# Evaluating a basic double integral

1. May 1, 2012

### NewtonianAlch

1. The problem statement, all variables and given/known data
Integrate the following function f over the given region Ω:

f(x,y) = xy; Ω bounded by y = 0, x = 2a, and x^2 = 4a

3. The attempt at a solution

The answer I got was (a^4)...ignore the answer I put in the attachment at the end, slight computational error.

http://img831.imageshack.us/img831/1600/dsc0017ye.jpg [Broken]

http://img225.imageshack.us/img225/7238/dsc0016xd.jpg [Broken]

Last edited by a moderator: May 6, 2017
2. May 1, 2012

### HallsofIvy

Staff Emeritus
So the boundaries are given by y= 0, x= 2a, and $x^2= 4ay$? (You forgot the "y" in the last equation.) Then x, over all, goes from 0 on the leftt ($x^2= 4ay$ and y= 0 intersect where $x^2= 0$) to x= 2a on the right. Then for each x y goes from y= 0 to $y= x^2/4a$. So your integral would be
$$\int_{x=0}^{2a}\int_{y=0}^{x^2/4} xy dydx$$

You may have misled yourself by miswriting $x^2= 4ay$ as $x^2= 4a$.

3. May 1, 2012

### NewtonianAlch

Ah. I mistakenly typed it like that, however I still calculated it with $x^2 = 4ay$ on paper, my mistake was with the bound of the second integral.

I don't understand how that upper bound of y is $y = (x^2)/4a$

I thought if the upper bound of x was 2a, then we substitute that into y-equation and hence get the upper bound of y. That's how I ended up getting "a" as the upper bound, which I can now see is incorrect, but I don't understand why.

4. May 1, 2012

### HallsofIvy

Staff Emeritus
No, what you are trying to do is wrong. In order that the result of a double integral be a number, the limits of integration on the "outer" integral must be constants but the "inner" integral may have the "outer" variable in its limits of integration. In fact, an integral like
$$\int_a^b \int_c^d f(x,y) dy$$
is over the rectangle with boundaries x= a, x= b, y= c y= d. To have an integration over any region other than a rectangle, limits of integration on the "inner" integral must be functions, not constants.

Yes,, x goes from 0 to 2a and, for each x, y goes from y=0 up to the upper boundary, $y=x^2/4a$.

5. May 1, 2012

Thank you!