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## Multivariable calculus

### 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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# Surface integral example part 2

Taking the cross product to calculate the surface differential in terms of the parameters. Created by Sal Khan.

## Want to join the conversation?

- Where can I find Sal's videos on determinants and cross-products, if any? Thanks!(3 votes)
- In the Linear Algebra playlist, click Vectors and Spaces and scroll down.(6 votes)

- Wait, the title of this video says it's a differential? I thought it would be an integral. What does differential even mean? Thanks(3 votes)
- The differential has more in common with a derivative than an integral. However, it is not exactly the same as a derivative. They are both rates of change, but the derivative of a function is the rate of change of the function with respect to its independent variable(s) (e.g. dy/dx, df/dx, df/dy, etc.). The differential of the function, on the other hand, is the rate of change of the function as a whole, without respect to its variables (e.g. df, dy, dz, etc.)(2 votes)

- Why is the paramaterization cos(t)cos(s)i+sin(t)sin(s)j+sin(t)k?

Shouldn't it be rsin(t)cos(s)i+rsin(t)sin(s)j+rsin(t)k?(3 votes) - At the end shouldn't it be sqrt(cos^4(t) + 1) not cos(t)?(2 votes)
- The t's and the +'s look very similar in the video, I think you meant that you end up with sqrt(cos^4(t)+cos^2(t)sin^2(t)), not sqrt(cos^4(t)+sin^2(t)+cos^2(t). Then you factor out a cos^2(t) and you get sqrt(cos^2(t) * (cos^2(t) + sin^2(t))) which simplifies to

sqrt(cos^2(t) * 1 which in turn becomes cos(t). Hope this helped(3 votes)

- you could've just told about jacobian which is more general than cross product(2 votes)
- in about1:30. why the order of the cross product is Rs *Rt. But not the reversed order? I dont know how to tell which order to use? Thank you very much.(0 votes)
- The order doesn't matter because either way the magnitude will be the same.

Essentially Rs*Rt = -Rt*Rs, if you reverse the order you get a minus sign, but when you take the magnitude the minus sign is irrelevant.(4 votes)

- you are saying a verbal misstake between 0.52-0.57 by saying the crossproduct of the parameterization with respect to the parameter whilst you really meant to say the partial with respect to the parametrization as you did in the first case just a couple of seconds before that(1 vote)
- at 12.29 sal has treated the second (sin^2+cos^2=1)as a coefficient, should it not be cos^4(t) +1?(1 vote)
- sorry my bad, forget that(1 vote)

- At10:45when he does the square root, can you take out the cos(t) and use it as a scalar? He did something similar to that with the torus example. Does leaving it in just make it simpler to, well simplify.

It would look like:

Cos(t) * sqrt(magnitude)(1 vote) - Why is cos t the magnitude of the cross product? Clearly, it can take on negative values.(1 vote)

## Video transcript

Now that we have our
parametrization right over here, let's get down to the
business of actually evaluating this surface integral. It's a little bit
involved, but we'll try to do it step by step. And 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 with respect to-- or a double
integral in the s, t plane. And remember, d sigma
right over here, it's just a little
chunk of the surface. It's a little area of the
surface right over there. And we saw in previous
videos, the ones where we learned what
a surface integral even is, we saw that d
sigma right over there, it is equivalent to the
magnitude of the cross product of the partial of our
parametrization with respect to one parameter crossed
with the parametrization with respect to
the other parameter times the differentials
of each of the parameters. So this is what we're
going to use right here. It's a pretty simple looking
statement, but as we'll see, taking cross products tend
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
the stuff with the t in it, you can just view
that as a constant. So cosine of t isn't
going to change. The partial of-- or the
derivative of cosine of s with respect to s is
negative sine of s. So this is going to
be equal to-- I'll put the negative out front--
negative cosine of t sine of s. I'm going to keep everything
that has a t involved purple. Sine of s. And let me make-- I don't know. I'll make the vectors orange--
i, and then you we'll plus and once again, we'll 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 it's going to be plus
cosine of t cosine of s 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 0. This is the same thing. It's just a constant
with-- this does not change with respect to s. So it does-- so our partial here
with respect to s is just 0. So we could write even 0k. Let me write 0-- I'll just
write 0k right over there. And that's nice to
see because that'll make our cross product a little
bit more straightforward. Now, let's take the
partial with respect to t. all right. And we get-- so the derivative
of this with respect to t-- now, cosine
of s is the constant. Derivative of cosine
t with respect to t is negative sine of t. So this is going to be negative
sine of t cosine of s i-- I'll do it in that-- I'll
use this blue-- i, and then plus-- now derivative of
this with respect to t. Derivative of cosine of
t is negative sine of t. So once again, so now we have
minus sine of t sine of s-- my hand is already
hurting from this. This is a painful problem. j plus derivative of sine
of t with respect to t. We're taking the partial
with respect to t. It's 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. And to take the cross products--
let me write this down. So we're going to
say, we're going to take the cross product
of that with that. Is going to be
equal to-- and I'm going to set up
this huge matrix. Or, it's really just a
three by three matrix but it's going to
be huge because it's going to take up a
lot of space to have to write down all this stuff. So maybe I'll take up
about that much space so that I have space to work in. And I'll write my
unit vectors up here. i, j, k. Or at least this is
how I like to remember how to take cross products
of three dimensional vectors. Take the determinant
of this three by three matrix-- the
first row are just our unit vectors-- the second
row is the first vector that I'm taking the
cross product of. So this is going to be negative. I'm just going to rewrite
this right over here. So it's going to be negative
cosine of t sine of s. And then you have
cosine of t cosine of s. And then you have 0
which will hopefully simplify our calculations. And then you have the next
vector, that's the third row. Negative sine of t cosine of s. And 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 it's done, try to
then do it again on your own because this is
one of those things you really got to do yourself
to really have it sit in. Negative sine of t sine of s. And then finally, cosine of t. So let's take the
determinant now. So first we'll think
about our i component. Are i component, you
would essentially ignore this column-- the first
column in the first row-- and then take the determinant of
this submatrix right over here. So it's going to
be i-- so this is going to be equal to
i times something. I'll put the something
in parentheses there. Normally you see the
something in front of the i but you can swap them there. So let me just-- so it's
going to be i times something. Ignore this column, that row. This determinant is cosine of t
cosine of s times cosine of t, which is going to be
cosine squared of t. Let me write it a little neater. Cosine squared of t cosine of s. And then from that we need
to subtract 0 times this. But that's just going to be 0
so we're just left with that. Now we're 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 will have a negative--
you write a negative j. A negative coefficient, I
guess, in front of the j times something. And so you ignore
j's column, j's row. And so you have negative
cosine of t sine of s times cosine of t. Well, that's going
to be negative cosine squared of t times sine of s. Let me make sure I'm
doing that right. Ignore that and that. It's going to be
that times that. So yep. Negative cosine sine of
s minus 0 times that. And so that's just going to
be 0, so we can ignore it. And you have a negative
times a negative here so they can both
become positive. And then finally, you
have the k component. And once again, you can
go back to positive there. Positive, negative, positive
on the coefficients. That's just evaluating
a three by three matrix. And then you have plus
k times-- and now this might get a little
bit more involved because we won't have
the 0 to help us out. Ignore this row,
ignore this column, take the determinant
of this sub two by two. You have negative
cosine t sine s times negative sine of t sine s. Well, that's going to be--
the negatives cancel out. So it's going to
be cosine of t sine of t times sine squared of s. And then, from that
we're going to subtract the product of these two things. But that product is
going to be negative so you subtract a
negative-- that's the same thing as
adding a positive. So plus-- and you're going to
have cosine t sine t again-- plus cosine t-- let me scroll
to the right a little bit-- plus cosine t sine t again. And that's times
cosine squared of s. Now this is already
looking pretty hairy but it already looks like
a simplification there. And that's where the
colors are helpful. Actually, now I have
trouble doing math in anything other than
multiple pastel colors because this actually
makes it much easier to see some patterns. And so what we can
do over here is we can factor out
the cosine t sine t. And 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 going
to be equal to 1. So that was a significant
simplification. And so now we get
our cross product being equal to-- let
me just rewrite it all. 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 cosine sine of s
times our j unit vector plus-- all we have left because
this is just 1-- 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 ds dt. So let's figure out what
the magnitude of that is. And this is really
the home stretch so I'm really
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-- and I'm just going to
square each of these terms and then add them
up-- the square root of the sum of the squares
of each of those terms. So the square of
this is going to be-- cosine squared squared is
cosine to the fourth-- cosine to the fourth t cosine
squared of s plus cosine to the fourth t sine
squared s-- and I already see a pattern jumping
out-- plus cosine squared t sine squared t. Now, the first pattern
I see is this-- is just this first
part right over here. We can factor out a
cosine to the fourth t. Then we get something
like this happening again. So let's do that. So 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 1. 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 this term
and this term both have a cosine squared t in them. So let's factor those out. So this is going to be
equal to-- everything I'm doing is under
the radical sign. So this is going to be
equal to a cosine squared t times cosine squared t. And when you factor out
a cosine squared t here, you just have a plus
a sine squared t. And that's nice because that,
once again, simplified to 1. All of this is under
the radical sign. Maybe I'll keep drawing
the radical signs here to make it clear that
all this is still under the radical sign. And then this is really,
really useful for us because the square root
of cosine squared of t is just going to be cosine of t. So all of that business
actually finally simplified to something pretty
straightforward. So all of this is just going
to be equal to cosine of t. So going back to what
we wanted before, we want to rewrite what d sigma. It's just cosine t ds dt. so let me write that down. So d sigma-- and then we can
use this for the next part-- d sigma is equal to
cosine of t ds dt. And I'll see you
in the next part.