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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 3: The home stretch

Using a few trigonometric identities to finally calculate the value of the surface integral. Created by Sal Khan.

## Want to join the conversation?

- What exactly is the x^2 doing here?

Is it basically just modifying the value (such as density) for each point of the sphere's surface?(15 votes)- yes.. if it wasnt there we would get the surface area of the unit sphere (4pi) ..(14 votes)

- So if it weren't for the x^2 the end result should be 4pi/3 right? Why did we get the same answer with the x^2? Is that just dumb luck or what?(5 votes)
- Actually, you ought to get 4π, since that would just be the surface area. No 1/3. If you took the volume instead, the 1/3 would show up again.(21 votes)

- Umm, at5:20mark, I didn't know that we could seperate double integrals as each others product. Is it a correct thing? Because I didn't see any other example of it before.(3 votes)
- What was done is based on the
*multiplication by a constant rule*you learned in the integral calculus course:`∫cf(x)dx = c∫f(x)dx`

In the case of the video's expression, we are integrating with respect to`t`

that is`∫cos³t dt`

and with respect to`s`

that is`∫cos²s ds`

.

Can you see that when we integrate with respect to one of the variables, say`s`

, the expression in`t`

is constant, so we can take it out of the integrand for`s`

via the rule`∫cf(x)dx = c∫f(x)dx`

.

In this case, when integrating with respect to`s`

the constant term is`∫cos³t dt`

, that is to say,`c = ∫cos³t dt`

He explains the process leading up to the "separation" starting around2:30.

Try watching again with this in mind.(6 votes)

- i didn't get why x^2 was required? all the dσ s should give me the surface area.(2 votes)
- If density fxn is 1, then mass=area. Choose a changing density function that describes the density over the surface, and you can get the mass of the surface. (density*area=kg/m^2 *m^2)(6 votes)

- So the purpose of surface integrals in a scalar field is to calculate area density?(3 votes)
- Not exactly--the "area density" function is given (in the video's example, it is x^2). Calculating the surface integral would lead to the
**mass**of the surface.(2 votes)

- When you take square root of a number, don't you get a positive and negative square root? So we must get +Cost and - Cost right? Why are we ignoring -Cost, even if Cost is always positive for this particular range of t? (At0:11)(2 votes)
- There are certain situations when we are trying to solve equations or simplify expressions with square roots where we only use the positive square root because the negative square root provides an invalid answer. For instance, consider a square with side length of one. If we draw the diagonal then use Pythagorean theorem to compute the length of the hypotenuse, we have c^2= 1^2+1^2. simplifying we end up with c= sqrt(2). so, c=-2,2. But we are looking for the length of the hypotenuse, and generally, we talk about length having a positive value. therefore the -2 is considered invalid in this instance since a negative distance is not considered valid in this instance.

Note: I am specifically avoiding saying distance cannot be negative. There are times considering a negative root has a purpose. When the negative is an indication of direction this can be very useful.(3 votes)

- The answer we got is 4/3 π, but the formula for the area of the sphere is 4π r^2. Even if we substitute 1 into r, its 4π. Why is that so?(1 vote)
- That is because we are not trying to find the surface of the unit sphere, we are evaluating the integral of
`x²`

over the surface of the unit sphere.

This means that for each point in the surface of the unit sphere we gather the`x`

value of that point, square it, and sum the resulting value. The sum of all those values is`4π/3`

, it has nothing to do with the area of the sphere.(5 votes)

- Why does it make sense that we get the value of the volume?(1 vote)
- We don't in general get volumes this way. It is an artifact of the symmetry of the sphere and the particular choice of integrand he chose, x^2. If we were to set up a parametrization for the solid sphere, requiring a third radial parameter, it happens that, for this case in particular, at some point in evaluating the relevant triple integral it could simplify to a step similar to the double integral Sal reaches in the end. So, while it happened so in this case, it needn't be expected, or make sense. Hope that helps!(4 votes)

- at10:48, did Sal mean to integrate (1/2+1/2cos2s) ds?(2 votes)
- yeah. (1+cos2s)/2 is a trigonometric identity for cos^2 (s)(1 vote)

- I understand how Sal calculated d"sigma," but how is that different from the 2-D Jacobian? I calculated the Jacobian as: -sin(t)*cos(t) for Sal's example.(2 votes)
- correct me if i wrong if i am not mistaken the jacobian is used when changed of coordinates such as from polar to rectangular but here we did not change regions(1 vote)

## Video transcript

- [Instructor] So now
that we have been able to express our d sigma, I think we're pretty close to evaluating the integral itself, and one thing I do wanna point out that might have been nagging you from the end of the last video, at the end of the last video, I took the principal root
of cosine squared of t, and I simplified that to
just being cosine of t. And you might have said,
"Wait, wait, wait, wait, "What if cosine of t evaluated
to a negative number? "If I then square it,
it would be positive, "and then if I took the
principal root of that, "I would then get the
positive version of it. "I would essentially
get the absolute value "of the cosine of t." And the reason why we were
able to do this in particular in this video or in this problem is because we saw t takes on values between negative pi over two
and positive pi over two. And so cosine of anything that's either in the first or the fourth quadrant, so this is t right over here, the cosine will always be
positive for our purposes, for the sake of this surface integral. Cosine of t is always
going to be positive, and so in this case,
we don't have to write absolute value of cosine of t. We can just write cosine of t, and so hopefully that makes you satisfied. That was just based on how
we parameterized the t. Now, that out of the way, let's actually evaluate the integral. Our original integral, the original integral, just to remind us, was the double, or I should
say the surface integral, of x squared, d sigma. We already know what d sigma is. Now we just have to write x squared in terms of the parameters. Well, we know the parameterization of x. The x, in terms of the parameters
right over here is cosine. This is our parameterization: x is going to be equal
to cosine t, cosine of s. Let me write that down:
x of s and t is equal to, I already forgot, I have a
horrible memory, is equal to, we have to go back to the
original parameterization, not these partial derivatives. Cosine t, cosine s. Cosine t, cosine s. Cosine t, and then cosine s, and we're taking the
integral of that squared. So let's think about this a little bit. So let's just do this
part right over here. If we square x, we're going to get cosine squared t, cosine squared s. Cosine squared s, that's the x squared
part right over there, and then you have the d sigma, which is this stuff,
which is times cosine, lemme do that same green, don't wanna confuse you with
different shades of colors, times cosine of t, ds, dt. And now that we have this in terms of the parameters, the
differentials of parameters, this essentially becomes a double integral with respect to these two parameters, and so, and the good thing is that the boundary is
pretty straightforward with respect to s and t:
s takes on all values, s takes on all values
between zero and two pi, t takes on all values
between negative two pi, and, sorry, negative pi over two and positive pi over two. So first, the way I wrote over here, we're gonna integrate
with respect to s first, s goes between zero
and two pi, and then t, lemme write and make clear, this is s, and then t will go between
negative pi over two and positive pi over two. And so let's see if we can
simplify this a little bit. This is equal to the double integral over that same region,
over that same area, I guess we could call it,
over that same area of, well now we have this cosine squared of t and then we have another
cosine of t right over there, so lemme just right it this way, as cosine to the third of
t times cosine squared... Cosine squared of s, and then ds, lemme color code it a little bit, ds, and so this is the
integral for the ds part, and then dt. And this is when we
integrate with respect to s, notice these two, the
t parts and the s part, they're just multiplied by each other, so when we're taking the
integral with respect to s, this cosine cubed of t
really is just a constant, we can factor it out, and it
could look something like this, so let me rewrite it, this
could be the integral from, t goes from, I'll rewrite the boundaries, negative pi over two,
to positive pi over two, cosine cubed of t, I just factor that out, and then I'll write the s part, times the integral, s is going to go between zero and two pi,
and I'll write this in blue, cosine squared of s, ds, and
then you have dt out there, you have dt, I'm gonna do the dt in green, gimme that same green, dt. And now, this outer sum we can view it, you essentially view it as the product, well, of all of this
business right over here, this thing has no t's
involved in it whatsoever, so we can rewrite this, and I'll write all the stuff
involving the t's as green. So we can rewrite this as pi over two, from negative pi over two, to pi over two, cosine cubed of t dt times the integral, and I'm really just rearranging things, I guess you could kind of view this as the associative property, or I guess the commutative property. Well those things always confuse me, times the integral of zero to two pi of cosine squared of s, ds, and you didn't have to do it this way, you could've just evaluated it while it was kind of mixed like this, but this'll help us kind
of work through the, the trigonometry a little bit easier. Now to solve these two integrals, we just have to resort
to our trigonometry. Cosine squared of s, we can
rewrite that as 1/2 plus one, actually let me do that
in that same blue color so we don't get confused. That is the same thing as
1/2 plus 1/2 cosine of two s, and cosine cubed t, well
that's the same thing, let's see, we can factor
out a cosine of t, so let me rewrite, ah let's just do it, well let me just do it,
both at the same time, just get all the trigonometry out of way. This right over here can
be rewritten as cosine of t times cosine squared of t,
and the intuition here is, if we can get a product
of a sine doing something with a cosine, because
cosine is sine's derivative, that's kind of, y'know, u-substitution, you see a function and its derivative, you can just kind of
treat it as a variable, so that's what we're trying
to get to right over here. So cosine squared of t can be rewritten as one minus sine squared of t, so this is cosine of t times
one minus sine squared of t. And so we can rewrite this as cosine of t minus cosine of t, sine squared of t, and you might say "Wait, "this looked a lot simpler
than this down here." That is true, it looks simpler, but it's easier to take
the antiderivative of this, easier to take the
antiderivative of cosine of t, and even over here, you have derivative of sine of t, which is cosine of t, and so essentially you
can do u-substitution, which you probably can
do in your head now. So let's evaluate each of these integrals. So this one, let me rewrite them just so we don't get too confused, so we have the integral
from negative pi over two, to pi over two, of cosine of t minus cosine of t, sine squared of t, dt, times the integral from zero to two pi of 1/2 plus 1/2 cosine of 2s, d s. Now we are ready to take
some antiderivatives, the antiderivative of this
right over here is going to be, the antiderivative of cosine
t, well that's just sine t, and then right over here, the derivative of sine t is cosine of t. So we can just essentially, if you wanna do
u-substitution, you would say, u is equal to sine of t, d u
is equal to cosine of t, dt, and you do all of that, but the, what we probably cannot
do in your head is, okay, I have the sine
t's derivative there, so I can treat sine t just
like I would treat a t, or I would treat an x. So this is going to be, you
still have this negative sign, minus sine to the third of t over three, if this was just a t squared, the antiderivative would be
t to the third over three, but now since we have a derivative, we can kind of treat it the same way, which is essentially doing
u-substitution in our head. So that's that, and we're
going to evaluate it from negative pi over two, to pi over two. And so this is equal to, if
you evaluate it at pi over two, sine of pi over two is one. So it's one minus 1/3, so that's just 2/3, actually lemme not write it that way, I don't wanna confuse people, and then minus sine of
negative pi over two, well that's going to
be negative one minus, sine of negative pi over two is negative one to the third
power is negative one, so this is negative 1/3. And so this is going to
be equal to, this is 2/3, and this is negative one plus
1/3, which is negative 2/3, but then you have a negative out front, so this is plus 2/3 again. So this part at least evaluates to 4/3. This part, all this, is
really the home stretch, that all evaluates to 4/3. Now this part right over here, antiderivative of 1/2 is just 1/2 t, antiderivative of cosine of 2s, well, ideally you would
have a two out front here, out front, lemme write
and make this clear, so if I were to take the antiderivative of cosine of 2s, ideally you
would want a two out here, so you have the derivative of the 2s, so you could put a two out front, but then you would have
to put a 1/2 out front so that you're not
changing the value of it, and of course you would
have a ds right over here. I'm just taking a general antiderivative, but once you have it like this, then this just like taking the antiderivative of cosine of s. This becomes, antiderivative
of cosine is sine. So this will become sine of s. So this right over here is just sine of s, and then you have the
1/2 out front, times 1/2, but then of course,
and then you would have plus a constant if you were
taking an indefinite integral, but we're taking a definite one, so you don't have to
worry about the constant, so just the antiderivative
of cosine of 2s, just the antiderivative of
cosine of 2s is 1/2 sine of s. And so you have this constant out front, 1/2 times 1/2 is 1/4. So it's going to be plus 1/4 sine of 2s. That's the antiderivative, and now we're going to evaluate
it from zero to two pi, and in either situation, this thing's going to evaluate to zero. Sine of zero is zero,
sine of four pi is zero, and so you're gonna have 1/2
times two pi, which is just pi, plus zero, 'cause sine of four pi is zero, minus 1/2 times zero, zero;
1/4 times sine of zero, zero, so you're essentially
just gonna end up with pi, so this whole thing right
over here evaluates to pi. And so we're done! You take the product of these
two things, 4/3 times pi, our entire surface integral
evaluates to: 4/3 pi. So this is equal to 4/3 pi, which is neat! If you have a sphere of radius one, its surface area, or actually
no, I shouldn't even go that, because, let me be very careful. I shouldn't make that statement, because this wasn't just
with respect to one. But, we have at least
evaluated the surface integral, and we deserve, I think,
a bit of a rest now.