## 17Calculus - Double Integrals - Rectangular Regions

##### 17Calculus

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

Problem Statement

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?

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

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

$$10$$

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.

### Michel vanBiezen - 3511 video solution

video by Michel vanBiezen

$$10$$

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

Problem Statement

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

$$10$$

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.

### Michel vanBiezen - 3512 video solution

video by Michel vanBiezen

$$10$$

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

Problem Statement

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

$$52/3$$

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.

### The Organic Chemistry Tutor - 3517 video solution

$$52/3$$

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

Problem Statement

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

$$2$$

Problem Statement

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

Solution

### Michel vanBiezen - 3513 video solution

video by Michel vanBiezen

$$2$$

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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

### The Organic Chemistry Tutor - 3521 video solution

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

Problem Statement

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

$$-8$$

Problem Statement

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

Solution

 $$\displaystyle{ \int_{-1}^{1}{ \int_{-2}^{2}{ x^2 - y^2 ~dy ~ dx } } }$$ $$\displaystyle{ \int_{-1}^{1}{ \left[ x^2 y - \frac{y^3}{3} \right]_{y=-2}^{y=2} ~dx } }$$ $$\displaystyle{ \int_{-1}^{1}{ \left[ 2x^2 - \frac{(2)^3}{3} \right] - \left[ -2x^2 - \frac{(-2)^3}{3} \right] ~dx } }$$ $$\displaystyle{ \int_{-1}^{1}{ 4x^2 - \frac{16}{3} ~dx } }$$ $$\displaystyle{ \left[ \frac{4x^3}{3} - \frac{16x}{3} \right]_{-1}^{1} }$$ $$\displaystyle{ \left[ \frac{4}{3} - \frac{16}{3} \right] - \left[ \frac{4(-1)^3}{3} - \frac{16(-1)}{3} \right] }$$ $$\displaystyle{ \frac{8}{3} - \frac{32}{3} = \frac{-24}{3} = -8 }$$

$$-8$$

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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

### Dr Chris Tisdell - 720 video solution

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

### Dr Chris Tisdell - 724 video solution

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

### PatrickJMT - 730 video solution

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 $$R = \{ (x,y) ~|~ 0 \leq x \leq 1,~ 0 \leq y \leq 2 \}$$

$$4/3$$

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

### The Organic Chemistry Tutor - 3518 video solution

$$4/3$$

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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 $$R = [0, 2] \times [0, 3]$$

$$3201/2$$

Problem Statement

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

Solution

### The Organic Chemistry Tutor - 3520 video solution

$$3201/2$$

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

Problem Statement

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

$$-1$$

Problem Statement

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

Solution

### Maths With Jay - 3527 video solution

video by Maths With Jay

$$-1$$

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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

### Krista King Math - 732 video solution

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

### Krista King Math - 733 video solution

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

### Krista King Math - 734 video solution

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

### Krista King Math - 735 video solution

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

### Krista King Math - 1282 video solution

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

### Krista King Math - 1292 video solution

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.

$$\sqrt{3}/8$$

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

### MIP4U - 1925 video solution

video by MIP4U

$$\sqrt{3}/8$$

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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

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

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

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

### Krista King Math - 2222 video solution

video by Krista King Math

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

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