# Double Integral question

1. Oct 30, 2007

### engineer_dave

1. The problem statement, all variables and given/known data

Evaluate the integral shown in the diagram

2. Relevant equations

3. The attempt at a solution

The first step to evaluating the integral is shown in the diagram (labelled as 2). They said they changed the order of integration. I was wondering what they mean by changing the order of integration and how did they come up with that first step. Cheers!

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2. Oct 30, 2007

### HallsofIvy

Staff Emeritus
The first integral is
$$\int_{y=0}^1\left[\int_{x= \sqrt{y}}^1 \sqrt{x^2+1}dx\right] dy$$
and the second is
$$\int_{x=0}^1\left[\int_{y= 0}^{x^2}\sqrt{x^2+ 1}dy\right] dx$$
I've included "x= " and "y= " in the limits of the integrals and put in braces, to make it clearer that, in the first line, they are first integrating with respect to x and then with respect to y, while in the second line, they are first integrating with respect to y and then with respect to x. That's signaled in two ways: the first line has "dx dy" while the second line has "dy dx" and the "inner" integral in the first line has a function of y as a limit of integration while the "inner" integral in the second line has a function of x as a limit of integration.

The way to see how to change the limits of integration when changing the order of integration is:

First draw a picture of the region you are integrating over: the "outer" integral in the first line has limits of integration y= 0 and y= 1 (We know they are "y= " because the "outer" integral is with respect to y) so we draw horizontal straight lines at y= 0 and y= 1. The "inner" integral has limits of $x= \sqrt{y}$ and x= 1 so we draw the graphs of $x= \sqrt{y}$ (that's the right half of y= x2) and x= 1 (a vertical line). For each y, (looking at the inner integral again) x goes from $\sqrt{y}$ up to 1 so we are looking at the area to the right of the parabola.

Now "reverse" the order. In order to cover the entire area it should be clear that x must go from 0 to 1. If the "outer" integral is with respect to x, then the limits must be x= 0 up to x= 1. For each x now, y must run from the lower line boundary, y= 0, up to the parabola boundary, y= x2. That gives the limits of integration for the "inner" integral.

Last edited: Oct 30, 2007
3. Oct 30, 2007

### engineer_dave

cheers mate. understand it loads better now!