# Double Integral

Discussion in 'Homework Help' started by u-will-neva-no, Feb 7, 2012.

1. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Hello everyone, I am having difficulty with this double integral.

I was given the following integral:
$I =\int ^1_0 dx \int ^{2x}_x 2xy dy$

I have to reverse the order of integration and so this was my approach:

I first plotted the graph of the above and is attached as 'integrate along y'.

So then I integrated along x so the lower limit is y and y/2. The second image to show what i was thinking is shown as 'figure 2'.

So the integral that I get is:
$I =\int ^1_0 dy \int ^{y}_{y/2} 2xy dx$

However, my solution comes to 3/16 whereas the answer should be 3/4.

Thanks fore reading and hopefully one could explain what I am doing wrong!

File size:
11.4 KB
Views:
35
File size:
14.8 KB
Views:
32
2. ### steveb Senior Member

Jul 3, 2008
2,433
469
The region you are integrating over is bounded by three lines, not two.

There is y=x, y=2x and x=1. They form a triangle shape. Recheck your limits and you will see that your new integral does not span over the same triangular region.

u-will-neva-no likes this.
3. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Thanks steveb. One other issue that I am having is the following. For the first integral that I wrote, the limits for the first part of the integral are 0 and 1. I used the second integral limits to arrive at the two equations (y=x and y=2x) and from previous examples, i would combine the two together to arrive at the limits for the first integral. This does not work in this instance.

The reason why I am asking this is because I am confused what to put for the first integral for when I reverse the order of integration.

4. ### u-will-neva-no Thread Starter Member

Mar 22, 2011
230
2
Actually, that was a silly question, because the lines do not cross!! Please ignore my above post