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 3 Example of calculating a surface integral part 3
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- In the last couple of videos we've been slowly moving
- towards our goal of figuring out the surface area
- of this torus.
- And we did it by evaluating a surface integral, and in order
- to evaluate a surface integral we had to take the
- parameterization-- take its partial with respect to s and
- with respect to t.
- We did that in the first video.
- Then we had to take its cross product.
- We did that in the second video.
- Now, we're ready to take the magnitude of the cross product.
- And then we can evaluate it inside of a double integral and
- we will have solved or we would have computed an actual surface
- integral-- something you see very few times in your
- education career.
- So this is kind of exciting.
- So this was the cross product right here.
- Now, let's take the magnitude of this thing.
- And you might remember, the magnitude of any vector is kind
- of a Pythagorean theorem.
- And in this case it's going to be kind of the distance
- formula to the Pythagorean theorem n 3 dimensions.
- So the magnitude-- this is equal to, just as a reminder,
- is equal to this right here.
- It's equal to the partial of r with respect to s cross
- with the partial of r with respect to t.
- Let me copy it and paste it.
- That is equal to that right there.
- Put an equal sign.
- These two quantities are equal.
- Now we want to figure out the magnitude.
- So if we want to take the magnitude of this thing, that's
- going to be equal to-- well, this is just a scalar that's
- multiply everything.
- So let's just write the scalar out there.
- So b plus a cosine of s times the magnitude of
- this thing right here.
- And the magnitude of this thing right here is going to be the
- sum, of-- you can imagine, it's the square root of this
- thing dotted with itself.
- Or you could say it's the sum of the squares of each of
- these terms to the 1/2 power.
- So let me write it like that.
- Let me write the sum of the squares.
- So if you square this you get a squared cosine squared
- of s, sine squared of t.
- That's that term.
- Plus-- let me color code it.
- That's that term.
- I'll do the magenta.
- Plus that term squared.
- Plus a squared cosine squared of s, cosine squared of t.
- That's that term.
- And then finally-- I'll do another color--
- this term squared.
- So plus a squared sine squared of s.
- And it's going to be all of this business to the 1/2 power.
- This right here is the same thing as the magnitude
- of this right here.
- This is just a scalar that's multiplying by
- both of these terms.
- So let's see if we can do anything interesting here.
- If this can be simplified in any way.
- We have a squared cosine squared of s.
- We have an a squared cosine squared of s here, so let's
- factor that out from both of these terms and
- see what happens.
- I'm just going to rewrite this second part.
- So this is going to be a squared cosine squared of s
- times sine squared of t-- put a parentheses-- plus cosine--
- oh, I want to do it in that magenta color, not orange.
- Plus cosine squared of t.
- And then you're going to have this plus a squared
- sine squared of s.
- And of course, all of that is to the 1/2 power.
- Now what is this?
- Well, we have sine squared of t plus cosine squared of t.
- That's nice.
- That's equal to 1, the most basic of trig identities.
- So this expression right here simplifies to a squared cosine
- squared of s plus this over here: a squared
- sine squared of s.
- And all of that to the 1/2 power.
- You might immediately recognize you can factor
- out an a squared.
- This is equal to a squared times cosine squared of s
- plus sine squared of s.
- And all of that to the 1/2 power.
- I'm just focusing on this term right here.
- I'll write this in a second.
- But once again, cosine squared plus sine squared of anything
- is going to be equal 1 as long as it's the same anything
- it's equal to 1.
- So this term is a squared to the 1/2 power.
- Or the square root of a squared, which is just
- going to be equal to a.
- So all of this-- all that crazy business right here just
- simplifies, all of that just simplifies to a.
- So this cross product here simplifies to this times
- a, which is a pretty neat and clean simplification.
- So let me rewrite this.
- That simplifies, it simplifies to a times that.
- And what's that? a times b, so it's ab.
- ab plus a squared cosine of s.
- So already, we've gotten pretty far and it's nice when you do
- something so beastly and eventually it gets to
- something reasonably simple.
- And just to review what we had to do, what our mission was
- several videos ago, is we want to evaluate what this thing is
- over the region from s-- over the region over with the
- surface is defined.
- So s going from 0 to 2 pi and t going from 0 to 2 pi.
- Over this region.
- So we want to integrate this over that region.
- So that region we're going to vary s from 0 to 2 pi.
- So ds.
- And then we're going to vary t from 0 to 2 pi-- dt.
- And this is what we're evaluating.
- We're evaluating the magnitude of the cross product of these
- two partial derivatives of our original parameterization.
- So this is what we can put in there.
- Things are getting simple all of a sudden, or simpler.
- ab plus a squared cosine of s.
- And what is this equal to?
- So this is going to be equal to-- well, we just take
- antiderivative of the inside with respect to s.
- So the antiderivative-- so let me do the outside
- of our integral.
- So we're still going to have to deal with the 0 to 2
- pi and our dt right here.
- But the antiderivative with respect to s right here is
- going to be-- ab is just a constant, so it's going to
- be abs plus-- what's the antiderivative of cosine of s?
- It's sine of s.
- So plus a squared sine of s.
- And we're going to evaluate it from 0 to 2 pi.
- And what is this going to be equal to?
- Let's put our boundaries out again or the t integral that
- we're going to have to do in a second-- 0 the 2 pi d t.
- When you put 2 pi here you're going to get ab times
- 2 pi or 2 pi ab.
- So you're going to have 2 pi ab plus a squared sine of 2 pi.
- Sine of 2 pi is 0, so there's not going to be any term there.
- And then minus 0 times ab, which is 0.
- And then you're going to have minus a squared sine
- of 0, which is also 0.
- So all of the other terms are all 0's.
- So that's what we're left with it, it simplified nicely.
- So now we just have to take the antiderivative of
- this with respect to t.
- And this is a constant in t, so this is going to be equal to--
- take the antiderivative with respect to t-- 2 pi abt and we
- need to evaluate that from 0 to 2 pi, which is equal to--
- so we put 2 pi in there.
- You have a 2 pi for t, it'll be a 2 pi times 2 pi ab.
- Or we should say, 2 pi squared times ab minus
- 0 times this thing.
- Well, that's just going to be 0, so we don't even
- have to write it down.
- So we're done.
- This is the surface area of the torus.
- This is exciting.
- It just kind of snuck up on us.
- This is equal to 4 pi squared ab, which is kind of a neat
- formula because it's very neat and clean.
- You know, it has a 2 pi, which is kind of the
- diameter of a circle.
- We're squaring it, which kind of makes sense because we're
- taking the product of-- you can kind of imagine the product
- of these 2 circles.
- I'm speaking in very abstract, general terms, but that
- kind of feels good.
- And then we're taking just the product of those
- two radiuses, remember.
- Let me just copy this thing down here.
- Actually, let me copy this thing because this is our new--
- this is our exciting result.
- Let me copy this.
- So copy.
- So all of this work that we did simplified to this,
- which is exciting.
- We now know that if you have a torus where the radius of the
- cross section is a, and the radius from the center of
- the torus to the middle of the cross sections is b.
- That the surface area of that torus is going to be 4 pi
- squared times a times b.
- Which I think is a pretty neat outcome.
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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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