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Topics You Need To Understand For This Page

Trig Identities and Formulas

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 (half-angle) 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

list of trigonometric identities - wikipedia

trig sheets - pauls online notes

17calculus trig formulas - full list

Related Topics and Links

Calculus Topics Listed Alphabetically

Single Variable Calculus

Multi-Variable Calculus

Differential Equations

Search Practice Problems

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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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.

MIP4U - Triple Integrals Using Cylindrical Coordinates [2mins-7secs]

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.

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

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

Before going on to spherical coordinates, work some practice problems involving cylindrical coordinates.

next: triple integrals (spherical) →

Practice

Conversion Between A-B-C Level (or 1-2-3) 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.

GOT IT. THANKS!

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

1929 video

video by MIP4U

Final Answer

\( 512 \pi/5 \)

close solution

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

Problem Statement

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

Final Answer

0

Problem Statement

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

Solution

1968 video

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

1970 video

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

1971 video

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 \( x^2 \).

1972 video

video by Krista King Math

Final Answer

\(2\pi/5\)

close solution

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

Problem Statement

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

Final Answer

\( 4 \pi/3 \)

Problem Statement

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

Solution

1973 video

video by MIP4U

Final Answer

\( 4 \pi/3 \)

close solution

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

Problem Statement

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

Final Answer

0

Problem Statement

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

Solution

1974 video

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 xy-plane.

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 xy-plane.

Solution

2307 video

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

1930 video

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

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

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.

1932 video

video by Dr Chris Tisdell

Final Answer

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

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

1934 video

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

1969 video

video by MIT OCW

Final Answer

\( \pi/2 \)

close solution
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