# Difficulty finding integral answer to set function

1. Dec 13, 2011

### wown

Note: as i was typing this post, i think i figured out the answer but i want to confirm.
Question:
let A denote the set of points that are interior to, or on the boundary of, a square with opposite verties at the points (0,0) and (1,1). let Q(A) = ∫∫dydx
if C, a subset of A, is the set {(x,y): 0 < x/2 ≤ y ≤ 3x/2 < 1}, compute Q(A)

Solution:
I kept getting 1/2 as the answer, but i think my limits of integration were wrong. is the following correct?

$^{2/3}_{0}$∫$^{3x/2}_{x/2}$∫dydx

this setup gives the answer as 2/9... unfortunately i do not have the correct answer which is why i needed to verify.

2. Dec 14, 2011

### mathman

The question is not completely clear. For x > 2/3 is anything included?
This would involve: x/2 < y < 1.

3. Dec 15, 2011

### wown

Unfortunately, i typed the question verbatim and not sure how to respond to you.

note that the 2/3 upper limit on x was defined by me. I figured that y <1 (from y ≤ 3x/2 < 1). so therefor x cannot be more than 2/3, since y must also be less than 3x/2. for x> 2/3, y>1. e.g. when x=1, y=3/2.

does that make sense?

4. Dec 15, 2011

### mathman

No. Your condition x/2≤y≤3x/2 for x > 2/3 restricts y to x/2≤y≤1, but does not remove it altogether.