I thought I understood double integrals until I saw this

[1MileCrash]

Homework Statement

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

Homework Equations

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!

 

[hqjb]

Homework Statement

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

Homework Equations

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.

 

[1MileCrash]

Can anyone clarify?

 

[hqjb]

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...

 

[1MileCrash]

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?

 

[LCKurtz]

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.

 

[Ray Vickson]

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

 

[1MileCrash]

*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.

Thank all for your input.

 

[LCKurtz]

*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.

Thank all for your input.

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.

 

[Ray Vickson]

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

 

[LCKurtz]

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 $\int _0^3\int_y^3\sin(x^2)dxdy$