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# Example of calculating a surface integral part 1

Video transcript

We saw several videos ago that
we can parameterize a torus or a doughnut shape as a position
vector-valued function of two parameters. And this is the
outcome that we had. I think I did it over
several videos because it was a bit hairy. And I'll write our position
vector-valued function first. So we have r as a function of
our two parameters s and t. And then I'll review a little
bit of what all the terms-- what the s, the t, and the
a's and the b's represent. But it's equal to b
plus a cosine of s. And once again, we saw
this several videos ago. So you might want to watch the
videos on parameterizing surfaces with two parameters to
figure out how we got here. Times the sine of t. I'm going to put the s
terms and the t terms in different colors. Times our i unit vector. I'll put the vectors,
or the unit vectors in this orange color. Plus-- I'll do it in
the same yellow. Plus b plus a cosine of s times
cosine of t times the j unit vector-- the unit factor
in the y direction. Plus a sine of s times the
k unit vector or the unit vector in the z direction. And in order to generate the
torus or the doughnut shape, this is true for our
parameters-- so we don't wrap multiple times around the
torus-- for s being between 0 and 2 pi and for t being
between 0 and 2 pi. And just as a bit of review,
we're all of this came from-- and I'm going to have to do
what my plan is for this video over several videos. But let's review where
all of this came from. Let me draw a doughnut. My best effort at a
doughnut right here. That looks like a
doughnut or a torus. And you can imagine a torus, or
this doughnut shape is kind of the product of two circles. You have the circle that's
kind of the cross section of the doughnut at any point. You could take it there. You could take it over there. And then you have the circle
that kind of wraps around all of these other circles or these
other circles wrap around it. And so, when we derived this
formula up here or this parameterization, a was the
radius of these cross sectional circles. That's a. That's what these a terms were. And b was the distance from the
center of our torus out to the center of these cross sections. So this was b. So you can imagine that b is
kind of the radius of the big circle up to the midpoint of
the, I guess, cross section. And a is the radius of the
cross sectional circles. And when we parameterized it,
the parameter s was essentially telling us how far-- s was
telling us how far or where are we wrapping around this circle. So it's an angle from 0
to 2 pi to say where we are on that circle. And t tells us how much
we've rotated around the larger circle. So if you think about it, you
can specify any point on this doughnut or on this surface or
on this torus by telling you an s or a t. And so that's why we picked
that as the parameterization. Now, the whole reason why I'm
even revisiting this stuff that we saw several videos ago is
we're going to actually use it to compute an actual
surface integral. And the surface integral we're
going to compute will tell us the surface area of this torus. So this surface right here is
sigma, like that and it's being represented by this position
vector-valued function. That is parameterized by these
two parameters right there. And if we wanted to figure out
the surface area, if we just kind of set it as the surface
integral we saw in, I think, the last video at least the
last vector calculus video I did that this is a surface
integral over the surface. Here this capital Sigma does
not represent a sum, it represents a surface of a bunch
of the little d sigmas-- a bunch of the little
chunks of the surface. And just as a review, you
can imagine each d sigma is a little patch of the
surface right there. That is a d sigma. It's a double integral here
because we want to add up all of the d sigmas
in 2 directions. You can imagine one kind of
rotating this way around the torus and then the other
direction is going in the other direction around the torus. So that's why it's
a double integral. And this is just going to give
you the surface area, which is the whole point of this video
and probably the next one or two videos. But if you wanted to also
multiply these sigmas times some other value-- there's some
scalar field that this is in that you cared about-- you
could put that other value right there. But here we're just
multiplying it by 1. And we saw in the last video
that it's a way of expressing an idea, but you really can't
do much computation with this. But a way that you can express
this so that you can actually take the integral, you say this
is the same thing-- and we saw this in the last
several videos. This is the same thing as the
double integral over the region over which our
parameters are defined. So it's this region over here
where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we
could just write a 1 if we like; it doesn't change much. Times-- and this is
what we learned. Times the magnitude of the
partial derivative of r with respect to s. The magnitude of that crossed
with the partial derivative of r with respect to t ds. You could take it in
either order, but ds dt. So we saw this in
the last video. What we're going to do now
is actually compute this. That's the whole
point of this video. We're going to take the cross
product of these two vectors. So let's figure out
these vectors. Then in the next video
we're going to take the cross product. And then the video after that
we'll actually evaluate this double integral. And you're going to see it's a
pretty hairy problem and this is probably the reason that
very few people ever see an actual surface integral
get computed. But let's do it anyway. So the partial derivative of
r with respect to s-- so this term right here. We'll do the cross product
in the next video. This term is what? We just want to hold t
constant and took the partial with respect to just s. So this up here, if you
distributed the sine of t times b-- that's just going to be a
constant in terms of s, so we can ignore that. Then you have sine of t
times this over here. So sine of t and
a is a constant. And you take the derivative
of cosine of s. That's negative sine of s. So the derivative of this with
respect to s or the partial with respect to s is going to
be minus a-- I'll write in green the sine of t, so you
know that's where it came from. Sine of t and then sine of s. The derivative of this
is negative sine of s. That's where that
negative came from. And then I'll write the
sine of s right there. Times the unit vector i. That's the partial of just this
x term with respect to s. And then we'll do the
same thing with the y term or the j term. So plus-- same logic-- b times
cosine of t with respect to s. When you take the
partial [INAUDIBLE] becomes 0 so you're left with
a-- well, it's going to be a minus a again because when you
take the derivative of the cosine of s it's going to
be negative sine of s. Let me do it. You're going to have a minus a. This cosine of t. Minus a cosine of t. That's the constant terms. Sine of s. Just taking partial
derivatives. Sine of s j. And then finally, we take
the derivative of this with respect to s. And that's pretty
straightforward. It's just going to
be a cosine of s. So plus a cosine of s k. Now hopefully you didn't
find this confusing. The negative sines because
the derivative of cosines are negative sines. So negative sine of s. That's why it's negative sine
of s times the constant. Negative sine of s times the
constant-- the constant cosine of t sine of t. So hopefully this makes some
sense just as a review of taking a partial derivative. Now let's do the same
thing with respect to t. And I'll do that in
a different color. So we're now going to take the
partial of r with respect to t. So the partial of r with
respect to t is equal to-- so now this whole term over here
is a constant, and so it's going to be that whole term
times the derivative of this with respect to t, which
is just cosine of t. So it's going to be b
plus a cosine of s times cosine of t i. And then, plus-- and it's
actually going to be a minus because when you take the
derivative of this with respect to t it's going
to be minus sine of t. So it's going to be negative
and then let me leave some space for this term right here. Negative sine of t. And you're going to have
this constant out there. That's a constant in t. b plus a cosine of s. That's just that
term right there. Derivative of cosine t is
negative sine of t times j. And then the partial of this
with respect to t-- this is just a constant in t. So the partial's going to be 0. So I'll write plus 0k. Let me do all my vectors
in that same color. Plus 0 times the unit vector k. So that gives us our
partial derivatives. Now we have to take their cross
product, then find the magnitude of the cross product,
and then evaluate this double integral. And I'll do that in the
next couple of videos.