Main content

### 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

© 2024 Khan AcademyTerms of usePrivacy PolicyCookie Notice

# Example of calculating a surface integral part 3

Example of calculating a surface integral part 3. Created by Sal Khan.

## Want to join the conversation?

- the surface area of the torus is the product of the circumferences of the two parameter circles and the volume is the area of the smaller times the circumference of the larger.A very satisfying result. Can this be generalized?(13 votes)
- If a bounded area is intersected by a continuous path through the centroid at a right angle and the area is moved along the parh so that it remains perpendicular, the volume generated is equal to the area timis the length of the path and its lateral area is the product of its circumference times the length of the path. I haven't proned this but I believe it is true.(5 votes)

- I don't know if I'm missing the point but I'm kind of playing devils advocate: Why couldn't you just snip the toroid along the cross section and then stretch it into a cylinder and then snip the cylinder along the length of the cylinder to make it a rectangle? Then you could use a single integral or solve it geometrically.(6 votes)
- If you snip the toroid along the cross section, it won't become a perfect cylinder. It will still have some intrinsic geometry, because the inside of the torus would be longer than the outside.(6 votes)

- If the torus was in some vector field so that the function was not 1 would there be additional steps involved, besides just multiplying by the function, in calculating the integral?(3 votes)
- If the function was a function of (x,y,z) you would have to change it to the parametrisation used and then just multiply, so no extra steps although the "hairiness" would most likely increase.(5 votes)

- It makes perfect sense when you read this:

"The first theorem states that the surface area A of a surface of revolution generated by rotating a plane curve C about an axis external to C and on the same plane is equal to the product of the arc length s of C and the distance d traveled by its geometric centroid."

https://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem#The_first_theorem

Not bad for someone who existed c. 290 AD. It's really really simple though, it's amazing.(4 votes) - How do we know that we should use "t" parameter the outside of the double integral, is there a specific rule?(3 votes)
- It's up to you. If you integrate with respect to s first, then you can do with t later or use it 'outside' of the double integral.(1 vote)

- I wanted to ask why at about2:45b + acoss didn't change. Why it didn't get raised to the 2nd power like the rest of the factors of the vectors. But then I thought maybe it did, but got cancelled out with the 1/2 power over the entire thing. I don't know if I'm right for sure though so could someone tell me for sure? Thanks in advance.(2 votes)
- You are correct, he just simplified it immediately!(2 votes)

- said surface area of the torus is 4*pi^2*a*b..

what if b=0 and a=r, it turns out 4*pi^2*r*0=0, but actually it is surface of sphere that is 4*pi*r^2..

I think something missing from the base..

but your teaching is very helpful and inspiring... thanks always.(1 vote)- This was a parameterization of a torus, not a sphere, they are topologically different. So why would the surface area of a sphere be the special case of a torus?(2 votes)

- is it possible to say that the surface area of the torus is the product of the circumference of the smaller, inner circle with the larger,outer circle?(1 vote)
- Indeed that is one way to view it. You can also view the volume of a torus as the product of the area of the smaller circle with the circumference of the outer circle.

Check out this page:

http://en.wikipedia.org/wiki/Pappus%27s_centroid_theorem(2 votes)

- What would happen if the inner integrand, integrated with respect to s, had terms involving both s and t, rather than just s? Would we treat any terms involving t as constants? (I'm not sure if 'partial integration' is a legitimate term, but that's how I think about the inverse of a partial derivative)(1 vote)
- Yes, the t would be treated as a constant.(2 votes)

- When i parameterized the torus a couple of videos back, I got a very similar answer of: r(s,t) = [(b+acos(s)) cos(t) ] i + [(b+acos(s)) sin(t) ] j + [asin(s)] k. The difference is the location of the cos(t) in the i component as opposed to the j component (and the same for the sin(t). This made sense due to the symmetry of the problem and had been mentioned elsewhere and been noted as equivalent.

However now when I get through all of this i come to a final answer of A = -4pi^2(ab).

Its very likely my arithmetic could be wrong somewhere or is this logical due to my differing parameterization. Thanks!(1 vote)

## Video transcript

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.