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basic trig identities 

\(\sin^2\theta+\cos^2\theta=1\)  \(1+\tan^2\theta=\sec^2\theta\) 
\(\displaystyle{\tan\theta=\frac{\sin\theta}{\cos\theta}}\)  \(\displaystyle{\cot\theta=\frac{\cos\theta}{\sin\theta}}\) 
\(\displaystyle{\sec\theta=\frac{1}{\cos\theta}}\)  \(\displaystyle{\csc\theta=\frac{1}{\sin\theta}}\) 
power reduction (halfangle) formulae 
\(\displaystyle{\sin^2\theta=\frac{1\cos(2\theta)}{2}}\)  \(\displaystyle{\cos^2\theta=\frac{1+\cos(2\theta)}{2}}\) 
double angle formulae 
\(\sin(2\theta)=2\sin\theta\cos\theta\)  \(\cos(2\theta)=\cos^2\theta\sin^2\theta\) 
links 
related topics 

external links you may find helpful 
Pauls Online Notes: Triple Integrals in Cylindrical Coordinates 
Single Variable Calculus 

MultiVariable Calculus 

Acceleration Vector 
Arc Length (Vector Functions) 
Arc Length Function 
Arc Length Parameter 
Conservative Vector Fields 
Cross Product 
Curl 
Curvature 
Cylindrical Coordinates 
Lagrange Multipliers 
Line Integrals 
Partial Derivatives 
Partial Integrals 
Path Integrals 
Potential Functions 
Principal Unit Normal Vector 
Differential Equations 

Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
 
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next: triple integrals in spherical coordinates → 

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On this page we cover triple integrals in cylindrical coordinates and several applications. Triple integrals in rectangular coordinates are covered on a separate page, as well as triple integrals in spherical coordinates.
Understanding of cylindrical coordinates is necessary for understanding the material on this page.
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.
video by MIP4U
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.
This video will not stop automatically at the 5min26sec 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
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.
video by Michel vanBiezen
Before going on to spherical coordinates, work some practice problems involving cylindrical coordinates.
next: triple integrals (spherical) →

Conversion Between ABC Level (or 123) and New Numbered Practice Problems 

Please note that with this new version of 17calculus, the practice problems have been relabeled but they are MOSTLY in the same order. So, Practice A01 (1) is probably the first basic practice problem, A02 (2) is probably the second basic practice problem, etc. Practice B01 is probably the first intermediate practice problem and so on. 
Instructions  Unless otherwise instructed, evaluate these integrals in cylindrical coordinates.
Basic Problems 

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 

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 

\( 512 \pi/5 \) 
Problem Statement 

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 

video by MIP4U
Final Answer 

\( 512 \pi/5 \) 
close solution 
\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4y^2}}^{ \sqrt{4y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \)
Problem Statement 

\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4y^2}}^{ \sqrt{4y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \)
Final Answer 

0 
Problem Statement 

\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4y^2}}^{ \sqrt{4y^2}}{\int_{\sqrt{x^2+y^2}}^{2}{ xz } } } }\) \( dz ~dx ~dy \)
Solution 

video by Krista King Math
Final Answer 

0 
close solution 
Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9  x^2  y^2 \).
Problem Statement 

Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9  x^2  y^2 \).
Final Answer 

\( 81 \pi/2 \) 
Problem Statement 

Calculate the volume of the solid bounded by \( z = 0 \) and \( z = 9  x^2  y^2 \).
Solution 

video by MIP4U
Final Answer 

\( 81 \pi/2 \) 
close solution 
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 

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 

\( 10 \pi/3 \) 
Problem Statement 

Calculate the volume of the solid bounded by \( z = 4  \sqrt{x^2+y^2} \), \( x^2 + y^2 = 1 \) and \( z = 0 \).
Solution 

video by MIP4U
Final Answer 

\( 10 \pi/3 \) 
close solution 
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 

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 

\(2\pi/5\) 
Problem Statement 

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 

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 \( x^2 \).
video by Krista King Math
Final Answer 

\(2\pi/5\) 
close solution 
\(\displaystyle{ \int_{1}^{1}{ \int_{\sqrt{1x^2}}^{\sqrt{1x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \)
Problem Statement 

\(\displaystyle{ \int_{1}^{1}{ \int_{\sqrt{1x^2}}^{\sqrt{1x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \)
Final Answer 

\( 4 \pi/3 \) 
Problem Statement 

\(\displaystyle{ \int_{1}^{1}{ \int_{\sqrt{1x^2}}^{\sqrt{1x^2}}{ \int_{0}^{2}{ \sqrt{x^2+y^2} } } } }\) \( dz ~dy ~dx \)
Solution 

video by MIP4U
Final Answer 

\( 4 \pi/3 \) 
close solution 
\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4x^2}}^{\sqrt{4x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\)
Problem Statement 

\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4x^2}}^{\sqrt{4x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\)
Final Answer 

0 
Problem Statement 

\(\displaystyle{ \int_{2}^{2}{ \int_{\sqrt{4x^2}}^{\sqrt{4x^2}}{ \int_{x^2+y^2}^{4}{ y~dz~dy~dx } } } }\)
Solution 

video by MIP4U
Final Answer 

0 
close solution 
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 xyplane.
Problem Statement 

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 xyplane.
Solution 

video by MIP4U
close solution 
Intermediate Problems 

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 

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 

\( 9 \pi/2\) 
Problem Statement 

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 

video by MIP4U
Final Answer 

\( 9 \pi/2\) 
close solution 
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 

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 

\(\displaystyle{ \int_{0}^{2\pi}{ \int_{0}^{a}{ \int_{r}^{a+\sqrt{a^2r^2}}{ } } } }\) \( r~dz~dr~d\theta \) 
Problem Statement 

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 

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, in rectangular coordinates.
video by Dr Chris Tisdell
Final Answer 

\(\displaystyle{ \int_{0}^{2\pi}{ \int_{0}^{a}{ \int_{r}^{a+\sqrt{a^2r^2}}{ } } } }\) \( r~dz~dr~d\theta \) 
close solution 
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 

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 

\( 9 \pi/2 \) 
Problem Statement 

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 

video by Dr Chris Tisdell
Final Answer 

\( 9 \pi/2 \) 
close solution 
Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Problem Statement 

Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Final Answer 

\( \pi/2 \) 
Problem Statement 

Find the volume between the paraboloid \( z = x^2 + y^2 \) and the plane \( z = 2y \).
Solution 

video by MIT OCW
Final Answer 

\( \pi/2 \) 
close solution 