Videos > Example of calculating a surface integral part 3
Example of calculating a surface integral part 3 Video Lecture
Text Transcript from Video
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.
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.
About this Video
Apr 22, 2026 Last updated
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