# Double Integrals, Rectangular Region

1. Jan 13, 2013

### ohlala191785

1. The problem statement, all variables and given/known data
Using ∫∫kdA = k(b-a)(d-c), where f is a constant function f(x,y) = k and R = [a,b]x[c,d], show that 0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32, where R = [0,1/4]x[1/4,1/2].

2. Relevant equations

∫∫kdA = k(b-a)(d-c)
0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32

3. The attempt at a solution

I tried to integrate ∫∫sin∏xcos∏ydA, but that didn't utilize the ∫∫kdA = k(b-a)(d-c) part and I didn't use the fact that ∫∫sin∏xcos∏ydA is between 0 and 1/32. I don't know how to incorporate all of these aspects to show that 0 ≤ ∫∫sin∏xcos∏ydA ≤ 1/32.

2. Jan 13, 2013

### haruspex

If you were to show that 0 ≤ sin∏xcos∏y ≤ k everywhere in R, what would you need k to be in order to prove the desired answer?

3. Jan 13, 2013

### ohlala191785

k=1/32.
Sorry, I still don't know how to proceed.

4. Jan 13, 2013

### haruspex

No, you're not taking into account the area of R. Try that again.

5. Jan 13, 2013

### ohlala191785

Ohh is it (1/32)(1/4-0)(1/2-1/4)?

6. Jan 13, 2013

### ohlala191785

Wait hold on is it ∫∫kdA = 1/32, for R = [0,1/4]x[1/4,1/2]? So k(1/4-0)(1/2-1/4)=1/32.

7. Jan 13, 2013

### haruspex

Yes. So what is k? Can you see why sin∏xcos∏y <= that everywhere in R?

8. Jan 14, 2013

### ohlala191785

k=0.5 I graphed sin∏xcos∏y on Mathematica and saw that for x in [0,1/4] and y in [1/4,1/2], z is between 0 and 0.5. And solving for k=0.5 agrees with the graph.

I'm also wondering if you can know that z is from [0,0.5] without graphing it on a computer or something?

9. Jan 14, 2013

### Dick

sin(pi*x) is less than sqrt(2)/2 on [0,1/4]. cos(pi*x) is less than sqrt(2)/2 on [1/4,1/2]. So?

10. Jan 15, 2013

### ohlala191785

So since the min of sin(pi*x) is 0 and max is sqrt(2)/2, and it's the same for cos(pi*x), then the min of their product is 0 and max is 0.5. Is that correct?

11. Jan 15, 2013

### Dick

Right, on those particular intervals.

12. Jan 15, 2013

### ohlala191785

Alright then. Thank you!