- #1

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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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- Thread starter DavidAp
- Start date

- #1

- 44

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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.

- #2

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- 0

- #3

- 44

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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?

- #4

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And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.

- #5

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Thank you so much, not just on this problem but on future problem to come! I never thought of visualizing it that way!

And just draw the lines [itex]y=0[/itex] and [itex]y=x[/itex] and you should be able to see where [itex]y[/itex] satisfies [itex]0{\leq}y{\leq}x[/itex]. Or think about it this way, where does [itex](x,y)[/itex] satisfy [itex]0{\leq}y[/itex] and where does it satisfy [itex]y{\leq}x[/itex]. Then you just find the intersection of those two areas.

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