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17Calculus - Double Integrals - Rectangular Regions

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This page covers double integrals in Cartesian coordinates. We start by discussing double integrals on rectangular regions. The next page discusses evaluating double integrals on non-rectangular regions.

Alternate Name For Double Integrals - Iterated Integrals

Double integrals are just integrals that are nested, i.e. there are two integrals, one inside of the other. The idea is to evaluate each integral separately, starting with the inside integral.
The discussion on this page is in two main parts based on the type of region described by the limits of integration.

If the limits of integration are constant, the region is rectangular.

If the limits of integration are NOT constant, the region is non-rectangular.

The basics of double integrals are discussed first in the context of rectangular regions. Before we get started on the details of evaluating double integrals, let's watch a video to help understand what double integrals represent.

Krista King Math - What does a double integral represent? [25mins-43secs]

video by Krista King Math

Rectangular Regions

Rectangular regions in the xy-plane are indicated by constants in the limits of integration, i.e. \( a \leq x \leq b \) and \( c \leq y \leq d\). \[\displaystyle{ \int_{c}^{d}{ \int_{a}^{b}{g(x,y) ~ dx} ~dy } = \int_{c}^{d}{ \left[ \int_{a}^{b}{g(x,y) ~ dx}\right] ~dy}}\] You can think about it this way.
First, evaluate the inside integral with respect to the inside variable, while holding the other variable constant. In the integral above, the inside variable is \(x\).
Second, take the result of the first integration and now integrate with respect to the outside variable. In the above integral, the outside variable is \(y\).
This integral, once evaluated, gives the area or volume above the xy-plane with height \(g(x,y)\). We have more discussion of area and volume with double integrals on the area and volume application page. For now, we will just discuss how to evaluate a double integral.

Notice the order \(dx~dy\) or \(dy~dx\) is critical in determining which limits of integration go with which variable. In the example above, we integrated on the interval \([a,b]\) in the x direction and we evaluated on the interval \([c,d]\) in the y direction.

Okay, so that is a very basic introduction on how to look at a double integral. Let's watch a video to get started. This video explains the motivation behind double integrals, explains what double integrals are and includes the basics on how to set them up and evaluate them, using some basic examples.

Dr Chris Tisdell - introduction to double integrals [29mins-20secs]

video by Dr Chris Tisdell

Are you ready for an example? Let's look at an easy double integral and take it step-by-step.

Evaluate \(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\)

42

Evaluate \(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\)

Solution

\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\)

\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } }\)

\(\displaystyle{ \int_{1}^{2}{ \left[ y^2 \int_{2}^{4}{3x ~ dx} \right] ~dy } }\)

\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 3x^2/2 \right]_{x=2}^{x=4} ~dy } }\)

\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 3(4)^2/2 - 3(2)^2/2 \right] ~dy } }\)

\(\displaystyle{ \int_{1}^{2}{ y^2 \left[ 18 \right] ~dy } = \int_{1}^{2}{ 18y^2 ~dy } }\)

\( \left[ 18y^3/3 \right]_{1}^{2} \)

\( 18(2)^3/3 - 18(1)^3/3 \)

\( 48 - 6 = 42 \)

Notice in this solution, that after evaluating the inside integral, all the x's are gone, so that the outside integral is just a basic single variable definite integral with integrand \(18y^2\). This is due to the fact that we have constants as the limits of integration for the inside integral. Makes sense, right?

Final Answer

42

Okay, let's watch a video tutorial. Here is another very good video explaining double integrals and how they work.

Dr Chris Tisdell - double integrals [1mins-11secs]

video by Dr Chris Tisdell

In the above example, we were able to use a special technique that works only in specific instances. Here is a great video that explains exactly when and how to use it. Although we do not usually encourage learning special cases, this is easy to use, works all the time ( when used correctly ), saves time and it is easy to learn. So go for it!

Dr Chris Tisdell - double integrals (special technique) [2mins-24secs]

video by Dr Chris Tisdell

Notation

As discussed in the videos above, a double integral can be thought of as an integral over a region in the plane. This can be written several different ways. If you are given a region in the plane defined by the region R, all of the forms below describe the same integral.

the integral of \(f(x,y)\) over the region \( R = \{ (x,y) ~:~ a \leq x \leq b, ~ c \leq y \leq d \} \)

\(\displaystyle{ \int_{c}^{d}{ \int_{a}^{b}{f(x,y) ~ dx} ~dy } }\)

\(\displaystyle{ \iint\limits_R {f(x,y) ~ dA} }\)

In the last form, \(dx~dy\) are combined to form \(dA\) and the integral signs are collapsed into the integral over R. This notation is very common and it is something you will see more of as you continue on in calculus.

Reversing Order of Integration

When the limits of integration of both integrals are constants (i.e. the region in the xy-plane is rectangular), we can switch the limits of integration with no other changes to the integral, without affecting the result. Stop and think about this. This holds only when the limits of integration are constants.
So in the example above, we could have integrated with respect to \(y\) first, then with respect to \(x\) and arrived at the same answer. The equivalent integrals are
\(\displaystyle{ \int_{1}^{2}{ \int_{2}^{4}{3xy^2 ~ dx} ~dy } = \int_{2}^{4}{ \int_{1}^{2}{3xy^2 ~ dy} ~dx } }\)
Notice that when we switched the integral signs on the left, we also switched \(dx~dy \to dy~dx\). This is required since the intervals do not change, i.e. \([2,4]\) is in the \(x\) direction and \([1,2]\) is in the \(y\) direction. So we need to keep the same association in an equivalent integral. [Try working this second integral to convince yourself of this.]

Let's work some practice problems. After that, you will be ready for double integrals over non-rectangular regions.

Practice

Unless otherwise instructed, evaluate these integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{1}{ (1+4xy) ~dx } ~dy } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{1}{ (1+4xy) ~dx } ~dy } }\)

Final Answer

\( 10 \)

Problem Statement

\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{1}{ (1+4xy) ~dx } ~dy } }\)

Solution

This is the same problem as 3512 except that the order of integration is reversed.

3511 video

video by Michel vanBiezen

Final Answer

\( 10 \)

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\(\displaystyle{ \int_{0}^{1}{ \int_{1}^{3}{ (1+4xy) ~dy } ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{1}{ \int_{1}^{3}{ (1+4xy) ~dy } ~dx } }\)

Final Answer

\( 10 \)

Problem Statement

\(\displaystyle{ \int_{0}^{1}{ \int_{1}^{3}{ (1+4xy) ~dy } ~dx } }\)

Solution

Notice that this is the same problem as 3511 except that the order of integration is reversed. The answers are the same, as should be the case.

3512 video

video by Michel vanBiezen

Final Answer

\( 10 \)

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\(\displaystyle{ \int_{0}^{2}{ \int_{1}^{3}{ xy^2 ~dy } ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{2}{ \int_{1}^{3}{ xy^2 ~dy } ~dx } }\)

Final Answer

\( 52/3 \)

Problem Statement

\(\displaystyle{ \int_{0}^{2}{ \int_{1}^{3}{ xy^2 ~dy } ~dx } }\)

Solution

In the second half of this video clip, he switches the limits of integration and shows that the answer will be the same.

3517 video

Final Answer

\( 52/3 \)

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\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{\pi/2}{ x \sin y ~dy } ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{\pi/2}{ x \sin y ~dy } ~dx } }\)

Final Answer

\( 2 \)

Problem Statement

\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{\pi/2}{ x \sin y ~dy } ~dx } }\)

Solution

3513 video

video by Michel vanBiezen

Final Answer

\( 2 \)

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\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{2}{ xy e^{x^2y} ~dy } ~dx } }\)

Problem Statement

\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{2}{ xy e^{x^2y} ~dy } ~dx } }\)

Solution

3521 video

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

Problem Statement

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

Final Answer

\( -8 \)

Problem Statement

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

Solution

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

\(\displaystyle{ \int_{-1}^{1}{ \left[ x^2 y - \frac{y^3}{3} \right]_{y=-2}^{y=2} ~dx } }\)

\(\displaystyle{ \int_{-1}^{1}{ \left[ 2x^2 - \frac{(2)^3}{3} \right] - \left[ -2x^2 - \frac{(-2)^3}{3} \right] ~dx } }\)

\(\displaystyle{ \int_{-1}^{1}{ 4x^2 - \frac{16}{3} ~dx } }\)

\(\displaystyle{ \left[ \frac{4x^3}{3} - \frac{16x}{3} \right]_{-1}^{1} }\)

\(\displaystyle{ \left[ \frac{4}{3} - \frac{16}{3} \right] - \left[ \frac{4(-1)^3}{3} - \frac{16(-1)}{3} \right] }\)

\(\displaystyle{ \frac{8}{3} - \frac{32}{3} = \frac{-24}{3} = -8 }\)

Final Answer

\( -8 \)

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\(\displaystyle{ \int_{1}^{3}{ \int_{2}^{3}{ x^2 - 2xy + 2y^3 ~ dy } ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{1}^{3}{ \int_{2}^{3}{ x^2 - 2xy + 2y^3 ~ dy } ~dx } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

720 video

video by Dr Chris Tisdell

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\(\displaystyle{ \int_{2}^{3}{ \int_{0}^{1}{ x^2 y ~dx } ~dy } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{2}^{3}{ \int_{0}^{1}{ x^2 y ~dx } ~dy } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

724 video

video by Dr Chris Tisdell

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\(\displaystyle{ \int_{1}^{3}{ \int_{0}^{2}{ xy + x^2y^3 ~dy } ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{1}^{3}{ \int_{0}^{2}{ xy + x^2y^3 ~dy } ~dx } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

730 video

video by PatrickJMT

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Evaluate \(\displaystyle{ \iint\limits_R { 2y - 3x^2y^2 ~ dA } }\) over the region \( R = \{ (x,y) ~|~ 0 \leq x \leq 1,~ 0 \leq y \leq 2 \} \)

Problem Statement

Evaluate \(\displaystyle{ \iint\limits_R { 2y - 3x^2y^2 ~ dA } }\) over the region \( R = \{ (x,y) ~|~ 0 \leq x \leq 1,~ 0 \leq y \leq 2 \} \)

Final Answer

\( 4/3 \)

Problem Statement

Evaluate \(\displaystyle{ \iint\limits_R { 2y - 3x^2y^2 ~ dA } }\) over the region \( R = \{ (x,y) ~|~ 0 \leq x \leq 1,~ 0 \leq y \leq 2 \} \)

Solution

3518 video

Final Answer

\( 4/3 \)

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Evaluate \(\displaystyle{ \iint\limits_R { (4x+y)^3 ~ dA } }\) over the region \( R = [0, 2] \times [0, 3] \)

Problem Statement

Evaluate \(\displaystyle{ \iint\limits_R { (4x+y)^3 ~ dA } }\) over the region \( R = [0, 2] \times [0, 3] \)

Final Answer

\( 3201/2 \)

Problem Statement

Evaluate \(\displaystyle{ \iint\limits_R { (4x+y)^3 ~ dA } }\) over the region \( R = [0, 2] \times [0, 3] \)

Solution

3520 video

Final Answer

\( 3201/2 \)

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\(\displaystyle{ \int_{1}^{4}{ \int_{\pi}^{3\pi/2}{ x \sin(xy) ~dx } ~dy } }\)

Problem Statement

\(\displaystyle{ \int_{1}^{4}{ \int_{\pi}^{3\pi/2}{ x \sin(xy) ~dx } ~dy } }\)

Final Answer

\( -1 \)

Problem Statement

\(\displaystyle{ \int_{1}^{4}{ \int_{\pi}^{3\pi/2}{ x \sin(xy) ~dx } ~dy } }\)

Solution

3527 video

video by Maths With Jay

Final Answer

\( -1 \)

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\(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ 3x + 4y ~dx } ~dy } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{0}^{2}{ \int_{0}^{4}{ 3x + 4y ~dx } ~dy } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

732 video

video by Krista King Math

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\(\displaystyle{ \int_{-1}^{2}{ \int_{1}^{3}{ 2x - 7y ~dy } ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{-1}^{2}{ \int_{1}^{3}{ 2x - 7y ~dy } ~dx } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

733 video

video by Krista King Math

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\(\displaystyle{ \int_{0}^{3}{ \int_{0}^{3}{ xy + 7x + y ~dx } ~dy } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{0}^{3}{ \int_{0}^{3}{ xy + 7x + y ~dx } ~dy } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

734 video

video by Krista King Math

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Integrate \(\displaystyle{ f(x,y) = 2xy - 3y^2 }\) over the region \( R=[-1,1] \times [-2,2] \)

Problem Statement

Integrate \(\displaystyle{ f(x,y) = 2xy - 3y^2 }\) over the region \( R=[-1,1] \times [-2,2] \)

Solution

735 video

video by Krista King Math

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\(\displaystyle{ \int_{1}^{4}{ \int_{1}^{2}{ \frac{x}{y} + \frac{y}{x} dy} ~dx } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{1}^{4}{ \int_{1}^{2}{ \frac{x}{y} + \frac{y}{x} dy} ~dx } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

1282 video

video by Krista King Math

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Evaluate the integral \(\displaystyle{ \iint\limits_R{ \frac{xy^2}{x^2+1} ~dA } }\) over the region \( R=\{(x,y)~:~ 0 \leq x \leq 1, ~ -3 \leq y \leq 3 \} \)

Problem Statement

Evaluate the integral \(\displaystyle{ \iint\limits_R{ \frac{xy^2}{x^2+1} ~dA } }\) over the region \( R=\{(x,y)~:~ 0 \leq x \leq 1, ~ -3 \leq y \leq 3 \} \)

Solution

1292 video

video by Krista King Math

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\(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Final Answer

\( \sqrt{3}/8 \)

Problem Statement

Evaluate the integral \(\displaystyle{ \int_{0}^{ \sqrt{2\pi/3}}{ \int_{0}^{\sqrt{7\pi/6}}{ xy ~ \sin(x^2+y^2) ~dx } ~dy } }\), giving your answer in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

Solution

1925 video

video by MIP4U

Final Answer

\( \sqrt{3}/8 \)

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\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)

Problem Statement

\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)

Hint

Use the trig substitution \( x = y\tan\theta \).

Problem Statement

\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)

Final Answer

\(\pi/4 - \arctan(1/2)\)

Problem Statement

\(\displaystyle{ \int_{1/2}^{1}{ \int_0^1{ \frac{y^2-x^2}{(x^2+y^2)^2} ~dx } ~dy } }\)

Hint

Use the trig substitution \( x = y\tan\theta \).

Solution

2222 video

video by Krista King Math

Final Answer

\(\pi/4 - \arctan(1/2)\)

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Trig Formulas

The Unit Circle

The Unit Circle [wikipedia]

Basic Trig Identities

Set 1 - basic identities

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\)

\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\)

\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\)

\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\)

Set 2 - squared identities

\( \sin^2t + \cos^2t = 1\)

\( 1 + \tan^2t = \sec^2t\)

\( 1 + \cot^2t = \csc^2t\)

Set 3 - double-angle formulas

\( \sin(2t) = 2\sin(t)\cos(t)\)

\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\)

Set 4 - half-angle formulas

\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\)

\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\)

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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Double Integrals Basics

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Practice Instructions

Unless otherwise instructed, evaluate these integrals, giving your answers in exact terms. If it is not possible to evaluate the integral in the given form, try changing the order of integration.

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