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### Course: Multivariable calculus > Unit 4

Lesson 11: Surface integrals- Introduction to the surface integral
- Find area elements
- Example of calculating a surface integral part 1
- Example of calculating a surface integral part 2
- Example of calculating a surface integral part 3
- Surface integrals to find surface area
- Surface integral example, part 1
- Surface integral example part 2
- Surface integral example part 3: The home stretch
- Surface integral ex2 part 1
- Surface integral ex2 part 2
- Surface integral ex3 part 1
- Surface integral ex3 part 2
- Surface integral ex3 part 3
- Surface integral ex3 part 4

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# Example of calculating a surface integral part 2

Example of calculating a surface integral part 2. Created by Sal Khan.

## Want to join the conversation?

- This video without cursor is a bit confusing for me, i don't know where sal talk about when he say:

"that guy" but without the cursor i don't know what he's talking about.(67 votes)- I was just about to complain about the same issue. I'm not sure if the problem is with the HTML player versus flash or if the video file is missing the cursor.(17 votes)

- Hello friends, I have an uneasy feeling about the way the computation is going. Sal created a parameterization of the torus in a left handed coordinate system and now he is taking a cross product which I think is defined only for a right handed coordinate system. In fact, the cross product essentially defines what is a right handed coordinate system. I am confident a better mathematician than I will suggest that it all comes out OK in the end, but I would be very reluctant to proceed in this way. Can anyone tell me why I should not be concerned? Best wishes.(4 votes)
- Good question, I didn't notice that he parametried it in a left handed coordinate system! Here's my thinking:

You can convert a left handed coordinate system to a right handed one by inverting the z-axis (imagine turning your left hand upside down). This transformation doesn't change the shape of the torus, it only flips it along the xy-plane. If we input some (s, t) into our left handed parametrization and get some point (x,y,z), this is equivalent to the (x,y,-z) in the right handed torus. The cross product is the infinitesmal area unit for an input (s,t). Since the torus is symmetrical over the xy-plane, the inifitesmal area unit in (x,y,z) and (x, y, -z) should be the same and therefore the orientation of the coordinate system doesn't matter for all inputs (s,t).

If this reasoning hold, this would only work for shapes symmetrical over the xy-plane and it seems like a much better idea to do right handed parametrizations when possible.

I've just started learning about this stuff though so I could be wrong!(1 vote)

- I thought that matrices had to be squared in order to take a determinant?(1 vote)
- It is, we let the unit vectors < i , j , k > be the first row, then the other two vectors to be the 2nd and 3rd rows. Thus, it becomes a 3x3 matrix.(3 votes)

- At2:00, he could have written (b+a*cos(t)) outside of determinant. Result would be the same, but there would be less writing, as you can divide one row (or a column) by some number, and than multiply determinant by the same number (in our case (b+a*cos(t)) )(1 vote)
- Is it just me or can no one see the cursor in this and the previous video cause without it, it gets kind of hard to understand which term Sal is referring to when he says "this guy" and "that guy"(2 votes)

- Wait so why do we have to take the cosine of i hat? is i hat supposed to be x vector?(1 vote)
- I find the approach of alternating + and - for sub-determinates confusing, and prefer to either actually copy the first column as an extra column to the right (I think that's how I was taught long ago), or to just think of it as being there, so the pattern of the operation always stays the same (upper left x lower right) - (upper right x lower left) in each case. I think that's also consistent with the Levi-Civita symbols, which I don't quite understand, but look like lower case epsilon, and through some symmetry rules that determine which indices are -1, 1 or 0, apparently prescribe the indices in a compact way so that the cross product can be represented as a simple summation, and also lead to some (apparently) useful identities when combined with the Kroneker delta operator for the dot product. So not a question I guess. More of a comment, but I would like to have the Levi-Civita symbols explained. The source I have is completely cryptic and basically fails, though i*(A2B3-A3B2)+j*(A3B1-A1B3)+k(A1B2-A2B2) is the outcome.(1 vote)
- Actually, I consider that method unnecessarily tedious and open to confusion, and that is the way my Calc III professor uses. It's easier just to know that the j vector component is negative, and evaluate a 3x3 matrix normally, which is the way I learned it in Engineering Statics.(1 vote)

- what are those, adidas or nike or what?(0 votes)

## Video transcript

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.