# Boundaries for Double Integral Area

• Oct 17th 2009, 05:55 PM
messianic
Boundaries for Double Integral Area
I'm trying to find the rea of the region bounded by:

y=x

y=2x

xy=2

I'm very confused about how to find the boundaries for x and y. Anyone?
• Oct 17th 2009, 06:02 PM
Krizalid
first make a sketch of the situation to identify the region.

due to symmetry, we duplicate our result. Now, we have a region which goes for $\displaystyle 0\le x\le1$ and $\displaystyle 1\le x\le\sqrt2.$

can you set up the double integral?
• Oct 17th 2009, 06:18 PM
messianic
so I will have two separate double integrals then?

what would the y values be?
• Oct 17th 2009, 06:31 PM
Krizalid
actually those $\displaystyle y$s are the functions, can you figure out how to set up the first double integral? if you can do that, then you'll be able to set up the second one.
• Oct 17th 2009, 06:34 PM
messianic
So, for the first double integral, the x range goes from 0 to 1, and the y is from x to 2x? Is that correct
• Oct 17th 2009, 06:38 PM
Krizalid
yes, so you're done!
• Oct 17th 2009, 06:50 PM
messianic
are the y-boundaries teh same for teh 2nd integral?
• Oct 17th 2009, 07:03 PM
mr fantastic
Quote:

Originally Posted by messianic
are the y-boundaries teh same for teh 2nd integral?

The y-boundaries define a region that must be broken up into two smaller regions. Did you draw a sketch of the situation as suggested?

The terminals for the double integral will depend on what the order of integration is. From post #5 it appears that you're integrating first with respect to y. So the first double integral you have is correct (and this has already been confirmed by Krizalid) and will give the area of the first smaller region.

Now you need to think about the second double integral that gives the second smaller region: $\displaystyle \int_{x = }^{x = } \int_{y = }^{y = } dy \, dx$

where a sketch of the situation (that I assume you drew) should show you what the missing integral terminals are.
• Oct 17th 2009, 07:06 PM
messianic
Well the y terminal for the 2nd integral is confusing because you have to factor in xy=2
• Oct 17th 2009, 07:07 PM
mr fantastic
Quote:

Originally Posted by messianic
Well the y terminal for the 2nd integral is confusing because you have to factor in xy=2

y = 2/x.
• Oct 17th 2009, 07:24 PM
messianic
so the y-paramenter for the 2nd double integral goes from y=x to y=2/x?
• Oct 17th 2009, 07:31 PM
mr fantastic
Quote:

Originally Posted by messianic
so the y-paramenter for the 2nd double integral goes from y=x to y=2/x?

Yes.