# I thought I understood double integrals until I saw this

• 1MileCrash
In summary, the conversation revolved around a problem involving a double integral. There was confusion regarding the limits of integration and the correct solution. It was suggested that there may have been a typo in the original problem, causing the confusion. Ultimately, it was determined that the correct solution was likely ##\int _0^3\int_y^3\sin(x^2)dxdy##.
1MileCrash

## Homework Statement

http://img710.imageshack.us/img710/4764/doubleintegral.png

## The Attempt at a Solution

Now, my understanding of the region is that x spans from the line x = y to x = 1, and that given that parameter, the applicable y's are 0 to 1.

In other words, my region looks like a right triangle with corners at the (0,0), (1,0), (1,1).

Since the region ends at the line x = 1 due to the parameters of x, my understanding is that the upper limit on y is effectively 1, and that running it to 3 doesn't do anything. So reversing the order of integration, I get that y will span from 0 to y = x, and x will span from 0 to 1.

However, when I click "GO tutorial:"

http://img714.imageshack.us/img714/6250/doubleintegral2.png

And their final answer is .99.

This has literally destroyed every ounce of comfort I had with the process. WHY are we running x from 0 to 3 when the order of integration is reversed? That's not in the region!

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1MileCrash said:

## Homework Statement

http://img710.imageshack.us/img710/4764/doubleintegral.png

## The Attempt at a Solution

Now, my understanding of the region is that x spans from the line x = y to x = 1, and that given that parameter, the applicable y's are 0 to 1.

In other words, my region looks like a right triangle with corners at the (0,0), (1,0), (1,1).

Since the region ends at the line x = 1 due to the parameters of x, my understanding is that the upper limit on y is effectively 1, and that running it to 3 doesn't do anything. So reversing the order of integration, I get that y will span from 0 to y = x, and x will span from 0 to 1.

However, when I click "GO tutorial:"

http://img714.imageshack.us/img714/6250/doubleintegral2.png

And their final answer is .99.

This has literally destroyed every ounce of comfort I had with the process. WHY are we running x from 0 to 3 when the order of integration is reversed? That's not in the region!

Im still learning double integrals and my intuition was that the upper limit on y is still 3 but you take the negative of that area and add to whatever's under y=1 so you get a negative value.
Thats what I get when I used the double integral calculator. H/e the solutions you shown on the next page are all positive. So I am confused too, any one else please enlighten, I want to know what's going on too.

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Can anyone clarify?

1MileCrash said:
Can anyone clarify?

Hey sry if I am wrong about this but I checked the calculation and I believe the answer is wrong it should be
$\int{^1_0}\int{^x_0}\sin(x^2)dydx - \int{^3_1}\int{^3_x}\sin(x^2)dydx$
I got the same value -0.4343...

I would agree, the region as is is a triangle that flips at x=1 and is upside down until x=3. They reversed integration as if it were just one big triangle.

Can anyone confirm?

I agree with hqib's answer. So 1milecrash, I think you understand perfectly well. I think almost certainly this problem has some typo in its statement. This type of problem usually appears in a calculus text to illustrate that the order of integration can make a real difference in the difficulty of calculating the integral. But here it doesn't help because in that second integral ##\int_1^3\int_x^3\sin(x^2)dydx## you are going to have a ##3\sin(x^2)## to integrate after you do the inner integral, which is as bad as the original problem. Almost surely a typo.

1MileCrash said:
I would agree, the region as is is a triangle that flips at x=1 and is upside down until x=3. They reversed integration as if it were just one big triangle.

Can anyone confirm?

If we let $$F(y) = \int_{x=y}^1 \sin(x^2)\, dx \text{ for } y \leq 1 \text{ and } F(y) = -\int_{x=1}^y \sin(x^2) \, dx \text{ for } y > 1,$$ then F(y) can be evaluated in terms of Fresnel functions. Evaluation of the integral
$\int_0^3 F(y) \,dy$ involves hypergeometric functions. Maple gets its numerical value as -0.4343175641. I, personally, do not understand the 'tutorial'.

RGV

*sigh*

I feel better, I guess, but this kind of thing happens too often. It sucks not having confidence in your textbook's info.

I have never run into a double integral where the two limits did not "end" at the same place graphically.

1MileCrash said:
*sigh*

I feel better, I guess, but this kind of thing happens too often. It sucks not having confidence in your textbook's info.

I have never run into a double integral where the two limits did not "end" at the same place graphically.

Actually, I think the typo was that the original problem should have been:$$\int _0^3\int_y^3\sin(x^2)dxdy$$That would also explain their answer.

LCKurtz said:
Actually, I think the typo was that the original problem should have been:$$\int _0^3\int_y^3\sin(x^2)dxdy$$That would also explain their answer.

If we let $$F(a) = \int_{y=0}^a \int_{x=y}^a \sin(x^2) \, dx \, dy = \int_{x=0}^a \int_{y=0}^x \sin(x^2) \, dx \, dy = \frac{1}{2}[1 - \cos(a^2)],$$ then F(1) = 0.22985 and F(3) = 0.95557. If we want F(a) = 0.99, we need to take a = 1.715009566.

RGV

I don't know about the .99 or where it came from but the third choice they have marked here:

http://img714.imageshack.us/img714/6250/doubleintegral2.png

could have come from ##\int _0^3\int_y^3\sin(x^2)dxdy##

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## What is a double integral?

A double integral is a type of mathematical operation that involves finding the total area under a surface in two dimensions. It is essentially a two-dimensional version of a single integral, which is used to find the area under a curve in one dimension.

## How is a double integral different from a single integral?

A double integral involves integrating over a two-dimensional region, whereas a single integral involves integrating over a one-dimensional region. In other words, a double integral calculates the volume under a surface, while a single integral calculates the area under a curve.

## What are some applications of double integrals?

Double integrals are used in many fields of science and engineering, including physics, engineering, and economics. They can be used to calculate the total mass, center of mass, and moments of inertia of three-dimensional objects. They are also commonly used in solving differential equations and modeling physical systems.

## What are some common techniques for solving double integrals?

There are several techniques for solving double integrals, including the method of rectangular regions, the method of polar coordinates, and the method of cylindrical coordinates. Each technique is useful for different types of integrals and can be applied depending on the shape and complexity of the region being integrated over.

## How can I improve my understanding of double integrals?

One of the best ways to improve your understanding of double integrals is to practice solving various types of integrals. You can also review the fundamental concepts and properties of integrals, such as the Fubini's theorem and the change of variables theorem. Additionally, seeking help from a teacher or tutor can also improve your understanding and provide valuable insights and tips for solving double integrals.

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