Reversing order of integration

  • Thread starter Fusi
  • Start date
  • #1
4
0

Homework Statement



Reverse the order of integration:

[tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]


This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?
 

Answers and Replies

  • #2
35,428
7,288

Homework Statement



Reverse the order of integration:

[tex]\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy[/tex]


This was on a test I took today. I was having trouble with it and was wondering how I should have gone about it?

Sketch a graph of the region over which integration is taking place. In the integral above, the inner integral runs from x = y2 to x = 2 - y. The outer integral runs from y = 0 to y = 1. When you reverse the order of integration for this problem you will end up with two iterated integrals.
 
  • #3
4
0
So x=2-y and x=[itex]y^{2}[/itex] so y=2-x y=[itex]\sqrt{x}[/itex]

intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

y at 0 is (0,2)

I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]

I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

[tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]
 
Last edited:
  • #4
35,428
7,288
So x=2-y and x=[itex]y^{2}[/itex] so y=2-x y=[itex]\sqrt{x}[/itex]

intersect at: [itex]2-x=\sqrt{x}[/itex] [itex]x=1[/itex] [itex]y=1[/itex]

y at 0 is (0,2)

I got confused because at y=1,x=1 each side can be explained differently so my first attempt at a solutions was:

[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx[/tex]
This isn't correct, but it's closer than what you have below. Here's what you should get:
[tex]\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{0}^{2 - x} f(x,y) \ dy\ dx[/tex]
I used some random logic in my head to get that and we've never seen anything like that in class before so I crossed it out and just wrote:

[tex]\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx[/tex]
 
  • #5
4
0
Yea I made a mistake on my last post. I actually did write my second bound as [itex]{0}\leq{y}\leq{2-x}[/itex]

We've never seen a problem like that before so I got so nervous I just crossed the entire answer out. The answer I decided to write down in the last couple seconds was actually:

[tex]\int_{0}^{2}\int_{\sqrt{x}}^{2-x} f(x,y) \ dy\ dx[/tex]

I kind of just tried to push the two integrals from my past answer together so I wouldn't have two iterated integrals.
 
Last edited:
  • #6
35,428
7,288
You understand why a single integral won't work, right?
 
  • #7
4
0
Yea, I have a bad habit of writing something that I know can't work when I get nervous.
 

Related Threads on Reversing order of integration

  • Last Post
Replies
1
Views
2K
Replies
9
Views
2K
Replies
2
Views
1K
Replies
17
Views
872
  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
8
Views
4K
Top