# Understanding the graph for reversing an integral

1. Jul 28, 2008

### lalah

1. The problem statement, all variables and given/known data
I'm trying to reverse the order of integration of
$$\int_0^{81}\int_{y/9}^{\sqrt{y}} dx dy$$
first integral is from 0 to 81
second integral is from y/9 to $$\sqrt{y}$$

The problem is, I identified the inequalities for the regions of integration, but I'm having problems understanding how the graph is formed.
2. Relevant equations

3. The attempt at a solution
I understand the inequalities are y/9 < x < $$\sqrt{y}$$
and 0 < y <81.

But I don't understand the graph.
x-axis is 0 to 9
y-axis is 0 to 81
y = x^2
y = 9x

And since according to the homework problem I'm doing, I can't progress the problem until I understand why the graph is so.

Also, can someone help me in formating the integrals in latex?

Last edited by a moderator: Jul 28, 2008
2. Jul 28, 2008

### HallsofIvy

Staff Emeritus
Draw a picture. In particular, graph x= 1/y and $x= \sqrt{y}$ which are the same as the hyperbola y= 1/x and the parabola y= x2. Now one problem you have here is that those two graph cross at (1,1). The area over which you are integrating is the area from the curve y= x2 on the left to the curve y= 1/x on the right up until y= 1, then from the curve y= 1/x on the left up to the curve y= x2 on the right.

Reversing the order of integration, you will need to divide this into four integrals. To cover the lower area, below y= 1, because of the change at (1,1), you will need to integrate from 0 up to y= x2 with x from 0 to 1, then from 0 up to y= 1/x, with x from 0 to 1/81. To cover the upper area, above y= 1, you need to integrate from y= 1/x up to y= 81, with x from 1/81 to 1, then from y= x2 up to y= 81, with x from 1 to up to 9:
$$\int_{x=0}^1\int_{y= 0}^{x^2} dy dx+ \int_{x=1}^\infty \int{y= 0}^{1/x} dy dx+ \int_{x= 1/81}^1 \int_{y=1/x}^81 dy dx+ \int_{x=1}^9\int_{y= x^2}^{81} dy dx$$

3. Jul 29, 2008

### lalah

I'm suck on that step. How do you determine which formulas to graph? I'm looking at the equations and inequalities, but I can't arrive to that step.

I can do reversing of integration, but this is the pothole in the bridge.

4. Jul 29, 2008

### Defennder

The graphs you should sketch are those determined by the limits of the integrands. The limits for x are x=y/9 and x=sqrt{y}. I think HallsOfIvy meant x=y/9 instead of 1/y.

5. Jul 29, 2008

### lalah

Oh yes, I get it now. Thanks! :)

6. Jul 29, 2008

### HallsofIvy

Staff Emeritus
Right! my eyes are going!