# Reversing order of integration

## Homework Statement

Reverse the order of integration:

$$\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy$$

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

Mark44
Mentor

## Homework Statement

Reverse the order of integration:

$$\int_{0}^{1}\int_{y^2}^{2-y} f(x,y) \ dx\ dy$$

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.

So x=2-y and x=$y^{2}$ so y=2-x y=$\sqrt{x}$

intersect at: $2-x=\sqrt{x}$ $x=1$ $y=1$

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:

$$\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx$$

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:

$$\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx$$

Last edited:
Mark44
Mentor
So x=2-y and x=$y^{2}$ so y=2-x y=$\sqrt{x}$

intersect at: $2-x=\sqrt{x}$ $x=1$ $y=1$

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:

$$\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{2-x}^{1} f(x,y) \ dy\ dx$$
This isn't correct, but it's closer than what you have below. Here's what you should get:
$$\int_{0}^{1}\int_{0}^{\sqrt{x}} f(x,y) \ dy\ dx + \int_{1}^{2}\int_{0}^{2 - x} f(x,y) \ dy\ dx$$
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:

$$\int_{0}^{2}\int_{2-x}^{\sqrt{x}} f(x,y) \ dy\ dx$$

Yea I made a mistake on my last post. I actually did write my second bound as ${0}\leq{y}\leq{2-x}$

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:

$$\int_{0}^{2}\int_{\sqrt{x}}^{2-x} f(x,y) \ dy\ dx$$

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:
Mark44
Mentor
You understand why a single integral won't work, right?

Yea, I have a bad habit of writing something that I know can't work when I get nervous.