Stuck on the last questions of my homework that's due in tomorrow. Somebody help. I think it's to do with integration but i can't do it. Can anyone give me a hint?
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John
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Hint: Join the centre to the $4$ corners of the orange (?) figure. That divides the figure into $4$ equal parts. Let's find the area of one of the parts, say the "top" one.
Draw horizontal and vertical axes through the centre. The top line of the square is $y=1$.
For reasonable $x$ and $y$, the point $(x,y)$ is on the top curvy part of the boundary if $1-y=\sqrt{x^2+y^2}$. This is the parabola $2y=1-x^2$.
The rest is calculation of a kind you are expected to be able to do,
André Nicolas
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It should be $2-y=\sqrt{x^2+y^2}$ – Francisco Mar 03 '14 at 22:20
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Thank you, I figured that you had to split it in 4 parts, tried everythin with cos 45 and other useless stuff, the formula is what it needed, i get it now – John Mar 03 '14 at 22:25
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@Francisco: Maybe, but I don't think so. I said to make the centre the origin, and we are looking at the "top" quarter. The equation of the top line of the square is $y=1$. So the vertical distance from $(x,y)$ to the top line is $1-y$. – André Nicolas Mar 03 '14 at 22:25
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@John: You are welcome. Because I don't like negative numbers, I would probably split into $8$ parts. – André Nicolas Mar 03 '14 at 22:28
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As you say, setting $x=0$ must give $y=1$. That's why it's $2-y$ – Francisco Mar 03 '14 at 22:29
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@Francisco: Setting $x=0$ must give $y=\frac{1}{2}$. – André Nicolas Mar 03 '14 at 22:33
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Ohh you are right sorry for insisting, I had another picture in my head – Francisco Mar 03 '14 at 22:35