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Lecture Comments (6)

1 answer

Last reply by: Professor Hovasapian
Thu Nov 6, 2014 1:31 AM

Post by Hayden Clark on November 3, 2014

How do we know when to convert from rectangular to cylindrical or spherical. Is there anything in the functions that can indicate when to shift?

0 answers

Post by Professor Hovasapian on May 24, 2013

Hi Josh,

The paraboloid itself is when z = x^2 + y^2. When z is less than or equal to this, it is all the points below the paraboloid -- therefore outside the paraboloid -- because you are just falling vertically down FROM the paraboloid to the xy-plane. Hope that makes sense. Let me know, if not....

Take Care.

Raffi

2 answers

Last reply by: Josh Winfield
Sun May 26, 2013 7:42 AM

Post by Josh Winfield on May 24, 2013

since z<=x^2=y^2 why is the bounded area outside the parabaloid not inside the parabaloid

Cylindrical & Spherical Coordinates

Convert (1,1,4) to cylindrical coordinates.
  • To convert rectangular coordinates to cylindrical coordinates (r,θ,z) we utilize r = √{x2 + y2} , tanθ = [y/x] and z = z for r0, 0 ≤ θ ≤ 2π and z ∈ ( − ∞,∞).
  • Since x = 1, y = 1 and z = 4 then r = √{12 + 12} = √2 , θ = tan − 1( [1/1] ) = [(π)/4] and z = 4.
Our cylindrical coordinates are ( √2 ,[(π)/4],4 ).
Convert ( − √3 ,3,1 ) to cylindrical coordinates.
  • To convert rectangular coordinates to cylindrical coordinates (r,θ,z) we utilize r = √{x2 + y2} , tanθ = [y/x] and z = z for r0, 0 ≤ θ ≤ 2πand z ∈ ( − ∞,∞).
  • Since x = − √3 , y = 3 and z = 1 then r = √{( − √3 )2 + ( 3 )2} = 2√3 and z = 1
  • For θ we have θ = tan − 1( [3/( − √3 )] ) = tan − 1( − √3 ) = − [(π)/3], but this is outside our range for θ. We compensate by adding 2π and obtaining θ = − [(π)/3] + 2π = [(5π)/3]
Our cylindrical coordinates are ( 2√3 ,[(5π)/3],1 ).
Convert (2,1, − 2) to spherical coordinates.
  • To convert rectangular coordinates to spherical coordinates (r,f,θ) we utilize r = √{x2 + y2 + z2} , f = cos − 1( [z/r] ) and θ = sin − 1( [y/(√{x2 + y2} )] ) for r0, 0 ≤ f ≤ πand 0 ≤ θ ≤ 2π.
  • Since x = 2, y = 1 and z = − 2 then r = √{22 + 12 + ( − 2 )2} = 3, f = cos − 1( [( − 2)/3] ) ≈ 2.30 and θ = sin − 1( [1/(√{22 + ( − 1)2} )] ) ≈ 0.46
Our spherical coordinates are ( 3,2.30,0.46 ).
Convert (0,0,3) to spherical coordinates.
  • To convert rectangular coordinates to spherical coordinates (r,f,θ) we utilize r = √{x2 + y2 + z2} for r0, 0 ≤ f ≤ π and 0 ≤ θ ≤ 2π .
  • Since x = 0, y = 0 and z = 3 then r = √{02 + 02 + 32} = 3.
  • Note that the point (0,0,3) has not moved any θ radians (from the x - axis) nor any f radians from the z - axis. Hence θ = 0 and f = 0.
Our spherical coordinates are (3,0,0).
Convert the cylindrical coordinates ( 2,[(π)/3], − 2 ) to rectangular coordinates.
  • To covert cylindrical coordinates to rectangular coordinates (x,y,z) we utilize x = rcosθ, y = rsinθ and z = z.
  • Since r = 2, θ = [(π)/3] and z = z we have x = 2cos( [(π)/3] ) = 1, y = 2sin( [(π)/3] ) = √3 and z = − 2.
Our rectangular coordinates are ( 1,√3 , − 2 ).
Convert the spherical coordinates ( 1,[(π)/6],π ) to rectangular coordinates.
  • To covert spherical coordinates to rectangular coordinates (x,y,z) we utilize x = rsinϕcosθ, y = rsinϕsinθ and z = rcosf.
  • Since r = 1, f = [(π)/6] and θ = π we have x = 1sin( [(π)/6] )cos( π) = − [1/2], y = 1sin( [(π)/6] )sin( π) = 0 and z = 1cos( [(π)/6] ) = [(√3 )/2]
Our rectangular coordinates are ( − [1/2],0,[(√3 )/2] ).
Integrate ∫ − 220 − 11 2r2 drdθdz
  • We evaluate the trile integral in respect to the differentials, in this case the order is dr, dθ and dz.
  • Integrating in respect to dr yields ∫ − 220 − 11 2r2 drdθdz = ∫ − 220 ( [2/3]r3 ) | − 11 dθdz = ∫ − 220 [4/3] dθdz
  • Integrating in respect to dθ yields ∫ − 220 [4/3] dθdz = ∫ − 22 ( [4/3]θ ) |0 dz = ∫ − 22 [(8π)/3] dz
Lastly, integrating in respect to dz yields ∫ − 22 [(8π)/3] dz = ( [(8π)/3]z ) | − 22 = [32/3]π
Find the integral of f(x,y,z) = √{x2 + y2} + z of the region bounded inside x2 + y2 = 4, outside x2 + y2 + z2 = 4 and outside x2 + y2 + (z − 6)2 = 4.
Use cylindrical coordinates. Do not integrate.
  • We set up a triple integral where R is the region to find our solution.
  • Note that our region is the inside of a cylinder of radius 2 which extends along the z - axis, capped by two spheres, one centered at the origin and the other centered at (0,0,6),of radius 2.
  • If we look at the xz - plane we have an idea of the bounds of the solid:
  • Now, z is bounded by the top of the sphere centered at (0,0,0) and the bottom of the sphere centered at (0,0,6) that is z ∈ [ √{4 − x2 − y2} ,6 − √{4 − x2 − y2} ]
  • In cylindrical coordinates x = rcosθ and y = rsinθ and so z ∈ [ √{4 − r2} ,6 − √{4 − r2} ]. Also, the cylinder transverses a circle, hecne θ ∈ [0,2π].
  • Lastly, r is the region transversed by the radius so r ∈ [0,2].
  • We must also convert f(x,y,z) = √{x2 + y2} + z into cylindrical terms and so f(r,θ,z) = r + z.
Our integral is ∫002√{4 − r2} 6 − √{4 − r2} ( r + z ) rdzdrdθ = ∫002√{4 − r2} 6 − √{4 − r2} ( r2 + zr ) dzdrdθ
Integrate ∫0ππ\mathord/ \protect phantom π4 4π\mathord/ \protect phantom π3 301 r2sinf drdfdθ
  • We evaluate the triple integral in respect to the differentials, in this case the order is dr, df and dθ.
  • Integrating in respect to dr yields ∫0ππ\mathord/ \protect phantom π4 4π\mathord/ \protect phantom π3 301 r2sinf drdfdθ = ∫0ππ\mathord/ \protect phantom π4 4π\mathord/ \protect phantom π3 3 sinf ( [1/3]r3 ) |01 dfdθ = ∫0ππ\mathord/ \protect phantom π4 4π\mathord/ \protect phantom π3 3 [1/3] sinfdfdθ
  • Integrating in respect to df yields ∫0ππ\mathord/ \protect phantom π4 4π\mathord/ \protect phantom p 3 3 [1/3] sinfdfdθ = ∫0π [1/3] ( − cosf ) |π\mathord/ \protect phantom π4 4π\mathord/ \protect phantom π3 3 dθ = ∫0π [(√2 − 1)/6] dθ
Lastly, integrating in respect to dθ yields ∫0π [(√2 − 1)/6] dθ = [(√2 − 1)/6]( θ) |0p = [(√2 − 1)/6]π
Find the volume of the solid bounded inside x2 + y2 + z2 = z and inside z = √{x2 + y2} .
Use spherical coordinates. Do not integrate.
  • To find the volume of the solid, we set up the triple integral where R is the region where the solid lays.
  • Note that our region is the inside of a cone with the upper half of a sphere on top. If we look at the xz - plane we have an idea of the bounds of the solid:
  • #
  • Now, r is bounded from 0 to the sphere, in spherical coordinates x2 + y2 + z2 = z is equivalent to r = cosϕ and so r ∈ [0,cosθ]
  • Also, ϕ is bounded by the cone, in spherical coordinates our cone z = √{x2 + y2} yields rcosf = rsinf. Solving for f gives f = [(π)/4], and so f ∈ [ 0,[(π)/4] ]
  • Lastly, we see that the solid transverses a circle, hecne θ ∈ [0,2π].
Our volume is ∫00π\mathord/ \protect phantom π4 40cosθ r2sinfdrdfdθ .

*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

Cylindrical & Spherical Coordinates

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
  • Cylindrical and Spherical Coordinates 0:42
    • Cylindrical Coordinates
    • When Integrating Over a Region in 3-space, Upon Transformation the Triple Integral Becomes..
    • Example 1
    • The Cartesian Integral
    • Introduction to Spherical Coordinates
    • Reason It's Called Spherical Coordinates
    • Spherical Transformation
    • Example 2

Transcription: Cylindrical & Spherical Coordinates

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

Today's topic is going to be cylindrical and spherical coordinates. It is just two alternative ways to describe points in 3 space. That is all.0004

Because again, just like polar coordinates, there are certain regions that are just expressed a little bit easier in cylindrical or spherical coordinates. That is all it is. 0012

Again, with the introduction of mathematical software, the real power behind this -- these coordinates systems, you do not really... they do not play as important a role as they used to.0022

But, of course, it is part of the mathematical curriculum. It is part of the history, so we study it.0035

Let us go ahead and get started. Okay. So, polar coordinates, which was the R, θ in two space, and the plane, that generalizes to 3-space two ways, cylindrical and spherical coordinates.0040

We are going to do cylindrical first. So, we have cylindrical coordinates.0057

Let us see. Given R, θ, z, again we need three numbers for 3 space, the relationship between the Cartesian coordinates and the cylindrical is x = Rcos(θ) which was the same as polar.0070

y = Rsin(θ) so the x and y are the same as polar, and z is just equal to z. So, z is just equal to z.0091

If you want to think about it, the polar coordinates gives you what is happening in the xy plane, and the z is just straight up and down, which is why they call it cylindrical coordinates, because Rcos(θ), Rsin(θ) defines the unit circle, a circle of radius R.0097

Then if you just take z infinitely up and down, what you get is a cylinder, that is why it is called cylindrical coordinates. We are not going to worry too much about going the other way.0115

Given xyz, how do you gate R, θ, z, well, I mean you can get that because again, this is the same as polar, this is the same as polar, and that is the same. 0125

Really what we are concerned with is going from R, θ, z to x, y, z.0137

So, let us go ahead and draw this out so we see what it is we are looking at. Just so we know if that is a point out there, drop a perpendicular down to the xy plane, this is y, this is x, this is z.0141

We have this thing right here, well this angle is the θ, okay? So, this is the point x, y, z, and this right here, that is R.0162

Well, again, this point is the polar coordinate of the xy, and then you have the z going up. That is it. So, R, θ z. That is all there is.0176

So, let us go ahead and write the transformation as -- oh, also, these, so R is greater than or equal to 0, and θ is > or = 0, < or = to 2pi, all the way around the circle... and z is > or = negative infinity and < or = positive infinity. z can take on any value it wants.0192

z is < or = positive infinity. z is arbitrary. Okay, we can write this transformation as c(r), θ, z = Rcos(θ), Rsin(θ), z.0222

You will see the transformation that way, and that is exactly what it is, a transformation. You are converting one coordinate system to another coordinate system.0243

In this case, given Rθz, you want to express -- this is your x, this is your y, and this is your z -- okay, now when integrating a region in 3-space, or integrating over a region in 3-space, when we want to do a change of coordinate system, that is where this is important.0250

So, let us go ahead and write the integral formula here. So, when integrating over a region in 3-space, upon transformation -- or conversion if you will -- upon transformation the triple integral becomes the following.0269

So, triple integral over a region R, of the function that is expressed as a function of x, y, z, in Cartesian... dx dy dz... it is equal to the triple integral of R under the transformation c... c is for cylindrical, f(c), x, y, z, -- so wherever you see x, y, z, you put Rcos(θ), Rsin(θ) and z.0311

So, you form the composite function f(c) and the differential volume element, R, dR, dθ, dz. It just depends on the region as to which one of these is going to go first. 0343

This dx dy dz, that is the volume element in Cartesian... R dR dθ dz is the volume element, the equivalent volume element under transformation for the cylindrical coordinate system. That is it, that is all that is going on here.0360

Okay. Let us do an example. Let us see, let me go ahead and start the example on the next page so that we have everything on one page. Let me do this in blue.0375

So, example 1... excuse me... let R be the region bounded by the circle of radius 2 in the xy plane.0390

Let z be > or = 0, < or = x2 + y2, and the plane x + y = 0.0426

Okay. let us see what this region looks like. Let us go ahead and draw out the xy plane here. It is often best to work in the xy plane here, and it is often best to work in the xy plane and then consider of course z being coming out and going down. It is the best way to think about it.0443

So, this is y -- oops, sorry, we are not drawing the 3-dimensional system -- this is x, and this is y, so the circle of radius 2 in the xy plane, well, let us just go ahead and draw this circle right here, so that is a circle of radius 2, and let us leave z alone for a second.0459

The plane x + y = 0. Well, at the plane x + y = 0, notice z does not show up here, so z can take on any value, so it is going to be infinitely up and infinitely down, but it is saying that z is > or = 0, so we are concerned with everything above the xy plane.0479

So if you are looking like this, where this is the z axis, this is the y, and this is the x, it is everything above that. So here, we get y = -x, well y = -x is this line right here.0496

So, imagine instead of just the line that that is the plane. That is going to be parallel to the z axis. So, now, z, x2 + y2, well, x2 + y2... z = x2 + y2 is just the paraboloid around the z axis, that is it.0515

That is all that is happening, so it is going like that. So, the base that you have is the circle of radius 2 and let us see here... we want -- so we have the circle of radius 2... good, that is one of the bounds... it is everything in here, and it is going to be everything above the xy plane.0538

That is going to be that one. So, our region, oh actually I think I forgot to take care of 1 other thing. Looks like there is one other constraint here. We are going to have y > or = 0.0567

What we are looking at is this region right here, in the xy plane, and then of course up above the xy plane, z is going to go up and it is going to go all the way up to x2 + y2. That is it.0586

So, if I were to draw out the zy axis, for example, so let us say this is z and this is y, what you are going to have is this paraboloid and there is some region under here in the xy plane and that is going to be this region.0605

It is going to be everything up, so this is like 2, it is going to be everything above that. So, that is it. Think of this as the base of R volume, R solid, and this is the z version, so this is the base, this is going to be the height. The height is going to go up to this point.0621

So, let us go ahead and see what our R, θ, and z is going to be. R is going to go from 0 all the way to 2.0645

It is going to go from 0 to 2, θ is going to go... so now we integrate along R and then we are going to sweep this out an angle of 135 degrees, or 3pi/4, so θ goes from 0 to 3pi/4, and z, well z is going to run from 0 all the way to x2 + y2.0658

But, we have to express x2 + y2 in terms of polar coordinates, so it needs to be in terms of R and θ. So, z actually is going from 0 all the way to R2.0691

That is it. That is our region, and these are our upper and lower limits of integration, so, let us go ahead and do this. So, let us see, now, what function are we integrating?0710

We will let the function of xyz be equal to x + y - z, so let us form our f(c).0724

Well, x is Rcos(θ), y is Rsin(θ) and then -z, just put the transformation, the cylindrical transformation into the function for xyz. 0739

Now we go ahead and form our integral. So, the integral is equal to... we will go ahead and do R first, so R goes from 0 to 2, so I will put that here... R.0757

R θ is going to go from 0 to 3pi/4, right? and then z is θ, and z is going to go from 0 all the way to R2.0770

Then the function is Rcos(θ) + Rsin(θ) - z.0785

Then we have R dz dR -- nope, sorry -- R dz dθ dR, because R θ z, z θ R, we are working out.0793

If we did everything correct, and the mathematical software works properly, the number that I got was -4pi + 32/5 + 32sqrt(2)/5. There we go.0814

So, again, the integral part is... what is important is being able to construct this integral. This is what we want. Being able to take a particular problem and convert it into an integral. Being able to find out what the region is... in 3-space.0834

Then finding out what the upper and lower limits of integration with respect to each variable are. 0851

In this particular case, the radius went from 0 to 2, and then we sweep out this radius... sweep it out this way, an angle of 135 degrees, which is 3pi/4, and then of course we take the z value.0859

z goes from 0 to R2, or x2 + y2, because you are going, the z value is changing. That is it. You just put everything in, make sure you form the composite function, make sure you have the proper volume element, and the rest is just a question of integration.0876

That is all. The important thing is being able to find what that region is.0893

Now, I am going to go ahead and actually express this in terms of the Cartesian coordinates, just to show you what the Cartesian integral would look like.0902

So, let me do this in blue... Now... the Cartesian integral would look like this.0912

Let me draw my region again, and my region was... so this was the y axis, this was the x axis, we had this circle... we had this thing, so we were looking at this region right here, right?0931

As it turns out in this particular case, I am going to break this up into 2 regions. I am going to call this region a, and this region b, so I am going to do 2 integrals.0950

Let us go ahead and the actual integral itself was going to be the integral over a and the integral over b, so let us talk about a first.0960

a, x is going from, well, this point, the x value of this point where you have this line which is y = -x and where it meets the circle of radius 2.0971

So, it is going to be -2/sqrt(2) all the way to 0, so x is going to go from here to here.0985

Now, the y is going to go from, well, you are going to have this little strip so it is going to go from -x all the way to 4 - x2, under the radical, because the equation for this circle is 4 - x2 under the radical, when expressed as a function... y as a function of x.0995

Because this circle is x2 + y2 = 22. Therefore, y = 4 - x2, all under the radical, and of course z is going to go from 0 all the way to x2 + y2.1021

Those are our first set of upper and lower limits of integration for this particular region a, this little triangular sector.1039

Now for b, well b, now x is going to go from 0 to 2, so x is going to run from 0 to 2, y is still going to run this time... this is our strip, so it is going to run from 0 all the way to 4 - x2 under the radical.1048

It goes from 0 to sqrt(4) - x2, and z is of course still... goes from 0 to x2 + y2.1067

When I put this together, the integral in Cartesian coordinates equals 0 -- oops, the integral of a -- so it is going to be -2/sqrt(2) to 0, the integral from -x to 4 -x2 under the radical, 0 to x2 + y2, and of course our function was x + y - z.1078

This is going to be... this is z, this is y, this is x, so we are going to do dz, dy, dx.1109

Then plus the integral from 0 to 2, the integral from 0 to 4 minus -- oops, this is all under the radical -- 4 - x2, and the integral from 0 to x2 + y2 of x + y - z.1118

Again, dz dy dx, so, this in Cartesian coordinates, the integral would look like this. Again, software can handle it, but again, as you can see, this looks reasonably complicated. Obviously you would not want to do this by hand.1141

I mean it might be okay, I am not exactly sure but clearly the cylindrical coordinates in this particular case work out a lot easier, so but ultimately it is a question of choice. 1159

If you have this software available, it is not really much of a problem. As long as you understand, again, this is the important part. Being able to describe the region, the upper and lower limits of integration. That is what is important.1171

Now, let us go ahead and start talking about the other coordinate system. The cylindrical... I am sorry, we just did the cylindrical, this is the spherical coordinate system. 1186

Again, it is just another way of expressing a point in 3-space, so let me go back to black for this one.1193

So, we have, let me put it over here, so this is our xyz coordinate system, this is y, this is z, this is x, and we have some point and when we a perpendicular to it, to the xy plane, we have this right here.1201

Okay. So, this one is the one that is most analogous to the polar coordinate. So, we have 3 variables, we have ρ, which they use the Greek letter ρ instead of R, because they reserve R for polar coordinates.1220

This ρ, is the length from the origin to the point. Okay. Now this angle, θ, is the same... it is the same θ that it makes in the xy plane, and then there is one more angle here.1241

It is this angle. We will go ahead and erase this. This angle right here that it makes with the z axis, this is the angle φ, and of course this right here is the ρ.1256

So, what you have is the following: a transformation given ρ, φ, and θ, if you want to transform them, the x value is equal to ρ, × sin(φ) × cos(θ).1274

The y value equals ρ × sin(φ_ × sin(θ), and the z value equals ρ × cos(φ). 1301

So, our three variables are ρ, φ, and θ. Now, ρ is going to be greater than or equal to 0, it is a length. φ is going to run from... it is going to be > or = to 0, and < or = to pi, 180 degrees.1314

θ is going to run > or = to 0, and < or = 2pi. This is the spherical coordinate system. Basically what is happening is this. What you are doing is you are starting on the x... well, you know what, let me write a few things and I will describe what is actually going on here.1329

So, this point, (x,y,z)... it consists of a length along the z axis and then a certain angle downward from the z axis, and then of course you are going to rotate this an angle θ away from the x axis.1350

So, let me write out the reason. It is called spherical, and I think it will make a lot more sense where this ρ, φ, θ stuff comes from... it is called spherical coordinates.1370

I am going to draw out the zy coordinate system, so this is z, and this is y.1390

z here and y here, so the x axis is coming out and going down. Here is what we do.1400

The first thing, here is how I think about it, I think it is the best way to think about it. Go up a distance ρ along the z axis.1409

We will go up -- I am going to do this one in blue -- so we will go up a distance ρ along the z axis.1426

Now, swing down... actually you know what, let me go ahead and make this the x -- so I am going to change that, I am going to make that the x axis.1436

So now the second thing I want to do, I want to swing down... in other words I am going to take this point, and I am going to swing it down in this direction.1453

Through an angle of 180 degrees. I am going to take this thing and I am going to swing it down so now it is down here.1465

That is our φ. Through φ = pi, 180 degrees.1476

Now, I am on the x axis, so now I am going to take this semi-circle and I am going to swing it around the z axis all the way around. A full circle, that is my θ, 2 pi. 1482

When I do that, I am going to end up sweeping out a sphere, that is where the spherical coordinate comes from.1498

So, sweep this semi-circle around the z axis, θ = 2pi, and you are going to get the sphere of radius ρ. 1504

This is the best way to think about it. If you are given a 3-dimensional coordinate system. Let us say... ρ is just go up along the z axis, and then swing down along the zx plane, a certain angle φ, and then θ is, you swing around this way and that will give you the point somewhere.1527

That is the whole idea. So, here or all the way down -- from your perspective that is the z axis -- this is the zx plane, you are going to get a semi circle, and then you are going to swing this semi circle around.1549

I do not know if you have ever held a chain, if you sort of swung it around this way it sweeps out a sphere. That is why it is called a spherical coordinate system. That is it, that is all that is going on here.1565

Now, let us go ahead and describe the transformation. So, given... yeah, that is fine I can do it on this page. 1574

Given, some function of f(x,y,z) and the transformation, the spherical transformation, we will call it s, which is just a change of coordinates... change of coordinates.1588

When we say transformation, we are just talking about a change of coordinates.1608

We will do it this way... s(ρ,φ,θ) = ... and I am going to write this in column vector form... ρ, sin(φ), cos(θ), ρ, sin(φ) sin(θ), and ρ cos(φ).1615

Given some function, hence the transformation which is this thing right here... that is the transformation, that is the change of coordinates.1645

Given that, the integral is going to be as follows: the triple integral over a region R of f(x,y,z), I am going to make this a little bigger and a little clearer... so the triple integral of f/R and f is a function of x,y,z dx dy dz is equal to the transformed region R under the transformation s, f(s) -- we form the composite function -- and then of course the volume element.1655

ρ2, sin(φ), ρ2, sin(φ), dρ, dφ, dθ, in the proper order... depending on the particular region that you are given. That is it.1700

So, this is our conversion. Very, very important. This f under the transformation becomes the f(s), that is the composite function. When I put the transformation into the x,y, and z.1720

This volume element in Cartesian three space becomes this column element in R, φ, θ space. Cannot forget this ρ2sin(φ).1733

In other words, when I take this volume element and I transform it, I have to multiply it by this factor. That is what this is.1746

Again, we will talk about this more formally when we talk about the change of variables theorem in a couple of lessons.1755

Okay. Now let us go ahead and do an example here. Let us do example 2.1765

Let f(x,y,z) = xyz. We will let R, the region, be the section of the unit sphere in the first octant. In the first octant.1777

In this case, the description that they gave you of the region... they did not give it to you as an equation, they did not tell you it is bounded by this, bounded by this, bounded by this... this time they just sort of told you qualitatively it is the part of the sphere in the first octant.1807

You have to figure out what the upper and lower limits of integration are.1819

So, our task is to find the integral of f over the region R. Okay. Let us just take a look at our region.1824

I am going to go ahead and draw out the full 8 octants. So, this is the first octant over here, this is z, this is y, this is x. this is z...1835

We have the unit sphere there, there, and there.1850

That is it. It is the section of the sphere that is just in the first octant, essentially 1/8 of the sphere.1861

Well, let us go ahead and find... so let us convert this... and since we are dealing with a sphere, a sphere is ideal for dealing with spherical coordinates because the equation of a sphere is just ρ... just some radius.1868

let us go ahead and find f(s) first... so, f(s) so we can put it into our integral formula here.1883

So, f(s), well, f is xyz, x is Rsin(φ)cos(θ), y is Rsin(φ)sin(θ), z is ρcos(φ), so we just put them in and multiply them out. 1891

So, xyz = ρsin(φ)cos(θ) × ρsin(φ)sin(θ) × ρcos(φ).1908

I end up with -- I want to simplify as much as possible -- so, ρ3sin2(φ), cos(θ)sin(θ)cos(φ).1924

Looks complicated but it is easily handle-able. Now we will go ahead and do the integral.1940

Well, let us see. First of all let us go ahead and talk about... so ρ, let us see, this is the unit sphere, so ρ is going to go from 0 to 1.1945

φ, well φ is in the first octant, so it is going to be from 0 to pi/2. It is going to swing down through pi/2, 90 degrees, and θ is going ot go from 0 to pi/2, so 0 to pi/2.1962

Now we are ready for our integral. Our integral = ... that is okay, I will go ahead and do it there... so 0 to 1, that is going to be our ρ, we have 0 to pi/2, let us make that our φ, so I will go ahead and write ρ and φ.1983

It is always good to write the variables underneath. Then our θ is going to be 0 again, to pi/2.2003

Then of course we have our function, so ρ3sin2(φ), just have to go slow, make sure everything is there, cos(θ)... and again, be very careful with these particularly because you are dealing with 2 angles, φ and θ, do not confuse them -- which I do all the time.2013

I can give you the advice, but I make the mistake all the time myself... cos(θ)sin(θ)cos(φ), that is the function, the f(s) part.2032

Now we have to do the transformed volume element, so ρ2sin(φ), and then we did θ first so we do dθ, dφ, and dρ. That is it.2047

When you put this into your mathematical software, you end up with 1/48.2062

That is all. Straight, simple, you just need to be able to take a look at a particular region, try to express it, that region in terms of... in this particular case... spherical coordinates, and decide this part right here.2067

What is going to be happening? ρ is going to go from what to what, φ is going to go from what to what, θ is going to go from what to what. That is all.2087

In this particular case, the spherical region, this is the best way to deal with it rather than delaying with Cartesian coordinates.2098

The integral for Cartesian coordinates is incredibly complicated. It has a lot of radicals in it, I do not even know if it is solved by elementary methods, you might just have to rely on numerical methods to solve it if you were actually going to use the Cartesian coordinate system.2105

So, that is it. That is spherical coordinates, cylindrical coordinates, hopefully we will get a chance to do more practice in the context of other concepts.2118

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2129