17Calculus - Triple Integrals in Cylindrical Coordinates
On this page we cover triple integrals in cylindrical coordinates and several applications.
Understanding of cylindrical coordinates is necessary for understanding the material on this page.
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Describing Surfaces and Volumes in Cylindrical Coordinates
So, you may ask, why do we need the cylindrical coordinate system when we describe surfaces and volumes in the rectangular coordinate system? Well, there are two reasons.
1. Some surfaces and volumes are more easily (simply) described in cylindrical coordinates. An example is given below.
2. When we get to triple integrals, some integrals are more easily evaluated in cylindrical coordinates and you will even have some integrals that can't be evaluated in rectangular coordinates but can be in cylindrical.
Setting Up and Evaluating Triple Integrals in Cylindrical Coordinates
Let's start by watching this short video clip, explaining cylindrical coordinates again and showing how to set up triple integrals in cylindrical coordinates.
MIP4U - Triple Integrals Using Cylindrical Coordinates [2mins-7secs]
video by MIP4U |
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A couple of clarifications are in order that he mentions in this video.
First, he mentions that there is an extra factor of r in the term \(dV\) in cylindrical coordinates but he doesn't explain where it comes from. Remember from your study of polar coordinates that \(dA = dx~dy\) in rectangular coordinates becomes \(dA = r~dr~d\theta\). This next video clip explains where \(dV\) comes from in more detail.
Larson Calculus - Triple Integrals in Cylindrical Coordinates [5mins-26secs]
This video will not stop automatically at the 5min-26sec mark. For the purposes of the current discussion, you can stop it there. However, there is some good content after that point that is okay to watch if you have time.
video by Larson Calculus |
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The second clarification involves the order of components of \(dV\). In the first video, he says that \(dV = r~dz~dr~d\theta\). This is only one of six possible representations for \(dV\). The order is determined by how the volume is described by the equations. To determine what order to integrate when setting up the integral, we use the same idea as we did when setting up double integrals.
In integral form, a triple integral in cylindrical coordinates looks like this.
\(\displaystyle{ \iiint\limits_V {f(x,y,z) ~ dV} = }\) \(\displaystyle{ \iiint\limits_V{ f(r\cos\theta, r\sin\theta,z)~r~dr~d\theta~dz } }\)
VERY IMPORTANT NOTE - - - Do not forget the \(r\) in \(\color{red}{r}~dr~d\theta~dz\) in the above equation. This is the most common mistake made by students learning this technique. Why do we have this extra \(r\)? Here is a video explaining this.
Michel vanBiezen - Cylindrical Coordinates: Small Volume Element dV [4mins-0secs]
video by Michel vanBiezen |
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Before going on to spherical coordinates, work some practice problems involving cylindrical coordinates.
Practice
Unless otherwise instructed, evaluate these integrals in cylindrical coordinates.
Basic |
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Integrate \( f(x,y,z) = x^2 + y^2\) over the solid region bounded by \( z = 0 \) and \( z = 4 - \sqrt{x^2+y^2} \) in cylindrical coordinates.
Problem Statement |
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Integrate \( f(x,y,z) = x^2 + y^2\) over the solid region bounded by \( z = 0 \) and \( z = 4 - \sqrt{x^2+y^2} \) in cylindrical coordinates.
Final Answer |
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\( 512 \pi/5 \)
Problem Statement |
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Integrate \( f(x,y,z) = x^2 + y^2\) over the solid region bounded by \( z = 0 \) and \( z = 4 - \sqrt{x^2+y^2} \) in cylindrical coordinates.
Solution |
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1929 video
video by MIP4U |
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Final Answer |
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\( 512 \pi/5 \) |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-y^2}}^{ \sqrt{4-y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \) in cylindrical coordinates.
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-y^2}}^{ \sqrt{4-y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \) in cylindrical coordinates.
Final Answer |
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\( 0 \)
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-y^2}}^{ \sqrt{4-y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \) in cylindrical coordinates.
Solution |
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1968 video
video by Krista King Math |
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Final Answer |
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\( 0 \) |
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Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9 - x^2 - y^2 \).
Problem Statement |
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Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9 - x^2 - y^2 \).
Final Answer |
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\( 81 \pi/2 \)
Problem Statement |
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Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9 - x^2 - y^2 \).
Solution |
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1970 video
video by MIP4U |
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Final Answer |
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\( 81 \pi/2 \) |
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Calculate the volume of the solid bounded by \( z = 4 - \sqrt{x^2+y^2} \), \( x^2 + y^2 = 1 \) and \( z = 0 \).
Problem Statement |
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Calculate the volume of the solid bounded by \( z = 4 - \sqrt{x^2+y^2} \), \( x^2 + y^2 = 1 \) and \( z = 0 \).
Final Answer |
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\( 10 \pi/3 \)
Problem Statement |
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Calculate the volume of the solid bounded by \( z = 4 - \sqrt{x^2+y^2} \), \( x^2 + y^2 = 1 \) and \( z = 0 \).
Solution |
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1971 video
video by MIP4U |
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Final Answer |
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\( 10 \pi/3 \) |
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Evaluate \(\iiint_{V}{ x^2 ~dV }\) where V is the solid that lies within the cylinder \( x^2 + y^2 = 1\) above the plane \(z=0\) and below the cone \( z^2 = 4x^2 + 4y^2 \).
Problem Statement |
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Evaluate \(\iiint_{V}{ x^2 ~dV }\) where V is the solid that lies within the cylinder \( x^2 + y^2 = 1\) above the plane \(z=0\) and below the cone \( z^2 = 4x^2 + 4y^2 \).
Final Answer |
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\(2\pi/5\)
Problem Statement |
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Evaluate \(\iiint_{V}{ x^2 ~dV }\) where V is the solid that lies within the cylinder \( x^2 + y^2 = 1\) above the plane \(z=0\) and below the cone \( z^2 = 4x^2 + 4y^2 \).
Solution |
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Note: Although she calls the result of this integral a volume, it is not truly a volume. If we were calculating the volume, then we would be integrating 1 over the solid region, not \( x^2 \).
1972 video
video by Krista King Math |
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Final Answer |
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\(2\pi/5\) |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-1}^{1}{ \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \) in cylindrical coordinates.
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-1}^{1}{ \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \) in cylindrical coordinates.
Final Answer |
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\( 4 \pi/3 \)
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-1}^{1}{ \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \) in cylindrical coordinates.
Solution |
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1973 video
video by MIP4U |
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Final Answer |
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\( 4 \pi/3 \) |
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close solution
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\) in cylindrical coordinates.
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\) in cylindrical coordinates.
Final Answer |
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\( 0 \)
Problem Statement |
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Unless otherwise instructed, evaluate the integral \(\displaystyle{ \int_{-2}^{2}{ \int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\) in cylindrical coordinates.
Solution |
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1974 video
video by MIP4U |
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Final Answer |
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\( 0 \) |
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Evaluate \( \iiint\limits_E e^z ~dV \) where E is enclosed by the paraboloid \( z = 3 + x^2 + y^2 \), the cylinder \( x^2 + y^2 = 2 \) and the xy-plane.
Problem Statement |
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Evaluate \( \iiint\limits_E e^z ~dV \) where E is enclosed by the paraboloid \( z = 3 + x^2 + y^2 \), the cylinder \( x^2 + y^2 = 2 \) and the xy-plane.
Solution |
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2307 video
video by MIP4U |
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Intermediate |
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Integrate \( f(x,y,z) = 1 \) over the solid region bounded by \( z = 0 \), \( x^2 + y^2 = 1 \), \( x^2 + y^2 = 4 \) and \( z = 4 - ( x^2 + y^2 ) \) in cylindrical coordinates.
Problem Statement |
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Integrate \( f(x,y,z) = 1 \) over the solid region bounded by \( z = 0 \), \( x^2 + y^2 = 1 \), \( x^2 + y^2 = 4 \) and \( z = 4 - ( x^2 + y^2 ) \) in cylindrical coordinates.
Final Answer |
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\( 9 \pi/2\)
Problem Statement |
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Integrate \( f(x,y,z) = 1 \) over the solid region bounded by \( z = 0 \), \( x^2 + y^2 = 1 \), \( x^2 + y^2 = 4 \) and \( z = 4 - ( x^2 + y^2 ) \) in cylindrical coordinates.
Solution |
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1930 video
video by MIP4U |
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Final Answer |
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\( 9 \pi/2\) |
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close solution
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Express as a triple integral, using cylindrical coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a > 0 \).
Problem Statement |
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Express as a triple integral, using cylindrical coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a > 0 \).
Final Answer |
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\(\displaystyle{ \int_{0}^{2\pi}{ \int_{0}^{a}{ \int_{r}^{a+\sqrt{a^2-r^2}}{ } } } }\) \( r~dz~dr~d\theta \)
Problem Statement |
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Express as a triple integral, using cylindrical coordinates, the volume of the region above the cone \( z = \sqrt{x^2+y^2} \) and inside the sphere \( x^2 + y^2 + z^2 = 2az, a > 0 \).
Solution |
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Note - This same problem is set up in rectangular coordinates on the rectangular coordinates page and in spherical coordinates on the spherical coordinates page. Also on the spherical coordinates page, the integral is evaluated. But you are not asked to evaluate it here.
1932 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ \int_{0}^{2\pi}{ \int_{0}^{a}{ \int_{r}^{a+\sqrt{a^2-r^2}}{ } } } }\) \( r~dz~dr~d\theta \) |
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Use cylindrical coordinates to find the volume outside the cylinder \( x^2 + y^2 = 1 \), (\(z \ge 0 \)) and inside the paraboloid \( z = 4 - x^2 - y^2 \).
Problem Statement |
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Use cylindrical coordinates to find the volume outside the cylinder \( x^2 + y^2 = 1 \), (\(z \ge 0 \)) and inside the paraboloid \( z = 4 - x^2 - y^2 \).
Final Answer |
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\( 9 \pi/2 \)
Problem Statement |
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Use cylindrical coordinates to find the volume outside the cylinder \( x^2 + y^2 = 1 \), (\(z \ge 0 \)) and inside the paraboloid \( z = 4 - x^2 - y^2 \).
Solution |
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1934 video
video by Dr Chris Tisdell |
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Final Answer |
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\( 9 \pi/2 \) |
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close solution
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Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Problem Statement |
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Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Final Answer |
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\( \pi/2 \)
Problem Statement |
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Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Solution |
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1969 video
video by MIT OCW |
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Final Answer |
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\( \pi/2 \) |
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You CAN Ace Calculus
Topics You Need To Understand For This Page
Trig Identities and Formulas
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basic trig identities |
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\(\sin^2\theta+\cos^2\theta=1\) | \(1+\tan^2\theta=\sec^2\theta\) |
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\(\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}\) | \(\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}\) |
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\(\displaystyle{\sec\theta=\frac{1}{\cos\theta}}\) | \(\displaystyle{\csc\theta=\frac{1}{\sin\theta}}\) |
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power reduction (half-angle) formulae |
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\(\displaystyle{\sin^2\theta=\frac{1-\cos(2\theta)}{2}}\) | \(\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}\) |
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double angle formulae |
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\(\sin(2\theta)=2\sin\theta\cos\theta\) | \(\cos(2\theta)=\cos^2\theta-\sin^2\theta\) |
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links |
Related Topics and Links
related topics |
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external links you may find helpful |
Pauls Online Notes: Triple Integrals in Cylindrical Coordinates |
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) |
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }\) | |
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) |
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }\) | |
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) |
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }\) |
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }\) |
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }\) | |
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) |
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }\) | |
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }\) |
Trig Integrals
\(\int{\sin(x)~dx} = -\cos(x)+C\) |
\(\int{\cos(x)~dx} = \sin(x)+C\) | |
\(\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C\) |
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\) | |
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) |
\(\int{\csc(x)~dx} = \) \( -\ln\abs{\csc(x)+\cot(x)}+C\) |
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Practice Instructions
Unless otherwise instructed, evaluate these integrals in cylindrical coordinates.
