# Evaluate the double integral

1. Oct 29, 2012

### Mdhiggenz

1. The problem statement, all variables and given/known data
∫∫x2dA; R is the region in the first quadrant enclosed by

xy=1, y=x, and y=2x.

First thing I did was notice that I had to find dydx, then

I graphed y=1/x, y=x, and y=2x.

Graphing I say that the limit of dy lie between x≤y≤2x

However I get confused as to how they want us to find dx.

My first approach was to use y=1/x and y=x and set them equal to each other, and solve for x. Doing so I get x2=1 or x = 1 getting only positive values since it lies in the first quadrant.

then I use set y=1/x, and y=2x equal to each other, and solve for x ; 1/x=2x thus giving me

x=√(2)/2.

Here is where I get confused the dx limit goes from 0≤x≤√(2)/2

I understand that dx would be the maximum value of x which is √(2)/2, but why wouldnt it be 1.

Also now giving it some thought, why did they choose to use 1/x and equal it to the other values, and not the graph of 2x or x?

Thanks

2. Relevant equations

3. The attempt at a solution

2. Oct 29, 2012

### Ray Vickson

The first thing you need to do is to describe the enclosed region correctly, and preferably, draw a picture with the enclosed region shaded in. The lines y = x, y = 2x and the curve xy=1 describe the boundary of the enclosed region. You need to know what are all the other points in the region. The inequality x ≤ y ≤ 2x that you gave above is only part of the story.

Within the interior of the region there is no relationship between dx and dy; I don't understand why you would think there was a connection between the two. (Again, maybe because you did not draw a picture?)

RGV