Double Integrals, Rectangular Region

ohlala191785
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Homework Statement


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].


Homework Equations



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

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.
 
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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?
 
k=1/32.
Sorry, I still don't know how to proceed.
 
ohlala191785 said:
k=1/32.
No, you're not taking into account the area of R. Try that again.
 
Ohh is it (1/32)(1/4-0)(1/2-1/4)?
 
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.
 
ohlala191785 said:
k(1/4-0)(1/2-1/4)=1/32.
Yes. So what is k? Can you see why sin∏xcos∏y <= that everywhere in R?
 
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?
 
ohlala191785 said:
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?

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
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
ohlala191785 said:
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?

Right, on those particular intervals.
 
  • #12
Alright then. Thank you!
 
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