How to Correctly Calculate This Basic Double Integral?

[NewtonianAlch]

Homework Statement

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

The Attempt at a Solution

The given answer is (a^4)/3

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

http://img225.imageshack.us/img225/7238/dsc0016xd.jpg

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

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.

 

[NewtonianAlch]

Thank you!