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 1 answerLast reply by: Professor HovasapianSat Aug 3, 2013 2:56 PMPost by michael Boocher on August 3, 2013Very profound, turns out all you can do is add things. (17:45) 1 answerLast reply by: Professor HovasapianSun May 26, 2013 4:30 PMPost by Josh Winfield on May 26, 2013Thanks raffi, example 2 should end with rd(theta)dr not other way around but the numerical answer is still correct

### Surface Area

Find the area of the parallelogram formed by a = (1,0,0) and b = (0,1,0).
• The area of the parallelogram formed by two vecotrs a and b is the norm of the cross product of the vectors, || a ×b ||.
• Hence a ×b = det(
 →i
 →j
 →k
 1
 0
 0
 0
 1
 0
) = 0i + 0j + 1k and so a ×b = (0,0,1).
Then || (0,0,1) || = 1 and the area of the parallelogram formed by a and b is 1.
Find the area of the parallelogram formed by a = ( − 4,0,2) and b = (2,0, − 4).
• The area of the parallelogram formed by two vecotrs a and b is the norm of the cross product of the vectors, || a ×b ||.
• Hence a ×b = det(
 →i
 →j
 →k
 − 4
 0
 2
 2
 0
 − 4
) = 0i + 12j + 0k and so a ×b = (0,12,0).
Then || (0,12,0) || = 2√3 and the area of the parallelogram formed by a and b is 2√3.
Find the area of the parallelogram formed by a = ( − 3,7,2) and b = (5,1, − 4).
• The area of the parallelogram formed by two vecotrs a and b is the norm of the cross product of the vectors, || a ×b ||.
• Hence a ×b = det(
 →i
 →j
 →k
 − 3
 7
 2
 5
 1
 − 4
) = − 30i − 2j − 38k and so a ×b = ( − 30, − 2, − 38).
Then || ( − 30, − 2, − 38) || = 2√{587} and the area of the parallelogram formed by a and b is 2√{587} .
Approximate the area of the parallelogram formed by a = ( [1/2],[1/2],[1/(√2 )] ) and b = ( − [1/(√3 )], − [1/(√3 )],[1/(√3 )] ).
• The area of the parallelogram formed by two vecotrs a and b is the norm of the cross product of the vectors, || a ×b ||.
• Hence a ×b = det(
 →i
 →j
 →k
 1 \mathord / phantom 1 22
 1 \mathord / phantom 1 22
 1 \mathord / phantom 1 √2 √2
 − 1 \mathord / phantom 1 √3 √3
 − 1 \mathord / phantom 1 √3 √3
 1 \mathord / phantom 1 √3 √3
) = 0i + 0j + 1k and so a ×b = ( [1/(2√3 )] + [1/(√6 )], − [1/(2√3 )] − [1/(√6 )],0 ).
Then || ( [1/(2√3 )] + [1/(√6 )], − [1/(2√3 )] − [1/(√6 )],0 ) || ≈ 0.9856 and the area of the parallelogram formed by a and b is approximately 0.99
Find the area of the triangle with vertices (1, − 1,2), (2,1,1) and ( − 1,2, − 2).
• We can find the area of the triangle by taking half the area of the parallelogram formed by the vertices.
• Let a = ( − 1,2, − 2) − (2,1,1) = ( − 3,1, − 3) and b = (1, − 1,2) − (2,1,1) = ( − 1, − 2,1) then a and b are vectors extending from the same vertex (2,1,1)
• To find the area of the parallelogram formed by a and b we compute || a ×b ||
• Now a ×b = ( − 3,1, − 3) ×( − 1, − 2,1) = det(
 →i
 →j
 →k
 − 3
 1
 − 3
 − 1
 − 2
 1
) = − 5i + 6j + 7k and so a ×b = ( − 5,6,7).
Then || ( − 5,6,7) || = √{110} and the area of the triangle with vertices (1, − 1,2), (2,1,1) and ( − 1,2, − 2) is [(√{110} )/2].
Find the surface area of a sphere of radius 5.
• The surface area of a sphere is found by computing SA = 4πR2 where R is the radius of the sphere.
Substitution yields SA = 4π(5)2 = 100π and so the surface area of a sphere of radius 5 is 100π.
Let C(t,u) = (3t2,4tu,2t). i) Find ∫ab || [dC/dt] || dt from 0 ≤ t ≤ 1. Do not integrate.
• We have [dC/dt] = (6t,4u,2) and so || (6t,4u,2) || = √{36t2 + 16u2 + 4} = 2√{9t2 + 4u2 + 1}
So ∫ab || [dC/dt] || dt = ∫01 2√{9t2 + 4u2 + 1} dt.
Let C(t,u) = (3t2,4tu,2t). ii) Find ∫ab || [dC/du] || du from 0 ≤ u ≤ 1. Do not integrate.
• We have [dC/du] = (0,4t,0) and so || (0,4t,0) || = √{02 + 16t2 + 02} = 4t
So ∫ab || [dC/du] || du = ∫01 4t du.
Let C(t,u) = (3t2,4tu).
iii) Find dtdu from 0 ≤ u ≤ 1, 0 ≤ t ≤ 1. Do not integrate.
• Since [dC/dt] = (6t,4u,2) and [dC/du] = (0,4t,0) then [dC/dt] ×[dC/du] = det(
 →i
 →j
 →k
 6t
 4u
 2
 0
 4t
 0
) = − 8ti − 0j + 24t2k and so [dC/dt] ×[dC/du] = ( − 8t,0,24t2)
• Then || ( − 8t,0,24t2) || = √{64t2 + 02 + 576t4} = 8t√{1 + 9t2} .
So dtdu = ∫0101 8t√{1 + 9t2} dtdu.
Find the surface area of the solid described by P(t,u) = (t,u,t2 + tu + u2) for − 2 ≤ t ≤ 2 and − 2 ≤ u ≤ 2. Do not integrate.
• The surface area of a solid described parametrically through P(t,u) is given by SA = dtdu where S is the region where the surface lies.
• Now [dP/dt] = (1,0,2t + u) and [dP/du] = (0,1,t + 2u) then [dP/dt] ×[dP/du] = det(
 →i
 →j
 →k
 1
 0
 2t + u
 0
 1
 t + 2u
) = − (2t + u)i − (t + 2u)j + k
• So [dP/dt] ×[dP/du] = ( − 2t − u, − t − 2u,1) and || ( − 2t − u, − t − 2u,1) || = √{5t2 + 8tu + 5u2} .
Hence SA = dtdu = ∫ − 22 − 22 √{5t2 + 8tu + 5u2} dtdu.
Find the surface area of the solid described by P(t,u) = (t + u,t − u,tu) for 0 ≤ t ≤ 1 and 0 ≤ u ≤ 1. Do not integrate.
• The surface area of a solid described parametrically through P(t,u) is given by SA = dtdu where S is the region where the surface lies.
• Now [dP/dt] = (1,1,u) and [dP/du] = (1, − 1,t) then [dP/dt] ×[dP/du] = det(
 →i
 →j
 →k
 1
 1
 u
 1
 − 1
 t
) = (t + u)i − (t − u)j − 2k
• So [dP/dt] ×[dP/du] = (t + u, − t + u, − 2) and || (t + u, − t + u, − 2) || = √{2t2 + 2u2 + 4} .
Hence SA = dtdu = ∫0101 √{2t2 + 2u2 + 4} dtdu.
Find the surface area of the solid described by P(f,θ) = (1 − sinf,1 + cosθ,1) for 0 ≤ f ≤ [p/2] and 0 ≤ θ ≤ p.
• The surface area of a solid described parametrically through P(f,θ) is given by SA = dfdθ where S is the region where the surface lies.
• Now [dP/df] = ( − cosf,0,0) and [dP/(dθ)] = (0, − sinθ,0) then [dP/df] ×[dP/(dθ)] = det(
 →i
 →j
 →k
 − cosf
 0
 0
 0
 − sinθ
 0
) = 0i − 0j + cosfsinθk
• So [dP/df] ×[dP/(dθ)] = (0,0,cosfsinθ) and || (0,0,cosfsinθ) || = cosfsinθ
• Hence SA = dfdθ = ∫0p0p \mathord/ phantom p 2 2 cosfsinθ dfdθ
Integrating yields ∫0p0p \mathord/ phantom p 2 2 cosfsinθ dfdθ = ∫0p sin θdθ = 2

*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.

### Surface Area

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 Area 0:27
• Introduction to Surface Area
• Given a Surface in 3-space and a Parameterization P
• Defining Surface Area
• Curve Length
• Example 1: Find the Are of a Sphere of Radius R
• Example 2: Find the Area of the Paraboloid z= x² + y² for 0 ≤ z ≤ 5
• Example 2: Writing the Answer in Polar Coordinates

### Transcription: Surface Area

Hello and welcome back to educator.com and multi-variable calculus.0000

So, today's lesson we are going to be talking about surface area. How to find the area of the surface given a particular parameterization.0004

From there we are going to jump off into subsequent lessons. We are going to be talking about integrals of functions and vector fields over given surfaces, but we want to start with the notion of surface areas.0013

So, let us just go ahead and jump in. Okay. So, in the last few lessons, we have introduced some concepts, so let us go ahead and recall real quickly.0026

We introduced this concept of a cross product, so given a vector a and a vector b, we can form a cross b and we used it to construct the normal vector to a surface.0035

Well, we also said that the magnitude of a cross b, well, was just the norm of a cross b.0052

I do not need to write that. The magnitude is equal to the norm of a cross b, so it's length, basically.0072

We also introduced another formula... it's the norm(a) × norm(b) × sin(θ), the sine of the angle between them.0081

I am not sure if we will have a lot of opportunity to use this particular formula. It may come up, it may not, depending on a particular example that we are doing, but mostly we are concerned with constructing the normal vector and then finding its norm.0094

That is what is important. Then, once we have the vector, finding a norm is really, really easy.0108

Okay. Now, we also said that a cross b, that this magnitude is the area... and we said it real quickly, but it will be really, really important here -- but we also said that this magnitude is the area of the parallelogram that is spanned by the vectors a and b.0112

So, if we are given a vector, given a couple of vectors like this, let us just say this is a and this is b, well, this... when we take those vectors and we close it out and form the parallelogram, well when we take a cross b, we end up with this vector which is perpendicular to both, right?0160

That is the a cross b, well this parallelogram that is actually spanned by a and b, the area of that parallelogram happens to be the norm of this vector. That is the thing.0181

So, there is an association between the vector itself and the area. So, there is some number we can attach to it and that number is the norm of the cross product.0195

Okay. So, now we can go ahead and jump into our notion of surface area.0207

So, now, given a surface in 3-space and the notions we are about to develop are complete analogs of the lower dimensional cases.0213

We work with curves, we define line integrals, we define the length of a curve, now we are dealing with a surface, we are going to define a surface area which is analogous to the length of a curve.0228

It is going to be the exact same thing. You are going to see that the integral is going to look the same, and that is what is nice.0239

So, if you ever lose your way, just sort of... if you know the lower dimensional case and feel comfortable with that, you can sort of extract the higher dimensional cases.0245

Okay. Given a surface in 3-space and a parameterization, which is the important part... and a parameterization... let us call that p, we found dp dt cross dp du.0254

So, now let me draw a surface here. This one... it is very, very important that we draw this one properly, so that is the surface, and we have a point -- I am going to draw one vector like that, and I am going to draw another vector like that.0287

Then, I am going to go that way, and I am going to -- oops, close this one out, let me just color this in here... so we have -- okay, so this point is our p(t,u) for some value of t and u.0312

This vector right here, let us call it dp du, and let us call this vector right here... that is the dp dt, okay?0349

Well dp dt and dp du, they span this particular parallelogram, okay?0365

So, dp dt and dp du span a parallelogram -- all of a sudden I do not know how to spell parallel -- parallelogram, whose area, we said, was the norm of that cross product. The area is... the norm of dp dt cross dp du, right? So this parallelogram has the area of a cross product.0373

Now, notice that this parallelogram right here, the reason I have drawn this little thing on the surface, what you have is this parallelogram which is the plane, so this parallelogram is part of the plane that is tangent to the surface at this point.0413

This colored is the shadow of this parallelogram if you happen to be shining a light from on top. So, we wanted to give the sense that there is this shadow, which is also in the shape of a rectangle on the surface.0428

Well, the idea is if you take these parallelograms small enough, what you end up actually getting is... you are adding up all of the surface area elements on the surface itself. That is essentially what we are doing.0444

I just wanted you to see this. This is just the projection, the shadow of the parallelogram on the surface.0458

Okay. So now we are ready to define our notion of surface area.0466

So, define surface area is equal to, we are going to use this symbol dΣ, that is the symbolic representation of that, and the double integral itself is the double integral over the entire surface of dp dt cross dp du with respect to t and with respect to u. This right here.0475

So, what we are doing is when we form dp dt and dp du, when we take their cross product we get the vector. When we take the magnitude of that vector, we get the area of the parallelogram.0515

Well, what we do is we take the area of the parallelogram and we basically add up all of the areas of the parallelogram, the differential elements and we end up getting the surface area.0529

This is what is important here. I am going to talk about this a little bit more, but we just want to be able to take a look at that.0539

So, the surface area given a parameterization, it is going to be the double integral of the normal vector, or the norm of dp dt cross dp du, integrated with respect to t and with respect to u. That is it. That is the definition of surface area.0547

Okay. Now, let us discuss this a little bit.0567

So, symbolically, ds, when we use the symbol ds, that is equal to dp dt cross dp du dt du.0573

This is a differential area element. Okay.0590

When we integrate, first with respect to t, then with respect to u, we have integrated over the entire surface, right? That is what we are doing.0608

When we do an integral, we can only go one variable at a time. We go along t first, then we go along u, and we have covered the entire surface. We have integrated over the entire surface.0635

Okay. Now, recall -- oops, let me do this one in blue -- recall the following. We said that curve length, we define curve length as the integral from a to b of the norm of the tangent vector, with respect to t.0652

We take the tangent vector and then we integrate it along the curve. Well, this is a complete analog. Except now, we are not dealing with a curve, we are dealing with a surface.0681

So, we are going to integrate along one curve, and then we are going to take that curve and we are going to integrate along the other parameter, and we are going to cover the entire surface. That is what we are doing.0694

This is the analog for a two-dimensional object, which a surface is, a two-dimensional object.0705

Okay. So, this area integral, I hope I am not belaboring too many points, I just want to make sure that there is a good intuitive sense of what is going on.0720

I mean, it is true ultimately that we are going to give you the definitions and we are going to lay these integrals out and then just sort of see if we can follow... take a look at what you need in the integrand, then fill in the blanks and integrate, but it is important to know where these integrands come from. Why it is that they take the form that they do.0732

So, we want to help develop a sense of intuition. We are still dealing with 2 and 3-space, so we still have the geometric intuition, we might as well use it.0751

Okay, so the area integral is a double integral. What we are doing, well not essentially, what we are doing explicitly, what we are doing is finding the length of p(t,u) holding one variable constant, then integrating this curve along the other variable -- I am going to draw this out in just a minute -- until we have covered the surface.0761

What we are doing, essentially, is finding the length of a curve and then we are going to integrate that curve, that whole curve. We are going to sweep it out along the other variable until we have covered the surface. What we have is something like this.0845

So if this is our surface, what we do when we take the first integral is we find essentially that curve length, and then we integrate in this direction, the other parameter.0858

What we do is we swing this, we sweep this curve out until we have covered the entire surface. So we move in that direction.0870

That is all that we are doing. It is like two curve integrals, one after the other. That is it.0878

Okay. So, let us go ahead and do an example, and again, when you are doing your examples, when you are doing your problems for your homework or whatever it happens to be, it is always best to write out what the integral is... the definition just to get use to writing it out symbolically.0885

Very, very important. So, let us do example 1.0901

Find the area of a sphere of radius r, find the surface area of a sphere of radius r. Well, you already know what that is from geometry, but the idea is where did that formula come from.0911

You know, when you learned it we just sort of dropped it on you, now we are going to show you where the formula actually comes from.0931

Okay. So, we have the parameterization, we have p(φ) and θ, so let us use the standard parameterization, and we know that this equals rsin(φ)cos(θ), rsin(φ)sin(θ), oops, that is sin(θ), and then we have rcos(φ), well we already found what the norm was.0939

We already, in the previous lesson, we did this, we already found the norm of dp dφ cross dp dθ.0969

That was just the norm of the normal vector, and that was r2 × sin(φ).0984

So, our ds element, dΣ, that is equal to this thing dφdθ = r2sin(φ)dφdθ.0994

Well, we know that φ, for a complete sphere is going to be > or = 0 and < or = pi, which is 180 degrees.1018

θ is going to run all the way from 0 to 2pi, which is 360 degrees, so, and like I just said a minute ago, let me go ahead and write the integral symbolically so we get used to writing it.1027

So, we are taking the integral over s of the differential area element. We are adding a differential area element and we are going to integrate, we are going to add all of the differential area elements over the surface.1042

Again, that is what integration is. Do not forget that. It is very important to remember that all you are doing is you are adding something. It is an infinite sum. The idea of integration is a particular technique, but it is just addition.1056

As it turns out in mathematics, all you can do to something is add things. That is it.1067

So, that is equal to the integral over s of -- alright -- dp dφ cross dp dθ, dφ dθ1073

Now we are actually going to write out the integral itself, and that is going to equal the integral from 0 to 2pi, so let us go ahead and do θ last.1090

It is going to be from 0 to pi, so this is going to be φ, and then we have this thing.1100

This thing is that, or ds is this whole thing.1107

So, we have r2, sin(φ), and we are going to be integrating with respect to φ first, then with respect to θ next, and when you go ahead and solve this integral... what you get is 4pi r2, which is exactly the formula you learned in geometry.1115

This is where it comes from. It is based on the definition of surface area given a particular parameterization which is given by this. That is the definition of surface area. That is it.1137

Okay. Let us do another example here... excuse me.1151

Alright, so, example number 2. We want to find the area of the paraboloid z = x2 + y2, so now it is given... this paraboloid is not given to us in parameterized form.1159

It is given to us in just a straight Cartesian fashion... z = some function of x and y... for z > or = 0, and < or = 5.1188

So, we are trying to find the area of... so we have this paraboloid and z is > or = 0, so obviously it is going to be above the xy plane, and up to a length... up to a height of 5.1211

So, we want the surface area of that paraboloid.1230

Real quickly before I get into this problem, I want to talk about the notion of length, area, and volume.1234

Length is -- you know what length is -- and there is a physical way of actually measuring length.1243

If I said what is the length of some curve, well what you can do is you can take a string, run it along a curve, make sure that it is there, and then pull the string taught and you can measure the length of that string.1250

There is a physical representation, a physical manifestation of length.1263

As far as volume is concerned, if you want me to measure the volume of some strange looking object, what you do is you have a cup of water, you measure the height... you measure how much... you read off the graduation and then what you do is you insert this object into the water until it is just -- you know -- under the surface, and of course the water level is going to rise.1267

Well, the difference in water level gives you the volume of the object. Unfortunately for area, for surface area, there is no physical manifestation, so in some sense, area is a very, very odd thing.1289

Length we can deal with physically, volume we can deal with physically, area we cannot really deal with physically.1302

Area is strictly a mathematical notion. I just wanted you to be aware of that. It is very, very curious that that is the case.1308

Okay, so let us go ahead and finish this problem. So we are given our equation that way, so let us go ahead and do our parameterization.1317

So, our parameterization, when we are given z = f(x,y), well that is equal to x and y and the function itself x2 + y2, so this is our parameterization.1326

Coordinate function 1, number 2, and number 3. Okay.1339

Let us find dp dx, so this is going to be a vector. This is going to be (1,0,2x), I am just differentiating this.1344

let us go ahead and find dp dy, so dp dy, that is going to equal (0,1,2y), excuse me.1355

Now we need to find the cross product of these two vectors, so dp dx cross dp dy, and we are going to go ahead and use our i,j,k, symbolism. We have (1,0,2x), and we have (0,1,2y).1366

When you expand this determinant you are going to end up with the following. You are going to end up with -2x,-2y, and 1.1387

In this particular case, I ended up choosing the opposite orientation. In this particular case, I ended up choosing such that this normal vector -- this is the normal vector -- is actually pointing into the paraboloid as opposed to out of it.1396

It is not really going to be a problem because when we take the norm of this, the norm is the same whether it is pointing in or pointing out, but it just so happens that I noticed that I have the incorrect orientation.1412

Later, that might be a problem, but for right now, it is not... and we will deal with it when we get to it.1424

Okay. So, this is our vector. Now we need the norm of this vector. So, the norm of dp dx cross dp dy = norm of this.1430

Well, it equal 4x2 + 4y2 + 1 under the radical.1448

So, our differential area element that we are going to integrate, ds is equal to the radical 4x2 + 4y2 + 1 dy dx.1457

That is what we are going to integrate. Now, for... now for z = x2 + y2, we said that we wanted z < or = 5, so x2 + y2 < or = 5.1474

Well, if that is 5, then the disc that we are looking at, this length right here is actually going to end up being sqrt(5), and this length right here is going to be sqrt(5).1499

So, what we are going to be doing is we are going to be integrating... this is x and this is y, the z axis is going like this.1514

We are integrating over a region of the xy plane, right? dy dx, so we need to know what x goes from and we need to know what y goes from.1522

In this particular case, z = x2 + y2 < 5.1534

Well, that means this circle is in the xy plane, that is this way, and of course the paraboloid is going up this way.1543

Well, if z itself is 5, well then this radius, our circle underneath has a radical of sqrt(5), simply based on the parameterization... x2 + y2.1552

I hope that that makes sense. So, in this particular case, our x is going to run from - sqrt(5) all the way to +sqrt(5).1567

It is going to go from this point to this point... excuse me... and our y value is going to go from... our y, the height is going to change... so y is going to run from -5 - x2 all the way to sqrt(5) - x2.1583

That comes from if I move this x2 over here, and I take this square root, I get y explicitly in terms of x, because remember, again, I am integrating with respect to y and I am integrating with respect to x, so I have to know what x does and what y does.1606

So, my area integral ends up being as follows: area = the integral from -sqrt(5) to sqrt(5) -- excuse me -- the integral - 5 - x2 under the radical, to sqrt(5) - x2 under the radical.1627

Of course I have my function, the norm of the cross product, which was 4x2 + 4y2 + 1 under the radical, dy dx.1653

Because this is y and this is x, so I am integrating that first.1667

When I put this into my mathematical software I get the number 49.864. There we go. That is the surface area of the paraboloid, that has a height of 5.1672

Okay. Now we did this in Cartesian coordinates. Now let us go ahead and run this integral in polar coordinates, just to see what it looks like. Just for the sake of some practice.1688

Okay, now, in polar coordinates -- forgive me, I seem to have something in my throat here -- in polar coordinates, it looks like this.1698

Again, we have a circle and we have this paraboloid on top and of course it is on top of this circle.1724

The radius is sqrt(5), so r is going to run from 0 to sqrt(5), because now we are going to be integrating r out this way, and then we are going to sweep this line around... all the way around the circle... to pi.1741

So θ is going to run from 0 to 2pi, so these are our lower and upper limits of integration. All I am doing is I am taking this region and I am expressing it in terms of r and θ. I am converting to polar coordinates.1765

When I convert to polar coordinates, I have to make the appropriate changes to the integral. Now, this is my -- let me go to red here -- this is my integrand.1780

I have to express this integrand in terms of r and θ.1794

Well, the conversion, the transformation is x = rcos(θ) for polar coordinates and y = rsin(θ), and when you put rcos(θ), rsin(θ) into this, you end up with the following.1798

So, 4x2 + 4y2 + 1 in polar form is.. .when you put these into here and solve, what you get is sqrt(4)r2 + 1.1813

That is our integrand. Okay. Now, dy dx... remember when we convert to polar coordinates, the differential area element is equal to r, dr, dθ, there is this extra factor of r.1836

Now let us go ahead and put everything together. Our area in terms of polar coordinates.1856

I am going to go ahead and do r last. Go from 0 to 2pi, this thing is that thing, it is 4r2 + 1, and now dy dx is equal to r dr dθ.1863

So I write r, dr, dθ. When I go ahead and put this into my mathematical software, I get the number 49.866. That is exactly write. This is .864, this is .866.1883

These had to be done numerically, there was no simply symbolic way of doing it to come up with nice closed form expression.1897

That is nice. The software will do that for us. That is it.1904

All I have done is, I was given a particular surface, I did the Cartesian xy, I found the integral this way, I wanted you to see what it looked like in polar coordinates, it is not necessarily easier, it is just a question of what is easier for you.1910

Or perhaps... and again, it is going to depend on the problem.1928

We cannot say do it this way or do it this way, but when you do make a conversion from a particular coordinate system, and you solve the integral in another coordinate system, you have to remember that the differential area element, or the differential volume element if you are dealing with triple integrals... it also undergoes a transformation.1931

That is it. For every object that you have in one integral, you are going to have a corresponding object in the other integral. Just keep things straight.1951

Okay. Well, thank you for joining us here at educator.com to discuss surface area, we will see you next time for a discussion of surface integrals. Take care, bye-bye.1960