Shouldn't the bounds be from 0 to 2 for r? The one part it is marked as 1 <= r <= 2.

Yes, you are correct. It should be from 0 to 2. These articles are great, but there are few typos.

The correct answer for this question should not be 1-pi/4-2ln2?

And I think it should be natural log (ln) not log base 10

Shouldn't there be a restraint making y greater than 0? Or make theta range from negative pi over two to positive pi over four?

We can say that *y ≥ 0* because we know that: 1.) x ≥ 0 2.) z ≥ 0 3.) z ≤ y/x _However_, in the 2D plot of the range of the x & y values, the author only mentions that *x≥0*, *y≤x*, & that *x^2 + y^2 ≤ 4* and _forgets_ to tell us how we know that *y≥0*. Good catch!

Howdy, Could someone please elaborate on how/when to use the Jacobian matrix with a "change" of variables or bounds. To be honest I didn't find sufficient learning material on "changing bounds" or "transformation" of variables. In the previous skill "Practice: Double integrals in polar", we need to use the Jacobian determinant to "scale" our transformation. I saw the videos KA had on the Jacobian determinant, but I'm afraid I was quite there when it came to using this for a transformation of variables. Is there a good external (non KA) source you'd recommend for me? Ever since KA made the skills for Multivariable Calculus I've been reinvigorated to learn some more math & I've been making a lot of notes but I've been slightly lost on some of the new skills. Regardless, I've been loving the new skills, so thanks KA!

Glad you like the new skills! Paul's Online Math Notes has some good material to learn about change of variables. Check out this link for example: https://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx Also Professor Leonard on YouTube has comprehensive material on multivariable calculus. Check out his video on change of variable: https://www.youtube.com/watch?v=VVPu5fWssPg Apologies that KA has no videos yet on change of variables. Maybe the hints can guide you a bit, but if you're stuck that's understandable. I hope there'll be videos soon!

how can I reach the formula: dV=dZdA=rdθdrdz I need a proof

If you go to MIT opencourseware there is great video with an explanation of triple integrals for cylindrical coordinates. You can either google "opencourseware triple integrals" or visit this link: https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-25-triple-integrals/ . The video is quite long but you can skip through to where he talks about cylindrical coordinates.

why is the bound for theta from 0 to pi/4 ?

This corresponds to the fact that y <= x which could be converted to r*sinΘ <= r*cosΘ, and sinΘ <= cos Θ, so theta must be smaller than pi/4, and it has to be greater than 0, hence the bound.

Main content

Multivariable calculus

Course: Multivariable calculus > Unit 4

Lesson 9: Polar, spherical, and cylindrical coordinates

Triple integrals in cylindrical coordinates

Q: Shouldn't the bounds be from 0 to 2 for r? The one part it is marked as 1 <= r <= 2.

Yes, you are correct. It should be from 0 to 2. These articles are great, but there are few typos.

Q: The correct answer for this question should not be 1-pi/4-2ln2?

And I think it should be natural log (ln) not log base 10

Q: Shouldn't there be a restraint making y greater than 0? Or make theta range from negative pi over two to positive pi over four?

We can say that *y ≥ 0* because we know that: 1.) x ≥ 0 2.) z ≥ 0 3.) z ≤ y/x _However_, in the 2D plot of the range of the x & y values, the author only mentions that *x≥0*, *y≤x*, & that *x^2 + y^2 ≤ 4* and _forgets_ to tell us how we know that *y≥0*. Good catch!

Q: Howdy, Could someone please elaborate on how/when to use the Jacobian matrix with a "change" of variables or bounds. To be honest I didn't find sufficient learning material on "changing bounds" or "transformation" of variables. In the previous skill "Practice: Double integrals in polar", we need to use the Jacobian determinant to "scale" our transformation. I saw the videos KA had on the Jacobian determinant, but I'm afraid I was quite there when it came to using this for a transformation of variables. Is there a good external (non KA) source you'd recommend for me? Ever since KA made the skills for Multivariable Calculus I've been reinvigorated to learn some more math & I've been making a lot of notes but I've been slightly lost on some of the new skills. Regardless, I've been loving the new skills, so thanks KA!

Glad you like the new skills! Paul's Online Math Notes has some good material to learn about change of variables. Check out this link for example: https://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx Also Professor Leonard on YouTube has comprehensive material on multivariable calculus. Check out his video on change of variable: https://www.youtube.com/watch?v=VVPu5fWssPg Apologies that KA has no videos yet on change of variables. Maybe the hints can guide you a bit, but if you're stuck that's understandable. I hope there'll be videos soon!

Q: how can I reach the formula: dV=dZdA=rdθdrdz I need a proof

If you go to MIT opencourseware there is great video with an explanation of triple integrals for cylindrical coordinates. You can either google "opencourseware triple integrals" or visit this link: https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-25-triple-integrals/ . The video is quite long but you can skip through to where he talks about cylindrical coordinates.

Q: why is the bound for theta from 0 to pi/4 ?

This corresponds to the fact that y <= x which could be converted to r*sinΘ <= r*cosΘ, and sinΘ <= cos Θ, so theta must be smaller than pi/4, and it has to be greater than 0, hence the bound.

Google Classroom

How to perform a triple integral when your function and bounds are expressed in cylindrical coordinates.

Background

[Need a quick refresher of cylindrical coordinates?]

What we're building to

The main thing to remember about triple integrals in cylindrical coordinates is that $d V$ ‍, representing a tiny bit of volume, is expanded as
$d V = r d θ d r d z$ ‍
(Don't forget to include the $r$ ‍)
Using cylindrical coordinates can greatly simplify a triple integral when the region you are integrating over has some kind of rotational symmetry about the $z$ ‍-axis.

The one rule

When performing double integrals in polar coordinates, the one key thing to remember is how to expand the tiny unit of area

d A

in terms of

d r

and

d θ

$\begin{array}{r} \iint_{R} f (r, θ) d A = \iint_{R} f (r, θ) r d θ d r \end{array}$ ‍

Note that the variable

r

is part of this expansion. Expanding the tiny unit of volume

d V

in a triple integral over cylindrical coordinates is basically the same, except that now we have a

d z

term:

$\begin{array}{r} ∭_{R} f (r, θ, z) d V = ∭_{R} f (r, θ, z) r d θ d r d z \end{array}$ ‍

Remember, the reason this little

r

shows up for polar coordinates is that a tiny "rectangle" cut by radial and circular lines has side lengths

r d θ

and

d r

The key thing to remember here is that $θ$ ‍ is not a unit of length, so

d θ

does not represent a tiny length in the same way that

d r

and

d z

do. It measures radians, which need to be multiplied by the distance

r

from the origin to become a length.

[Thinking more deeply about this expansion]

Example 1: Volume of a sphere

Problem: Find the volume of a sphere with radius

1

using a triple integral in cylindrical coordinates.

First of all, to make our lives easy, let's place the center of the sphere on the origin.

Next, I'll give the sphere a name,

S

, and write the abstract triple integral to find its volume.

$\begin{array}{r} ∭_{S} d V = ∭_{S} r d θ d r d z \end{array}$ ‍

As always, the hard part is putting bounds on the integral. However, doing this with cylindrical coordinates is much easier than it would be for cartesian coordinates. In particular,

r

and

θ

will just live within the unit disc, which is very natural to describe in polar coordinates:

Concept check: Which of the following sets of bounds for

r

and

θ

should we use to integrate over the unit disc?

Choose 1 answer:
(Choice A)
$0 \leq r \leq 1$ ‍
$0 \leq θ \leq 2 π$ ‍
(Choice B)
$- 1 \leq r \leq 1$ ‍
$0 \leq θ \leq 2 π$ ‍
(Choice C)
$0 \leq r \leq 1$ ‍
$0 \leq θ r \leq 2 π$ ‍

[Answer]

Since the bounds of

z

will depend on the value of

r

, we let the innermost integral handle

z

, while the outer two integrals take care of

r

and

θ

. Writing down what we have so far, we get

$\begin{array}{r} \int_{0}^{2 π} \int_{0}^{1} \int_{?}^{?} r d z d r d θ \end{array}$ ‍

Remember, it's important to make sure the order of the differential terms

d z

d r

and

d θ

matches up with the appropriate integral.

This next question is a little trickier.

Concept check: For a given value of

r

, which of the following shows the right range of values for

z

Choose 1 answer:
(Choice A)
$- \sqrt{1 - r^{2}} \leq z \leq \sqrt{1 - r^{2}}$ ‍
(Choice B)
$- \sqrt{1 + r^{2}} \leq z \leq \sqrt{1 + r^{2}}$ ‍
(Choice C)
$- r^{2} \leq z \leq r^{2}$ ‍
(Choice D)
$- r \leq z \leq r$ ‍

[Answer]

Applying this bound to our innermost integral, we get something that can be worked out.

Concept check: Solve this triple integral.

$\begin{array}{r} \int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2}}}^{\sqrt{1 - r^{2}}} r d z d r d θ = \end{array}$ ‍

[Answer]

And with that, you just found the volume of a unit sphere!

Moreover, this tool is powerful enough to do more than just find the volume of the sphere. For example, you could integrate a three-variable function

f (r, θ, z)

inside the sphere,

$\begin{array}{r} \int_{0}^{2 π} \int_{0}^{1} \int_{- \sqrt{1 - r^{2}}}^{\sqrt{1 - r^{2}}} f (r, θ, z) r d z d r d θ \end{array}$ ‍

The hard part of finding the bounds is no different, but the computation of the integrals (done by either you or a computer) will change.

Example 2: Integrating over a pie slice

For this example, we will integrate over a region which looks kind of like a slanted pie slice:

Khan Academy video wrapper

See video transcript

In a problem, this region might be described to you using the following list of properties:

$x \geq 0$ ‍
$y \geq 0$ ‍
$z \geq 0$ ‍
$y \leq x$ ‍
$x^{2} + y^{2} \leq 4$ ‍
$z \leq \frac{y}{x}$ ‍

This time, we will not just be finding the volume of this region. Instead, our task is to integrate the following three-variable function:

$f (x, y, z) = z - x^{2} - y^{2}$ ‍

This might seem out of place in an article about integrating in cylindrical coordinates, since everything here is given in cartesian coordinates. Indeed, you could setup the triple integral using cartesian coordinates if you wanted. However, there's one key fact suggesting that our lives can be made dramatically easier by converting to cylindrical coordinates first:

The expression $x^{2} + y^{2}$ ‍ shows up in the function $f$ ‍, as well as in the description of the bounds. This suggests some rotational symmetry around the $z$ ‍-axis, which cylindrical coordinates are well-suited for.

For example, look at the range for our

x

and

y

values:

$x \geq 0$ ‍
$y \geq 0$ ‍
$y \leq x$ ‍
$x^{2} + y^{2} \leq 4$ ‍

Describing this with a pair of integrals over

d x

and

d y

is a real pain. However, in polar coordinates, this becomes very simple:

$0 \leq θ \leq \frac{π}{4}$ ‍
$0 \leq r \leq 2$ ‍

This means the bounds on the integrals handling

d θ

and

d r

will be constants. You can't do better than that!

What about the other criteria, such as

$z \leq \frac{y}{x}$ ‍

Since converting to polar coordinates involves the property

$\tan (θ) = \frac{y}{x}$ ‍

The bounds on

z

can be translated to

$0 \leq z \leq \tan (θ)$ ‍

Putting this together, our triple integral looks like this:

$\begin{array}{r} \int_{0}^{π / 4} \int_{0}^{2} \int_{0}^{\tan (θ)} f d V \end{array}$ ‍

Notice how simple the bounds are. If you are up for a little pain, you can try finding the appropriate triple integral bounds in cartesian coordinates to see just how much uglier they are.

We now write the function

f

using polar coordinates.

$\begin{aligned} f (x, y, z) & = z - x^{2} - y^{2} \\ ⇓ \\ f (r, θ, z) & = z - r^{2} \end{aligned}$ ‍

And of course, we incorporate the main takeaway of this article, which is how to write

d V

in polar coordinates:

$d V = r d θ d r d z$ ‍

Putting this all together, we get our triple integral in its final solvable state.

More practice: Solve this integral

$\begin{array}{r} \int_{0}^{π / 4} \int_{0}^{2} \int_{0}^{\tan (θ)} (z - r^{2}) r d z d r d θ = \end{array}$ ‍

[Answer]

Summary

The main thing to remember about triple integrals in cylindrical coordinates is that $d V$ ‍, representing a tiny bit of volume, is expanded as
$d V = r d θ d r d z$ ‍
(Don't forget to include the $r$ ‍)
Using cylindrical coordinates can greatly simplify a triple integral when the region you are integrating over has some kind of rotational symmetry about the $z$ ‍-axis.

Want to join the conversation?

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shed412
Posted 8 years ago. Direct link to shed412's post “Shouldn't the bounds be f...”
Shouldn't the bounds be from 0 to 2 for r? The one part it is marked as 1 <= r <= 2.
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(23 votes)
Answer
- T H
  Posted 8 years ago. Direct link to T H's post “Yes, you are correct. It ...”
  Yes, you are correct. It should be from 0 to 2.
  These articles are great, but there are few typos.
  Button navigates to signup page
  (19 votes)
Marin Dumitru
Posted 7 years ago. Direct link to Marin Dumitru's post “The correct answer for th...”
The correct answer for this question should not be 1-pi/4-2ln2?
Button navigates to signup pageComment on Marin Dumitru's post “The correct answer for th...”
(6 votes)
Answer
- Christine Chesley
  Posted 7 years ago. Direct link to Christine Chesley's post “And I think it should be ...”
  And I think it should be natural log (ln) not log base 10
  Button navigates to signup page
  (4 votes)
Roman Burridge
Posted 7 years ago. Direct link to Roman Burridge's post “Shouldn't there be a rest...”
Shouldn't there be a restraint making y greater than 0? Or make theta range from negative pi over two to positive pi over four?
Button navigates to signup pageButton navigates to signup page
(5 votes)
Answer
- SteveSargentJr
  Posted 5 years ago. Direct link to SteveSargentJr's post “We can say that *y ≥ 0* b...”
  We can say that y ≥ 0 because we know that:
  
  1.) x ≥ 0
  2.) z ≥ 0
  3.) z ≤ y/x
  
  However, in the 2D plot of the range of the x & y values, the author only mentions that x≥0, y≤x, & that x^2 + y^2 ≤ 4 and forgets to tell us how we know that y≥0.
  
  Good catch!
  Button navigates to signup page
  (6 votes)
Iron Programming
Posted 3 years ago. Direct link to Iron Programming's post “Howdy, Could someone ple...”
Howdy,

Could someone please elaborate on how/when to use the Jacobian matrix with a "change" of variables or bounds.

To be honest I didn't find sufficient learning material on "changing bounds" or "transformation" of variables.

In the previous skill "Practice: Double integrals in polar", we need to use the Jacobian determinant to "scale" our transformation.

I saw the videos KA had on the Jacobian determinant, but I'm afraid I was quite there when it came to using this for a transformation of variables.

Is there a good external (non KA) source you'd recommend for me?

Ever since KA made the skills for Multivariable Calculus I've been reinvigorated to learn some more math & I've been making a lot of notes but I've been slightly lost on some of the new skills.

Regardless, I've been loving the new skills, so thanks KA!
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- lakern
  Posted 3 years ago. Direct link to lakern's post “Glad you like the new ski...”
  Glad you like the new skills! Paul's Online Math Notes has some good material to learn about change of variables. Check out this link for example:
  https://tutorial.math.lamar.edu/Classes/CalcIII/ChangeOfVariables.aspx
  
  Also Professor Leonard on YouTube has comprehensive material on multivariable calculus. Check out his video on change of variable:
  https://www.youtube.com/watch?v=VVPu5fWssPg
  
  Apologies that KA has no videos yet on change of variables. Maybe the hints can guide you a bit, but if you're stuck that's understandable. I hope there'll be videos soon!
  Button navigates to signup page
  (2 votes)
Jonathan
Posted 7 years ago. Direct link to Jonathan's post “In the xy diagram for pie...”
In the xy diagram for pie slice, why doesn't the shaded area include points under the X-axis? For example, wouldn't the point (1,-1) satisfy the conditions?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
David
Posted 8 years ago. Direct link to David 's post “why is the bound for thet...”
why is the bound for theta from 0 to pi/4 ?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Sam
  Posted 8 years ago. Direct link to Sam's post “This corresponds to the f...”
  This corresponds to the fact that y <= x which could be converted to r*sinΘ <= r*cosΘ, and sinΘ <= cos Θ, so theta must be smaller than pi/4, and it has to be greater than 0, hence the bound.
  Button navigates to signup page
  (5 votes)
marceil.alaraj
Posted 7 years ago. Direct link to marceil.alaraj's post “how can I reach the formu...”
how can I reach the formula: dV=dZdA=rdθdrdz
I need a proof
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Harrison
  Posted 6 years ago. Direct link to Harrison's post “If you go to MIT opencour...”
  If you go to MIT opencourseware there is great video with an explanation of triple integrals for cylindrical coordinates. You can either google "opencourseware triple integrals" or visit this link: https://ocw.mit.edu/courses/mathematics/18-02-multivariable-calculus-fall-2007/video-lectures/lecture-25-triple-integrals/ . The video is quite long but you can skip through to where he talks about cylindrical coordinates.
  Button navigates to signup page
  (3 votes)
STrumbore88
Posted 7 years ago. Direct link to STrumbore88's post “Has nobody else pointed o...”
Has nobody else pointed out that this result is negative? There are still problems with the way the integral is being set up
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(2 votes)
Answer
- Jonathan Turner
  Posted 7 years ago. Direct link to Jonathan Turner's post “Since we are integrating ...”
  Since we are integrating some function over the volume, our result can be negative. Its similar to taking a surface integral to solve for the flux through an area. Your result can be positive or negative. Sorry, Im failing to think of a 3-d analog. Maybe someone else has a thought?
  Button navigates to signup page
  (1 vote)
apndolby297
Posted 4 years ago. Direct link to apndolby297's post “I think there is a mistak...”
I think there is a mistake in the second example as y can also take negative values,therefore,it represents a quarter of a circle having the theta varying from -pi/2 to +pi/2
(p.s. correct me if i am wrong....)
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(1 vote)
Answer
26cperezserna
Posted 2 months ago. Direct link to 26cperezserna's post “Can you Add A Video for t...”
Can you Add A Video for this
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(1 vote)
Answer