Double Integral problem What am I suppose to do? Related to polar coordinates.

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SUMMARY

The discussion centers on solving a double integral problem involving polar coordinates, specifically the limits of integration for θ. The user initially set θ from 0 to cos(θ), which was corrected to range from 0 to π/4. The user also struggled with visualizing the region defined by 0 ≤ y ≤ x, which is crucial for setting the correct limits. The correct answer to the integral is 3/64 π², but the user mistakenly calculated it as 3/16 π² due to an error in the limits of integration.

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DavidAp
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The problem and my work is shown in the image below. However, I feel like I did something horrible wrong but I'm not sure where!

24xjrl5.jpg


I'm sorry if my handwriting is illegible. If you're having difficulties please leave a comment and I will not hesitate to type it out as a response. Any guidance is greatly appreciated.

Thank you for taking the time to review my question.
 
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You've got \theta ranging from 0 to cos(\theta), that doesn't make any sense. I'm fairly sure you should have \theta going from 0 to \frac{\pi}{4}. Which you'll be able to see if you draw a picture of R.
 
Stimpon said:
You've got \theta ranging from 0 to cos(\theta), that doesn't make any sense. I'm fairly sure you should have \theta going from 0 to \frac{\pi}{4}. Which you'll be able to see if you draw a picture of R.

I don't know how to draw 0 <= y <= x. I think that might be part of my problem...

I did something wrong though when using pi/2! The answer is 3/64 pi^2 but I keep getting 3/16 pi^2! Why is my denominator 4x less than the answer? Here's my work.

∫(1 -> 2) ∫(0 -> pi/2) θr drdθ
1/2 ∫(1 -> 2) θ(r^2)(0 -> pi/2) dθ
1/2 ∫(1 -> 2) θ(pi^2)/4 dθ
(pi^2)/8 ∫(1 -> 2) θ dθ
(pi^2)/16 (θ^2)(1 -> 2)
(pi^2)/16 (2^2 - 1^1)
(pi^2)/16 (4 - 1)
(pi^2)/16 (3)
3/16 pi^2

Any insight on what I did wrong?
 
Ah I think I didn't edit my post in time then, the upper limit should have been \frac{\pi}{4}, sorry. Your method is fine.

And just draw the lines y=0 and y=x and you should be able to see where y satisfies 0{\leq}y{\leq}x. Or think about it this way, where does (x,y) satisfy 0{\leq}y and where does it satisfy y{\leq}x. Then you just find the intersection of those two areas.
 
Stimpon said:
Ah I think I didn't edit my post in time then, the upper limit should have been \frac{\pi}{4}, sorry. Your method is fine.

And just draw the lines y=0 and y=x and you should be able to see where y satisfies 0{\leq}y{\leq}x. Or think about it this way, where does (x,y) satisfy 0{\leq}y and where does it satisfy y{\leq}x. Then you just find the intersection of those two areas.
Thank you so much, not just on this problem but on future problem to come! I never thought of visualizing it that way!
 

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