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Lecture Comments (6)

1 answer

Last reply by: Professor Hovasapian
Wed May 11, 2016 3:29 AM

Post by Tram T on May 7 at 05:44:10 PM

Dear prof, Hovasapian,

So vector F dot with vector N is the "component" of vector F along the direction of vector N.
1/It was said that because vector N is not unit vector so F.N is not really "component" of vector F on N, but it is still the component is it?

2/Is F.N with N not being unit vector make the component of F on N longer or shorter in length compared to F dot unit vector n?

3/ What is the differenc of 2 vectors dotted with each other in the case of one vector being unit vector and in the case of no unit vector?

Thank you!!!

1 answer

Last reply by: Professor Hovasapian
Sun May 26, 2013 11:48 PM

Post by Josh Winfield on May 26, 2013

In example 3 Raffi why when you take the composite function is it sin(y) not sin(y^2)?

1 answer

Last reply by: Josh Winfield
Sun May 26, 2013 9:31 PM

Post by Hector Herrera on May 11, 2013

Hello, first of all, thanks for the vid lectures. They are a big help when I have to miss lecture.

It is not a big deal but I think you may have forgotten the negative in front of cos(theta) for the solution to the cross product of dP/dt x dP/d(theta). 27:00

I believe it should be: note m = theta for simplicity

dP/dt x dP/dm = <sin(m), -cos(m), t>

Surface Integrals

Find a parametrization P(x,y) for x + y = z − 3xy.
  • We can find a parametrization P(x,y) for a multivariable function f by setting f(x,y) = z.
  • Solving for z yields z = 3xy + x + y.
Our parametrization becomes P(x,y) = (x,y,f(x,y)) = (x,y,3xy + x + y).
Find a parametrization P(y,z) for √{x − y2} = z − 1.
  • We can find a parametrization P(y,z) for a multivariable function f by setting f(y,z) = x.
  • Solving for x yields x = (z − 1)2 + y2.
Our parametrization becomes P(y,z) = (y,z,f(y,z)) = ( y,z,(z − 1)2 + y2 ).
Find the surface area of the solid described by f(x,y) = y2 − x2 for 0 ≤ x ≤ 1, 0 ≤ y ≤ 1. Do not integrate.
  • For an explicit function f, the surface area is found by computing SA = dydx where S is the region where the surface lays.
  • Now [df/dx] = − 2x and [df/dy] = 2y and so √{1 + ( [df/dx] )2 + ( [df/dy] )2} = √{1 + ( − 2x)2 + (2y)2} = √{4x2 + 4y2 + 1}
Hence SA = ∫0101 √{4x2 + 4y2 + 1} dydx
Find the surface area of the region formed by the intersection of the cylinder 1 = x2 + y2 and the plane x + y − z = 1. Do not integrate.
  • For an explicit function f, the surface area is found by computing SA = dydx where S is the region where the surface lays.
  • To find S, we note that when the cylinder and plane intersect, we obtain an elliptic surface bounded by the cylinder's boundaries, that is x ∈ [ − 1,1] and y ∈ [ − √{1 − x2} ,√{1 − x2} ]. See image.
  • The surface is described by x + y − z = 1 or z = f(x,y) = x + y − 1.
  • Now [df/dx] = 1 and [df/dy] = 1 and so √{1 + 12 + 12} = √3
Hence SA = ∫ − 11 − √{1 − x2} √{1 − x2} √3 dydx
Find the surface area of the region of the paraboloid z = x2 + y2 that is below the plane z = 4. Integrate using polar coorindates.
  • For an explicit function f, the surface area is found by computing SA = dydx where S is the region where the surface lays.
  • To find S, we note that when the paraboloind and plane intersect, we obtain a circle of radius 2, in polar coordinates r ∈ [0,2] and θ ∈ [0,2π].
  • The surface is described by z = f(x,y) = x2 + y2 and so [df/dx] = 2x and [df/dy] = 2y.
  • Then √{1 + ( [df/dx] )2 + ( [df/dy] )2} = √{1 + 4x2 + 4y2} . Since x = rcosθ and y = rsinθ, √{1 + 4x2 + 4y2} = √{1 + 4r2} .
Hence SA = ∫002 r√{1 + 4r2} drdθ = [1/12]∫0 ( 17√{17} − 1 ) dθ = [(π)/6]( 17√{17} − 1 )
Find the surface integral of f(x,y,z) = x + y over the surface given by P(t,u) = (t,u,t + u) from 0 ≤ t ≤ 1, 0 ≤ u ≤ 1.
  • We can compute the integral of a function f over a surface S (the surface integral) through dtdu where P is a parametrization of the surface S and D is the region where S lays.
  • Since x = t, y = u and z = t + u then f(P(u,t)) = t + u.
  • Now, [dP/dt] = (1,0,1) and [dP/du] = (0,1,1) so [dP/dt] ×[dP/du] = ( − 1, − 1,1).
  • Then || ( − 1, − 1,1) || = √3 and have dtdu = ∫0101 (t + u)( √3 ) dtdu = √3 ∫0101 (t + u) dtdu . Note that f(P(u,t))|| [dP/dt] ×[dP/du] || is just a product.
Integrating yields √3 ∫0101 (t + u) dtdu = √3 ∫01 ( [1/2] + u ) du = √3
Find the surface integral of f(x,y,z) = x2 − y2 + z over the surface given by z = y − x from − 1 ≤ x ≤ 1, − 1 ≤ y ≤ 1.
  • We can compute the integral of a function f over a surface S (the surface integral) through dtdu where P is a parametrization of the surface S and D is the region where S lays.
  • We can parametrize z = f(x,y) = y − x through P(t,u) = (x = t,y = u,f(x,y) = z) that is P(t,u) = (t,u,u − t).
  • Then f(P(u,t)) = t2 + u2 + (u − t) = t2 + u2 + u − t.
  • Now, [dP/dt] = (1,0, − 1) and [dP/du] = (0,1,1) so [dP/dt] ×[dP/du] = (1, − 1,1).
  • Then || (1, − 1,1) || = √3 and have dtdu = ∫ − 11 − 11 (t2 + u2 + u − t)( √3 ) dtdu = √3 ∫ − 11 − 11 (t2 + u2 + u − t) dtdu . Note that f(P(u,t))|| [dP/dt] ×[dP/du] || is just a product.
Integrating yields √3 ∫ − 11 − 11 (t2 + u2 + u − t) dtdu = √3 ∫ − 11 [2/3] du = [(4√3 )/3]
Find the surface integral of f(x,y,z) = xy − z over the surface given by √z + 5y = x from 0 ≤ x ≤ 1, − 1 ≤ y ≤ 0. Do not integrate.
  • We can compute the integral of a function f over a surface S (the surface integral) through dtdu where P is a parametrization of the surface S and D is the region where S lays.
  • We can parametrize √z + y = x by solving for z = f(x,y), so z = (x − y)2.
  • Then for P(t,u) = (x = t,y = u,f(x,y) = z) we obtain P(t,u) = (t,u,(t − u)2).
  • Then f(P(u,t)) = tu − (t − u)2 = − t2 + 3tu − u2.
  • Now, [dP/dt] = (1,0,2t − 2u) and [dP/du] = (0,1, − 2t + 2u) so [dP/dt] ×[dP/du] = ( − 2t + 4u,2t − 2u,1).
Then || ( − 2t + 4u,2t − 2u,1) || = √{8t2 − 24tu + 20u2 + 1} and have dtdu = ∫01 − 10 ( − t2 + 3tu − u2)( √{8t2 − 24tu + 20u2 + 1} ) dtdu.
Find the surface integral of the vector field F = (x − 1,y + 1,z) over the surface given by P(t,u) = (t,u,t + u) from 0 ≤ t ≤ 1, 0 ≤ u ≤ 1.
  • We can compute the integral of a vector field F over a surface S (the surface integral) through dtdu where P is a parametrization of the surface S and D is the region where S lays.
  • Since x = t, y = u and z = t + u then F(P(u,t)) = (t − 1,u + 1,t + u).
  • Now, [dP/dt] = (1,0,1) and [dP/du] = (0,1,1) so [dP/dt] ×[dP/du] = ( − 1, − 1,1).
  • Then dtdu = ∫0101 (t − 1,u + 1,t + u) ×( − 1, − 1,1) dtdu = ∫0101 0 dtdu . Note that f(P(u,t)) ×( [dP/dt] ×[dP/du] ) is a scalar product.
Integrating yields ∫0101 0 dtdu = 0
Find the surface integral of integral of the vector field F = (x2 + 1,y2 − 1,z) over the surface given by z = y − x from − 1 ≤ x ≤ 1, − 1 ≤ y ≤ 1.
  • We can compute the integral of a vector field F over a surface S (the surface integral) through dtdu where P is a parametrization of the surface S and D is the region where S lays.
  • We can parametrize z = f(x,y) = y − x through P(t,u) = (x = t,y = u,f(x,y) = z) that is P(t,u) = (t,u,u − t).
  • Then F(P(u,t)) = (t2 + 1,u2 − 1,u − t).
  • Now, [dP/dt] = (1,0, − 1) and [dP/du] = (0,1,1) so [dP/dt] ×[dP/du] = (1, − 1,1).
  • Then dtdu = ∫ − 11 − 11 (t2 + 1,u2 − 1,u − t) × (1, − 1,1)dtdu = ∫ − 11 − 11 (t2 − u2 + u − t + 2) dtdu . Note that f(P(u,t)) ×( [dP/dt] ×[dP/du] ) is a scalar product.
Integrating yields ∫ − 11 − 11 (t2 − u2 + u − t + 2) dtdu = ∫ − 11 ( [2/3] − 2u2 + 2u + 4 ) du = 8

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Surface Integrals

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Surface Integrals 0:25
    • Introduction to Surface Integrals
    • General Integral for Surface Are of Any Parameterization
    • Integral of a Function Over a Surface
    • Example 1
    • Integral of a Vector Field Over a Surface
    • Example 2
    • Side Note: Be Very Careful
    • Example 3
    • Summary

Transcription: Surface Integrals

Hello and welcome back to educator.com and multivariable calculus. 0000

Today's topic is going to be surface integrals. Last lesson we introduced the notion of surface area, and now we are just going to add onto that.0004

However, before we discuss surface integrals, I just wanted to say one final word regarding surface area.0012

Especially with respect to the last problem that we did. Let me go ahead and talk about that for just a minute or two and then we will jump into surface integrals.0018

Okay, so, we had this... so the last example in the previous lesson... we had this function and it was given to us as z = some function of x and y. It was given explicitly as a function of two variables. This surface in 3-space.0027

Now, if you have that, if you are given z = f(x,y), of course you know that the parameterization that... for that... you can form the parameterization of x,y and that is going to be x, y, and of course the function of x, y.0045

That is the basic parameterization when you are given this thing explicitly. Well, if you were to just take this parameterization and run a surface area problem, run through it generically, without actually having a specific function in mind.0078

If you just took the partial with respect to x, the partial with respect to y. If you took the cross product, you are going to end up with this.0095

You are going to end up with an area equal to sqrt(1+df/dx2 + df/dy2) dy dx.0103

If you actually run this problem generically with this parameterization, this is the surface area integral that you get. This is the integral that you actually learn in single variable calculus, and it is also going to be one of the formulas that shows up in your book for multi variable calculus.0130

This is a very specific formula for a specific way that a function is given to you. If your function is given to you explicitly as a function of x and y, you can go ahead and use this formula directly.0149

However, I think it is better not to have to learn specific formulas for specific situations. What is the best is the learn it the way that we did learn it, the way that we defined it. Given a particular parameterization, you want to be able to form that integral.0160

Therefore, you will always end up with this if in the particular case that you are dealing with, you are given a function explicitly as opposed to given a parameterization.0173

So, it is better to have the general integral for surface area, that way you will always be right. You will always end up where you need to end up... surface area for any parameterization, and that is what is important... parameterization, because it is the parameterization that actually controls how the integral looks.0185

Oh, again, the area, as a recap, is equal to the double integral over s, and we use Σ for the area -- that is the symbol -- and the actual mathematics is the norm of dp dt cross dp du dt du. Where this ds is this thing.0223

So, this is the symbolic representation of it. This is the actual mathematics, that says take dp dt, the parameterization, take dp du, take the cross product of those things and then you are going to get a vector... take the norm of that vector which is a number and then integrate that number over the surface and that will give you the surface area. That is it. Nothing more.0257

You will always... and then for a particular situation... if you do this, you will end up with this when you are given that and when you have a parameterization like that.0277

Okay. So, let us go ahead and start our surface integrals. Let us start on the next page. So, let us see what we have.0287

We previously defined... well, I do not need to write all of this out... we previously defined two types of line integrals.0297

We did an integral -- excuse me -- of a function over a curve, so we did the integral of a function over a curve and we also did the integral of the vector field over a curve... a path, a line... something like that.0309

So, now we are going to do the same for surfaces. So, now we are going to just move one up. A line is a one dimensional object, a surface is just a two dimensional object. We are just moving up one dimension.0348

Now, we just do the same for surfaces. We are going to define a... the integral of a function over a given surface... and we are going to define the integral of a vector field over a given surface and just see that they are perfectly analogous in terms of actual notation.0362

okay. So, now let us go ahead and start with some basic presumptions.0386

Let s be a surface parameterized by some parameterization p, t, and u, and of course t and u are generic variables. 0392

Now, let f(x,y,z) be a function defined on s. So, real quickly, in case you need to see what a picture looks like, so you have got some surface in 3-space, and it is going to be parameterized by some parameterization, and since it is in 3-space, the surface itself, the points on that surface are in 3-space, therefore we can form if we have some function of three variables, x,y,z, it can take that function on that point, on that surface.0413

That is all that is happening. Exactly, exactly the same for line integrals. Okay. So, now let us go ahead and do our definition. Then, the integral, so this one is the integral of a function defined over a surface, f(x,y,z), it is a function of 3 variables.0458

It takes a vector, 3 variables, and it spits out a number.0480

The integral of f, small f, over s is, well, symbolically we have f dΣ, and mathematically it is equal to f(p(t,u)).0484

In other words we have formed the composite function of f of the parameterization p, just like in a line integral we formed f(c(t)).0509

× dp dt cross dp du, dt du. Notice this is exactly the same as the surface area except you know you have this extra component f. That is the symbolic -- and really what you are doing -- is you are taking f(p), f of the parameterization.0524

The composite function of the function of the parameterization and you are multiplying it by the norm of the cross product of the normal vector.0546

Then, this is a number, and of course you are integrating this number over 1 variable, over the other variable, in other words over the surface. That is all that is happening here.0555

When f = 1, when f is identically 1, well, we simply recover the integral for surface area integral. That is it0568

We defined surface area, now we are going to stick a function in there. After this, we are going to stick a vector field in there. Let us do an example.0589

So, example 1. Well, let us let f(x,y,z) = xsin(z). Okay? and s be the surface z = x3 + y, and we are going to specifically say x > or = -1, and less than or = 3, and y is > or = 0, and < or = 3.0597

We are going to specify the x and y, we are going to specify the domain. This is our function, it is given explicitly in terms of a function of x and y, so we know how to form the parameterization when we need to in a minute.0643

We have the function of 3 variables, the y does not show up here but it still is a function of 3 variables. It is a surface, y is arbitrary, so now let us go ahead and solve for the integral of this function over this surface.0657

Okay, we are going to use this thing right here.0670

So, let us go ahead and I think I am going to actually write down one more time up here in the corner... that is going to be the integral of f(p), so I am going to write it this way, f(p) × norm(n), and of course n is the cross product of the partial of p with respect to t, that big long cross product thing.0675

So dt du, just so I have it as a reference. Okay.0707

Well, so let us go ahead and list our parameterization. These are actually pretty easy in the sense that you just have to run through the process.0713

Again, running through the process is easy enough to do, it is just a question of keeping track of everything that is going on.0721

There are going to be a lot of symbols, a lot of numbers floating around, just be careful with what you do.0730

So, our parameterization, it is going to be in terms of x and y, so we have x, we have y, and then we have z, which is the function x3 + y. So that is that.0734

Now, let us go ahead and calculate dp dx, so dp dx is going to be (1,0,3x2) and dp dy is going to be (0,1,1)... there we go.0749

Now we need to form the cross product of these two, so dp dx cross dp dy, and we are going to use our symbolic determinant i,j,k... (1,0,3x2) -- excuse me -- (0,1,1). We are going to form the determinant of that. 0780

When I actually expand -- I am not going to go through the expansion, but I hope you are going to confirm for me, because I could very well have made some arithmetic errors -- 3x2,-1, and 1. This is n, right?0804

This is the normal vector the surface at a given point. That is what the cross product of the two partials is. That is n.0823

Now, what we want to do. We need the norm(n), so let us go ahead and take the norm of this, so the norm(n), well, it is just going to equal this squared + this squared + this squared all under the radical.0830

So it is going to be 9x4 + 1 + 1 under the radical, which equals 9x4 + 2 under the radical, okay?0846

Let me go ahead and put a little circle around that... we are going to need that, that is this one.0857

Okay. Now, we also need f(t), okay, well f is -- let me go back to black -- so f, that was equal to x × sin(z), well the parameterization p of x,y, we said was xy and the function x3 + y.0863

So f(p), f of the parameterization... well f is just x, z is x3 + y, so it is x × sin(x3 + y). There you go.0893

Now we have this... and now let us go ahead and put it all together. So, the integral is equal to, well, let us see, let us go ahead and put... so we are going to do dy dx, so let us go ahead and do x as the outer integral, and we said that x is going to be > -1, < or = 3, so it is going to be from -1 to 3.0909

The y value is going to go from 0 to 3, right? We said y was going to be > 0 and < 3.0943

f(p), well we have x × sin(x3 + y), that is that, and then if we multiply that by the norm, which is sqrt(9x4 + 2), and then of course we decided to do y here, then x, so it is going to be dy dx. There you go. You are just literally plugging it in.0955

Find this, you find this, you put them next to each other, you put it into your math software, or if you want to do it by hand that is fine.0983

You do not want to do this by hand. When you solve this integral, you get the number 3.8527, to five decimal places, or to 4 decimal places... 5 digits. There you go. That is it.0990

The important thing here, as always, is being able to construct the integral. Not being able to solve the integral. You need to be able to construct the integral. Run through it piece by piece.1005

You have the parameterization, find the normal vector, find the norm of the normal vector, and then form f(p), and then just put it into the integral and solve the integral, and be able to extract the upper and lower limits of integration.1014

The surface over which you are actually integrating. The domain of the parameterization. The t and the u. In this case the x and the y. Okay.1032

So, now that we have seen the definition of a function, the integral of a function over a surface, let us go ahead and do the integral of a vector field over a surface.1041

For this one, let us see, we have the integral of a vector field over a surface.1052

So, let me go ahead and draw a surface real quickly. So, we have something like this, like that, like that, like that. So this is some surface and of course there is some point p here, and there is some -- you know -- again, this is a vector field, now you have a surface in 3-space.1072

Now the vector field is a 3-vector field. In other words, for every point, this particular vector field is going to be a function from R3 to R3.1090

In other words, for every point, in 3-space, there is some vector that has 3 components emanating from that point. That is what this is. This is f right here.1103

Well, a surface is in 3-space, so all of the points on that surface are subject to the vector field. You can form the composite function f of the parameterization. That is what is happening here.1115

So, we know that we can form the normal vector, right? This is the normal vector and we said that the normal vector is equal to dp dt cross dp du, and we have a function, a vector field which has composite functions f1, f2, f3.1128

Well, again, just like the vector field over a curve, you can think of what we are going to be doing is we are going to be integrating... well, let me just go ahead and write the definition and then we will tell you what this is.1158

So, the definition of the integral of a vector field over a surface is the following.1175

We have the integral over s -- f -- · ds, this is the symbol for it, here is the mathematics.1182

The integral over s... f(p), it is the composite function dotted with n dt du.1193

This right here, that is the quote component of f along n. It is the component of f in the direction of n. "Component" in quotes, because n is not usually a unit vector.1207

I can go ahead and make it a unit vector by dividing by its norm, but that is not it. This is how you want to think of it.1233

When you are putting this integral together, what you want to do is you want to take your parameterization, you want to form the composite f(p), that is going to be a vector because this is a vector field.1241

f(p) is going to be a vector, well n is also going to be a vector. You want to take the dot product of those two vectors, and that is going to be a number.1252

After you have a number, you can integrate. You can only add numbers. You cannot -- you can add vectors, but you are not really adding vectors, what you are doing is you are adding components.1257

Here, when we integrate something, a function, a vector field, whatever it is, we are adding numbers. That is why this dot product shows up.1266

We need to add a number. That is it. This is the definition of the integral of a vector field over a surface. Let me give you the expanded version.1273

s f(p) dotted with dp dt cross dp du dt du. Notice, in this case, the norm does not show up for the integral of a vector field..1285

The reason the norm does not show up is because this is a vector, therefore I need it to dot it with the vector itself, not its norm and this dot product is what gives me the number.1304

This is the symbol f · ds. This ds, that is this part right here.1314

It is different than the Σ. The dΣ, that had the norm of this. The ds, the capital s, that has the actual cross product, the vector itself, not the norm of the vector. Okay.1323

Let us go ahead and do an example. Example 2.1337

Okay. We will let f, our vector field of x, y, z... we will let it equal to xy, xz, and yz.1346

We will let s be the surface parameterized as follows: p(t) Θ = tcos(Θ), tsin(Θ), Θ.1362

Where t is going to run from 0 to 1, and Θ is going to run from 0 to 3pi.1401

Okay. What we want you to do is to find the integral of f over this surface. Great. Should be no problem at all.1416

So, let us go ahead and -- I actually should of started this on another page, it is not a problem -- I will just rewrite everything that I need to rewrite.1431

Let me go ahead and rewrite the formula itself. Always a good idea. So, I need to go -- let me do it over here -- so the integral of f(p) dotted with n dt du. Okay.1439

Well, let us go ahead and form f(p). Actually, you know what, I am going to go ahead and write f and p over again.1462

f = xy, xz, and yz... and my parameterization of t and Θ is going to be tcos(Θ), tsin(Θ), and Θ.1470

Therefore my f(p), so I am just going to put... and this is my x, this is my y, this is my z... let me go ahead and go to blue ink here.1490

x,y,z... wherever I see x I put this, wherever I see y I put this, wherever I see z I put this.1504

So, I end up with the following. I end up with t2sin(Θ)cos(Θ), Θtcos(Θ), and Θtsin(Θ).1510

I certainly hope that you are all confirming this for me. Now, I go ahead and I take dp dt.1533

Again, just running through it. I need this, I need this, I just have to go through this step to get what I need.1540

So, I take the partial of the parameterization with respect to t, and I end up with cos(Θ), sin(Θ), and 0.1549

I am going to take the partial of the parameterization with respect to Θ, and that is going to be -tsin(Θ), it is going to be tcos(Θ), if I am not mistaken, and it should be 1.1559

Okay. Now of course I have to form the cross product of these 2, because that it going to give me n.1574

So, dp dt cross dp d(Θ), that is going to be my symbolic determinant.1581

So I have got cos(Θ), sin(Θ), 0, -tsin(Θ), tcos(Θ), and 1.1592

When I form this determinant, I end up with the following: I end up with sin(Θ), cos(Θ), and t. Good.1605

Okay. So this is going to equal n. This is my n. So now I have n and I have f(p), now I am ready to go ahead and construct my integral.1620

So, my integral is equal to, and of course they gave me the range. They said t goes from 0 to 1, and the Θ goes from 0 to 3pi, so I go ahead and I have my upper and lower limits of integration.1630

Let me go ahead and do t first, that is going to go from 0 to 1, and let me go ahead and do Θ next, that is going to go from 0 to 3pi.1646

Now, I have my f(p), which is right here, and I have my n, which is right here.1659

I need to form the dot product of f(p) · n, so I have to form... I could do this × this + this × this + this × this. That is what I am doing.1669

So, my integrand is going to look like this. t2sin2(Θ)cos(Θ) + this × this + Θtcos2(Θ) + Θt2sin(Θ).1682

I did Θ first, so this is going to be dΘ, and dt is going to be last. 1712

When I solve this I end up with 9/8 pi2 + pi. That is the integral of that particular vector field over this particular surface with that particular parameterization. That is it. It is that simple. 1719

Okay. Here is where we have to give a caveat. Now, let me go ahead and actually write this out... "Be very careful and vigilant."1739

Here is why. You have both these problems, surface integrals, particular vector fields with surface integrals, you have both dot products and cross products in the same problem.1756

You have both dot products and cross products in this integral.1774

Right? The definition of the integral... it contains a cross product, which is the definition of n, the normal vector, and the integral itself contains a dot product. You are dotting f(p) with n.1788

So, just keep track of everything very carefully. I actually made the mistake... I still make the mistake all the time. I did when I was doing this problem the first time.1800

I have to make sure that if I am taking a cross product and if I have two vectors and I need their cross product, to do their cross product, do not dot product them1810

If I need a dot product, take their dot product, do not take the cross product. The dot product gives you a scalar, a number. The cross product gives you a vector. Very important to keep track of those.1818

Yes, so keep track of everything very carefully. Keep track of things carefully, that is the take home lesson here.1831

Okay. Well, let us go ahead and do another example and we will finish off with this. So, let us see. Example 3.1842

Alright. Let f(x,y,z) = let us do something slightly more complicated here. We will do sin(x), we will do sin(y2), and we will do sin(z). Maybe it just looks complicated, we will see how it actually turns out.1855

So, that is our vector field. We want to find the integral of f over the surface z = x2 + y2, where z runs from 1 all the way to 3.1880

Let us go ahead and draw a picture of this first to see what this looks like. So, we know what this surface looks like, z = x2 + y2, that is just a paraboloid.1911

So, that is just the paraboloid that is going around the z axis, for z from 1 to 3, that means... so z1 is going to be right about here, and 3 is going to be right about here.1921

We are looking at that surface right about there. We are looking at this region right here. Basically, a piece of the paraboloid. That is it.1939

Okay. Well, let us go ahead and see what we can do. So, let us go ahead and see what this actually looks like, well, let us do the parameterization first.1949

So, we are given this function explicitly, so we know our parameterization is going to be a function of x and y, so we are going to have x, we are going to have y, we are going to have the function itself... x2 + y2. That is our parameterization. 1960

Well, let us see what this looks like in the xy plane. Well, z = 1, that is going to be a circle of radius 1, and here it equals 3, for z =3, let us see if I go ahead and project this down, and I go ahead and project this down, I am looking at this region right here, as far as the parameterization domain, that is the whole idea.1977

So, here my... this radius right here is equal to 1. This radius here is equal to sqrt(3), and the reason is because if this is sqrt(3), sqrt(3)2 is what gives me the 3.2006

They said z < or = to 3, so when you project this down onto the x-y plane, the circle that forms the boundary, the shadow here when you drop this perpendicular, it is sqrt(3), not 3. z is 3. It goes from 1 to 3.2025

So, the parameterization runs from 1 to sqrt(3). Just wanted to make sure that you guys get that.2043

We have our parameterization, so dp dx, well that is going to equal (1,0,2x).2053

Well we go ahead and form dp dy, that is going to equal (0,1,2y).2065

Well, we go ahead and form n, which is of course n which is the cross product of dp dx cross dp dy. We are doing a cross product here.2074

Now the cross product involves the symbolic determinant, I am not going to go ahead and do it here, but I hope that you will actually confirm this.2084

What you are going to end up with here is (-2x,-2y,1), good.2090

Okay. Now, let us go ahead and do... so we have n, now we are going to have to form f(p), and so f(p), the composite function.2104

Well, the composite function, the parameterization is x, y, x2 +y2, so f(p) is going to equal the sin(x), the sin(y), and the sin(x2 + y2).2119

Of course we need to form the dot product now, f(p) dotted with n, so it is going to be this thing dotted with n, and we said that n was -- let me go ahead and write it again here -- n = (-2x,-2y,1).2144

Okay. So, this dotted with that, you are going to end up with -2x × sin(x), -2y × sin(y), and 1 × sin(z), which is x2 + y2, there you go. That is my integrand.2169

Let us do this in polar coordinates. The reason you want to do this in polar coordinates is instead of in terms of x and y is because we have a nice circular region in the x, y, plane. In our domain of parameterization the x y, we have a nice circular region.2205

In fact, we have a region between two circles, which is even more important that we do it in terms of polar coordinates. Right?2218

So when we look at this in the x y plane, this is y, this is x, we have this region right here, where r is going from 1 to sqrt(3), and Θ is going from 0 to 2pi, all the way around. This actually goes all the way around.2226

I just did a... this is our domain of parameterization -- the x and y.2247

Now, let us go ahead and see what we can do here. So, in polar form, well, this is our integrand wherever we see x we need to put rcos(Θ), wherever we see y we need to put rsin(Θ), right?2257

Because, that is the whole polar coordinate transformation. So, go ahead and write that down.2276

So x = rcos(Θ), y = rsin(Θ), so whenever we convert something in x, y to polar coordinates, we are going to have to... so let us go ahead and write f(p) · n2284

In polar form it looks like this, it is going to be -2rcos(Θ), sin(Θ) -- no, sorry, let me get this right, -2rcos(Θ) × sin(rcos(Θ)) - 2rsin(Θ) × sin(rsin(Θ) + sin(x2 + y2), when I put these in I am going to get r2 sin(r2).2308

Well, r runs from 1 to sqrt(3), from 1 to sqrt(3), that is the radius and Θ) is going to run from 0 to 2pi, so r is going to run from 1 to sqrt(3), and we are going to sweep this line out all the way around 2pi.2354

Therefore, you get... so, of course we have to remember that dy dx, whenever we are doing a change of coordinates R dr dΘ, right? The differential area element. That is the conversion, so we have to put that in.2379

So, here we go. The integral is equal to... let us see if we can fit it all in here... let us go ahead and do it over here.2399

The integral from 1 to sqrt(3), just going to make this a little bit bigger, the integral from 1 to sqrt(3), the integral from 0 to 2pi, of this whole thing right here.2413

Okay, you know what, I am just going to write it all in again... -2rcos(Θ)sin(rcos(Θ)) - 2rsin(Θ)sin(rsin(Θ)) + sin(r2) × r dΘ dr, because this is Θ, this is r -- oops, I definitely do not want these random lines here.2439

Up there, dΘ dr, because we are doing the Θ first and the dr last... we are going inside to out.2484

Of course when you solve this integral, when you put it into your math software, you are going to get -... = -4.6667. There you go.2491

Now, you probably noticed -- you know -- why is this a negative number? Well, the reason this is a negative number, if you caught it earlier on, when I form the dp dt cross dp du, remember we said that a surface has 2 normals.2510

It has one normal that actually sticks away from the surface, and it has another one that goes towards the surface. It just depends on how you do your cross product. I did dp dx cross dp dy, that ended up giving me this one.2524

Well, this one, if I did dp dy cross dp dx, I would have gotten this normal.2539

I would have gotten the positive one. The only difference between the two normals is that one is positive and one is negative.2547

Ultimately, it does not really matter. I mean, yes, the integral here is... you are going to get the same number, it is just going to be a positive 4.6667 instead of a negative 4.6667.2555

If you want to keep track of that and at the end just sort of drop this negative sign, if you can sort of realize what is going on, this is... I would not say it is a minor point.2567

I mean definitely the way that this integral is defined, the way the whole idea of a surface integral is defined or surface area is defined is with that normal vector pointing away from the surface... away... in other words, the surface curves away from that normal. 2577

Because the idea is you want to be able to form a tangent plane here, right? A tangent plane this way, not that way.2594

So... but again, the difference is really just one of a negative sign, so if you can keep track of that, that is what I mean when I say it is a minor point. It is not a minor point, we still want this one instead of this one.2601

Time and experience and doing some problems getting accustomed to what the surfaces look like, to what 3-dimensional space looks like, to how certain curves and surfaces behave... that will give you a sense of when you can recognize when something is a certain way.2614

Other than that, if you want, the mathematics is still correct... I would not worry so much about the sign of the number.2629

Okay. So, as a quick recap here, so let us recap what we have done.2638

Alright. The integral of f(d(Σ)) = the integral of f(p) × norm(n) dt du, this is the integral of a function over a surface and let me go ahead and write where it... n = dp dt cross dp du.2647

Again, dp du, they just differ by... it is a negative sign.2692

Now, the integral of f · ds... recognize these, these are the symbolic representations of these integrals.2699

The integral of a function over a surface, the integral of a vector field over a surface. Okay.2709

That is equal to the integral over s of f(p), this part is the same, f(p) F(p), small f, capital F, · n.2715

Not times the norm of n, dot n, dt du.2726

This happens to be the integral of a vector field a surface. These are the things that you want to know. This is the definition of the integral. You can solve every problem by using this right here.2736

You find n, you find dp dt, you find dp du, you have to work from a parameterization, you form the cross product, that is n.2755

In the case of a function, you take the norm of n. If it is not a function, if it is a vector field, you just keep it as regular n. You form f(p), you form F(p), here you multiply the two, here oyu take the dot product of the two, and then you integrate depending on what t and y are, what particular domain of parameterization you are dealing with.2766

That is it. Integral of a function over a surface, integral of a vector field over a surface. Entirely analogous to the same definitions for line integrals.2785

The integral of a function over a curve, the integral of a vector field over a curve. You have just moved up one dimension, and as you can imagine, you can keep moving up one dimension at a time. It is true in any number of dimensions.2797

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2808