as a cosine of s cosine of t-- that's that
- right there, just putting it in the same order as that--
- times the unit vector j.
- And then when we factored this guy out-- so we're not going
- to see that or that anymore.
- When you factor that out, we can multiply this
- out, and what do we get?
- So in green, I'll write again.
- So if you multiply sine of t times this thing over here--
- because that's all that we have left after we factor out this
- thing-- we get a sine of s, sine squared of t, right?
- We have sine of t times sine of t.
- So that's that over there.
- Plus-- what do we have over here?
- We have a sine of s times cosine squared of t.
- And all of that times the k unit vector.
- And so things are looking a little bit more simplified,
- but you might see something jump out at you.
- You have a sine squared and a cosine squared.
- So somehow, if I can just make that just sine squared plus
- cosine squared of t, those will simplify to 1.
- And we can.
- And this term right here, we can-- if we just focus on that
- term-- and this is all kind of algebraic manipulation.
- If we just focus on that term, this term right here can be
- rewritten as a sine of s-- if we factor that out-- times sine
- squared of t plus cosine squared of t times
- our unit vector, k.
- Right?
- I just factored out an a sine of s from both of these terms.
- And this is our most fundamental trig identity
- from the unit circle.
- This is equal to 1.
- So this last term simplifies to a sine of s times k.
- So, so far we've gotten pretty far.
- We were able to figure out the cross product of these 2, I
- guess, partial derivatives of the vector valued,
- or our original parameterization there.
- We were able to figure out what this thing right here, before
- we take the magnitude of it, it translates to this
- thing right here.
- Let me rewrite it-- well, I don't need to rewrite it.
- You know it.
- Well, I'll rewrite it.
- So that's equal to-- I'll rewrite it neatly and we'll use
- this in the next video-- b plus a cosine of s times open
- bracket a cosine of s sine of t times i plus-- switch back to
- the blue-- plus a cosine of s cosine of t times j plus--
- switch back to the blue-- this thing-- plus-- this simplified
- nicely-- a sine of s times k.
- Times the unit vector k.
- This right here is this expression right there.
- And I'll finish this video, since I'm already
- over 10 minutes.
- And in the next video, we're going to take
- the magnitude of it.
- And then, if we have time, actually take
- this double integral.
- And we'll all be done.
- We'll figure out the surface area of this torus.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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