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### 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 ex3 part 3

Parametrizing the top surface. Created by Sal Khan.

## Want to join the conversation?

- Now I know that dS = r*√(2)

But at12:04don't we use "r dr dø", instead of just dr dø.

If not, why?(11 votes)- dS represents an infinitesimally small area of the the surface, S. You're correct that in polar coordinates, we use r dr dθ to represent this area. However, when we calculate surface integrals, the area of the surface element is already encoded in the magnitude of the cross-product |d
**s**/ dr x d**s**/ dθ|, where**s**(r,θ) is the position vector from the origin to the surface. Each of these derivatives represent a side of dS. Recall that the magnitude of the cross-product of two vectors is the area of the parallelogram formed by them.(9 votes)

- I've been wondering if it is alright to just use x and y as variables, since z=1-x, and the boundaries of x is

-1<=x<=1, y is -1<=y<=1

is that correct? Why or why not?(7 votes)- well if you integrated from -1 to 1 for both x and y you'd get a square base, which is why he uses the cos and sin.(6 votes)

- When parameterizing x and y, why do we say that r is not fixed at 1 for Surface 3? Is it to account for the fact that the plane z=1-x is cutting into our cylinder at a slant? Wouldn't r be variable when leaving the xy-plane and entering the z-dimension only? While on the xy-plane, wouldn't r always equal 1 since it's a unit circle? Sorry, I'm finding the parametrization to be the hardest part of these.(6 votes)
- z is a function of x. so wherever x is defined, z is also defined. "Wouldn't r be variable when leaving the xy-plane and entering the z-dimension only?" So they go together. Don't know if this makes sense to you.(1 vote)

- Just wondering why is the r not considered 1? isn't the circle have a radius of 1? thanks!(3 votes)
- Keeping the r value at 1 would only let you parameterize the edge of the circle. We want to parameterize the entire area of the circle, and so we need to vary the radius from 0 to 1. This allows us to go to every point inside the circle.(2 votes)

- wasn't there a mistake in the determinant j component?

becuase the order was wrong on the j component ( i think)

etheirway it gives 0 for the j component too(1 vote) - What grade do we generally learn this? (•_•)(1 vote)
- Single variable calculus is generally taught in 12th grade, the last year of highschool, so multivariable calculus is usually taught in the first year of college. Then again, there are plenty of exceptions. I, for one, am in 11th grade. There are plenty of people learning this even younger than me ¯\
*_(ツ)_*/¯. Go at your own pace!(1 vote)

- What grade is this generaly learned in(1 vote)
- If you are pursuing engineering, they teach you this in the first semester and maybe in second year when you are learning pure mathematics.(4 votes)

- Are we trying to sum up the values of all the z coordinates on all three parts of the surface? On S1, this results in a zero since it's on the x-y plane?(1 vote)
- It looks like you are left-handed, yet the little pointer icon is a right hand. Just bugs me a bit!(0 votes)

## Video transcript

- [Instructor] We're
now in the home stretch. We just need to evaluate
this third surface integral, which is over this top part
of our little chopped cylinder right over here. So let's try to think
of a parameterization, and let me just copy and
paste this entire drawing, just so that I can use it down below as we parameterize it. So let me copy it, copy, and
then go all the way down here and let me paste it. Let me paste it, okay, that is
our shape again, our surface. And then we go to the layer
that I wanted to get on. And then let me start,
let's start evaluating it. So what we wanna care
about is the integral over surface three of z d s. And surface three here, we
see that the x and y values, and since you take on all
the possible x and y values, inside of the unit circle,
including the boundary, and then the z values
are gonna be a function of the x values. We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. Z is equal to one minus x. It's the subset that's kind
of above the unit circle in the xy-plane, or kind of the subset that intersects with our
cylinder and kind of chops it. So let's think about x and y's first. Let's think about it in
terms of polar coordinates because that's probably the
easiest way to think about it. I'm gonna redraw kind of a top view. So I'm gonna redraw a top
view, so that is my y-axis and this is my x-axis. X-axis. And the x's and y's can take on any value, they said you have to
fill the unit circle. So if you were to kind of
project this top surface down onto the xy-plane, you would
get this orange surface, that bottom surface,
which looked like this. It was essentially the
unit circle just like that. And I'm gonna redraw it a
little bit neater than that. I can do a better job, so
that would be, all right. So let me draw the unit
circle as neatly as I can. So there's my unit circle. And so we can have one parameter. We can have one parameter
that essentially says how far around the unit
circle we're going. Essentially that would be our angle and let's use theta, 'cause
that's, well just for fun. We haven't used theta as a parameter yet. That's theta, but if we had x's and y's as just a function of theta
and we had a fixed radius, that would essentially
just give us the points on the outside of the unit circle. But we need to be able
to have all of the xy's that're outside and
inside the unit circle. So we actually have to
have two parameters. We need to not only vary this angle, but we also need to vary the radius. So we would wanna trace out
the outside of that unit circle and maybe we'd wanna
shorten it a little bit. And then trace it out again. And then shorten it some more. And then trace it out again. And so you wanna actually
have a variable radius as well so that you can have how
far out you're going. You could call that r. So for example if r is fixed and you change your ranges of theta, then you would essentially
get all of those points right over there and you would
do that for all of the r's and from r zero all the way to r one and you would essentially fill
up the entire unit circle. Let's do that, so r is going
to go between zero and one. R is going to be between zero and one. And our theta, our theta's
gonna go all the way around. So our theta is going to
go between zero and two pi. This is, lemme write this down, I wrote zero instead of theta. Our theta is going to be
greater than or equal to zero, less than or equal to two pi. And now we're ready to parameterize it. X of r and theta is going to be equal to, so whatever r is, it's
going to be r cosine theta. So x is going to be r cosine theta. Y is going to be r sine theta. That's going to be the y value. R sine theta. And now z is essentially
just a function of x. Z is going to be equal to one minus x, but x is just r cosine theta. So there you have it. We have our parameterization of this surface right over here. The x's and y's can take all
the values on the unit circle, but then the z is up here
based on a function of, well based on really on a function of x. It's one minus x. So that would give us all
of the possible points right over here on the surface. You pick an x and y and then
the z is going to pop us right here someplace on that surface. And we can write it as a
position vector function. Instead of calling that
position vector function r, since we've already used r for radius, I will call it, I don't
know, let's call it, I'm just gonna pick a random letter here, let's call it p for
position vector function. And so p, our surface p
is, we can write it as, actually no why don't I
just call it surface three. So surface three, surface three, I'll do it in that same purple color too, just so we know we're talking about this, surface three as a
position vector function, as a function of theta and r,
maybe I'll write r and theta 'cause that's how I think of things. R and theta is going to be
equal to r cosine theta i plus r sine of theta j plus
one minus r cosine theta, need to get some real estate here, one minus r cosine of theta k. One minus r cosine of theta k. And now we are ready to start
doing all of the business of evaluating the actual surface integral. So the first thing we need to
do is take the cross product of this, the partial of
this with respect to r, and the partial of this
with respect to theta. Let's just get down to business. Let's take the cross product,
let's take the cross product and so we have our i unit vector,
we have our j unit vector, and we have our k unit vector. And this might get a little bit involved but we'll try our best
to just work through it. Give myself a little bit more space. And so the partial of
this with respect to r, so let's take the partial
of this with respect to r, I'll do it in blue, the partial
of this with respect to r, the partial of this with respect
to r is just cosine theta i so this is just cosine theta, I said I was gonna do it in
blue and that's not blue, so this is going to be
cosine, cosine theta. The partial of this with
respect to r is sine theta. Sine theta. And the partial of this
term right over here with respect to r is
negative cosine theta. Negative cosine theta. Now let's take the partial
with respect to theta. The partial of this with respect to theta is negative r sine of theta. Negative r sine of theta. The partial of this with respect
to theta is r cosine theta. R cosine theta. And the partial here, this is zero, and then this would be
negative r sine theta. Negative r, oh no let me be careful. This is going to be, you have a negative r so the derivative of cosine
theta with respect to theta is negative sine theta. So the negatives are gonna cancel out and so you're going to have
r, r sine, r sine theta. And now we can actually
evaluate that this determinant to figure out the cross product of the partial of this with respect to r and the partial of this
with respect to theta. I'm not writing it down, just to kind of save
some real estate here. And so we have, actually
maybe I will write it down just to be clear what we're doing. The partial of s three
with respect to r crossed with the partial of s three with respect to theta is equal to, now our i component is gonna be sine theta times r sine theta, so that's going to be
r sine squared theta, minus r cosine theta times
negative cosine theta. So that's plus r cosine squared theta. All of that times i, and you already, a simplification might be
popping out here at you, and then you have minus the j component. The j component is gonna be
cosine theta times r sine theta. So it's r cosine theta sine theta. And then we're gonna subtract from that. See the negative signs cancel out. So you're gonna subtract
r sine theta cosine theta, or r cosine theta sine theta. Well this is interesting because these are the
negatives of each other. R cosine theta sine theta minus
r cosine theta sine theta. This just evaluates to zero. So we have no j component. And then finally for our k component, we have cosine theta times r cosine theta. So we have r cosine squared theta minus r sine theta times sine theta. Or minus negative r sine
theta times sine theta. So this would give you a negative but we're gonna have to subtract until it gives you a positive. So plus r sine squared theta k. And so this simplifies quite nicely because this is going to be equal to, this term up here you can factor out an r. This is r times the sine squared theta plus cosine squared theta, which is just, that part simplifies to one
so that's just r times i. So this is equal to r times i. And we do the same thing over here. This also simplifies. This is actually the same thing. This also simplifies to r. So this whole thing simplifies,
lemme write it this way, this is also r sine squared
theta plus cosine squared theta. Also simplifies to r. So you have r times k, and
so if you want the magnitude of this business, so let me make it clear, so the magnitude, we'll
go back to the magenta, the magnitude of, I don't
feel like rewriting it all, just copy and paste it. Edit copy and paste. The magnitude of all of
this business is going to be equal to the square root of this squared, which is r squared, plus this
squared, which is r squared, which simplifies to,
this is two r squared. So you take the square
root of both of those, you get the square root of two times r. So this is equal to the
square root of two times, it would be the absolute value of r but we know that r only
takes on positive values. R only takes on positive values so it's the square root of two times r, which is very nice because
now we can evaluate d s. D s is going to be this
business times d r d theta. So let's do it. So our surface integral, the thing that we were dealing
with from the beginning, that thing right over there. Our surface integral, the
surface s three, of z d s is now equal to, so I'm gonna
use different colors for the different
variables of integration, so one on the outside, and
then I'll do one on the inside, I'll do the inside one in pink. Z is equal to one minus r cosine theta. So the z is equal to one
minus r cosine theta. It involves both so I'll
use a different color. Minus r cosine theta. And then I just have to integrate relative to both of the variables. One minus r cosine theta, oh
no I have to do the times d s. D s is this thing, it's this thing, times d theta d, d r. So let's see, let's write this out. Times square root of two r,
so let's write that down, so this times the square root of two r, times the square root of two, and we can write the square
root of two out front since it's a constant
so lemme just do that, simplifies things. Square root of two times
r, and now d theta d r. Or we can write d r d theta, either way. So let's do that. D, let's do d r d theta. So d r d theta, we could do it either way, it's gonna be about the
same level of complexity. And so first we're going to
integrate with respect to r, and lemme do the colors
the same way, actually. So d r, d theta. If I've got colors I
might as well use them. And actually I just realized
that I'm way out of time. So actually, let me continue
this in the next video. We'll set up the boundaries of integration and then just evaluate it.