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LogicalApple

  • 2 years ago

Ok the attached image is from a calculus textbook ( http://www.stewartcalculus.com/media/11_home.php). The question is simple enough -- it says that this figure contains all points inside the square that are closer to the center than to the sides of the square. Find the area of this region. Anyone have an idea on how to get started? The image will be uploaded shortly.

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  1. LogicalApple
    • 2 years ago
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  2. KingGeorge
    • 2 years ago
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    I think, you'll need to do 2 integrals, and then using symmetry you should be able to apply these to the rest of the diagram. However, I'm not sure of the best way to set up the integrals. |dw:1356844485378:dw| This picture very roughly imitates your shape. The areas I've shaded need to be integrated, each one separately. The area on the left you can find by looking at the center of the square and the left side of the box. The area in the middle can be found by looking at the center of the square and the upper part of the square. You just need to find the equation of the line that is equidistant from the corresponding sides of the square and the center of the square. If I remember correctly, these should take the form of hyperbolas.

  3. KingGeorge
    • 2 years ago
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    Alright, I've got to go. Hopefully this was enough to get you started. I'm pretty sure that the lines you're looking for will be hyperbolic. Then you just have a couple definite integrals to solve.

  4. LogicalApple
    • 2 years ago
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    Thank you for your time!

  5. LogicalApple
    • 2 years ago
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    I will work on this right away

  6. wio
    • 2 years ago
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    It also might be interesting to note that a parabola is defined as a curve where all points are equidistant to a focus (point) and a directrix (line). so you can tell that these lines are parabolas.

  7. LogicalApple
    • 2 years ago
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    Hm. I was thinking parabolas, too. I think that might be just what I am looking for.

  8. wio
    • 2 years ago
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    From my calculus book: An equation of a (horizontal) parabola with focus \((0, p)\) and directrix \(y = -p\) is: \(x^2 = 4py\).

  9. wio
    • 2 years ago
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    Let the very bottom line be the directrix and let the very center be the focus...

  10. wio
    • 2 years ago
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    The center and directrix are 1 unit apart, so \[ |p-(-p)| = 1 \\ 2|p| = 1 \\ |p| = 1/2 \] Obviously \(p\). can't be negative

  11. wio
    • 2 years ago
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    \[ x^2 = 4(1/2)y = 2y \implies y = x^2/2 \]So this gives us the equation of the bottom line when the x-axis is the very center, and the y axis is halfway between the center (focus) and the bottom (directrix). @LogicalApple I think you can do the rest from here.

  12. LogicalApple
    • 2 years ago
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    Thank you so much!

  13. wio
    • 2 years ago
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    No problem, it was a bit of a learning experience for me as well.

  14. LogicalApple
    • 2 years ago
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    Would you agree with the points I drew in and labeled ?

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  15. wio
    • 2 years ago
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    Actually, let me think about it for a moment...

  16. wio
    • 2 years ago
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    I'm not sure of the logic used to get those points so I can't really confirm or deny... I am thinking perhaps you could find where the two parabolas would intercept.

  17. LogicalApple
    • 2 years ago
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    I based my calculations on these points. I had enough points to generate parabolas. One parabola was x = (2 - sqrt 2)y^2 - 1 This would he the parabola on the far left side. Since I'm only interested in a portion of it (namely from when x = -1 to when x = -sqrt (2) / 2) I could integrate y = sqrt( (1 + x) / (2 - sqrt (2)) from -1 to -sqrt (2)/2 Multiply this result by 4 and that would take care of the 4 triangular regions (I only drew two of them in the figure). For the parabola hanging overhead from x = -sqrt(2)/2 to sqrt(2)/2 I calculated y = (sqrt 2 - 2)x^2 + 1. Integrate this from -sqrt(2)/2 to sqrt(2)/2 and multiply the result by 2. This takes care of the remainder of the region I think. My end result was ( 8 sqrt(2) - 2)/3 But I have no way of verifying this

  18. LogicalApple
    • 2 years ago
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    be the*

  19. LogicalApple
    • 2 years ago
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    The boundary of this figure is equally as far from the center as it is from the perimeter of the square. So along the horizontal axis, the point (-1, 0) is as far from (0, 0) as it is from (-2, 0). I suppose I could have used (0, 1) as another point.... Instead I determined the point of the corner of the boundary as half of the diagonal of a square of hypotenuse 4 (2^2 + 2^2).

  20. LogicalApple
    • 2 years ago
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    Wait now that I think about it..

  21. LogicalApple
    • 2 years ago
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    Those numbers should be cut in half

  22. wio
    • 2 years ago
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    |dw:1356849810908:dw| I know it's not a circle but think of it this way.

  23. LogicalApple
    • 2 years ago
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    You're absolutely right. Back to the drawing board for me.

  24. wio
    • 2 years ago
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    |dw:1356849937491:dw|

  25. wio
    • 2 years ago
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    |dw:1356850006505:dw|

  26. wio
    • 2 years ago
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    In this case, it was just finding the right algebra.... lol

  27. wio
    • 2 years ago
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    \[ \sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2} \]

  28. LogicalApple
    • 2 years ago
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    Right, sqrt(2) . Hm.

  29. wio
    • 2 years ago
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    Yeah, well, just solve for \(x\)...

  30. wio
    • 2 years ago
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    \[ \sqrt{x^2+x^2}+x = \sqrt{2x^2}+x = \sqrt{2} \cdot x+x \]Thus \[ (1+\sqrt{2})x = \sqrt{2} \]@LogicalApple Make sense... convinced? Don't assume I'm a master, either. Look at what I did to check for mistakes.

  31. wio
    • 2 years ago
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    @LogicalApple Hmm?

  32. LogicalApple
    • 2 years ago
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    I'm still trying to figure out why x + y = sqrt(1^2 +1^2) instead of x + y = sqrt(2^2 + 2^2)?

  33. wio
    • 2 years ago
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    More stuff:\[ (\sqrt{2}+1)x = \sqrt{2} \\ (\sqrt{2}-1)(\sqrt{2}+1)x=(\sqrt{2}-1)\sqrt{2} \\ (2-1)x = 2-\sqrt{2} \\ x = 2-\sqrt{2} \]

  34. wio
    • 2 years ago
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    Because the total length is 2, but we're only doing halfway... which is 1.

  35. LogicalApple
    • 2 years ago
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    Yeah scratch that

  36. LogicalApple
    • 2 years ago
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    I am in agreement that x = 2 - sqrt 2

  37. LogicalApple
    • 2 years ago
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    I will modify my equations

  38. wio
    • 2 years ago
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    Okay, so what we can do if first just find the area of the inner square. Then we can find the area between one of the parabola and the inner square... and by symmetry just multiply that by 4.

  39. wio
    • 2 years ago
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    |dw:1356850703108:dw|

  40. wio
    • 2 years ago
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    4x as in 4 times

  41. LogicalApple
    • 2 years ago
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    I like that idea a lot actually

  42. wio
    • 2 years ago
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    The area of the square is easy. We just found half the length of its side. So just double that and square it.

  43. LogicalApple
    • 2 years ago
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    4(2 - sqrt 2)^2

  44. wio
    • 2 years ago
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    We also already found the equation of the parabola, and we know its limits.

  45. wio
    • 2 years ago
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    @LogicalApple Got an idea for finding the area under that parabola?

  46. LogicalApple
    • 2 years ago
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    I'm still uncertain of the equation of the parabola. I calculate something like: y = (sqrt 2 - 1) / (4 sqrt 2 - 6) * x^2 + 1

  47. wio
    • 2 years ago
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    I don't know how you're getting that...

  48. LogicalApple
    • 2 years ago
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    This parabola contains the points (-2 + sqrt 2, 2 - sqrt 2), (0, 1), and (2 - sqrt 2, 2 - sqrt 2) The top two corner points of the square and the vertex of the parabola

  49. LogicalApple
    • 2 years ago
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    I use a system of 3 equations. Where the general form is f(x) = ax^2 + bx + c I let x = -2 + sqrt 2, f(x) = 2 - sqrt 2 x = 0, f(x) = 1 x = 2 - sqrt 2, f(x) = 2 - sqrt (2)

  50. wio
    • 2 years ago
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    We could just ignore that it a parabola altogether and just try using the distance formula.

  51. wio
    • 2 years ago
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    Distance of some point \((x, y)\) to the line \(y=1\) \[ \sqrt{(x-x)^2(y-1)^2} \]Distance of some point \((x, y)\) to the origin \[ \sqrt{x^2+y^2} \]Set them equal \[ \sqrt{(x-x)^2(y-1)^2} = \sqrt{x^2+y^2} \]

  52. wio
    • 2 years ago
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    Should be a plus in there.

  53. LogicalApple
    • 2 years ago
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    Now I think I'm getting a rational answer of 8/3 for the area of the region.

  54. wio
    • 2 years ago
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    \[ \sqrt{(y-1)^2} = \sqrt{x^2 + y^2} \\ (y-1)^2 = x^2 + y^2 \\ y^2 - 2y + 1 = x^2 + y^2 \\ -2y + 1 = x^2 \\ y = -\frac{x^2}{2} +\frac{1}{2} \]We got a different equation because we had a different origin.

  55. wio
    • 2 years ago
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    But notice how our slope is pretty much the same.

  56. wio
    • 2 years ago
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    Just reflected because this time we took the concave down parabola at top.

  57. LogicalApple
    • 2 years ago
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    I'm following.. and I'm curious. I am going to review this entire post when it's all said and done.

  58. wio
    • 2 years ago
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    To find the area under parabola \(y = -x^2/2 + 1/2\) and above the line \(y = 2 - \sqrt{2} \), we take the integral: \[ \int_{1(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx \]

  59. wio
    • 2 years ago
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    \[ \int_{-(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx \]

  60. wio
    • 2 years ago
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    Sorry, been having quite a few typos today.

  61. LogicalApple
    • 2 years ago
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    Don't sweat it

  62. wio
    • 2 years ago
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    That's a pretty easy integral... but it just has a lot of sqrts and stuff that I am a bit reluctant to do it.

  63. LogicalApple
    • 2 years ago
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    Right but I can just use symmetry again. Plus there is wolfram alpha

  64. wio
    • 2 years ago
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    No it's an extremely easy integral it's just:\[ \int_{-(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx \\ = \left[ \frac{-x^3}{3\times 2} + \frac{x}{2} - (2-\sqrt{2})x \right] _{-(2-\sqrt{2})}^{2-\sqrt{2}} \]

  65. wio
    • 2 years ago
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    plugging that pellet in that is gonna be a pain in the retrice..

  66. wio
    • 2 years ago
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    Anyway, I know you can do the rest from here. Just look things over a couple times if you don't get anything I said. There maybe some typos but the general methodology is correct as far as I can see.

  67. LogicalApple
    • 2 years ago
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    \[\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} \frac{ \sqrt{2} - 1 }{ 4\sqrt{2} - 3} x^{2}dx -\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} (2 - \sqrt{2}) dx = 4\sqrt{2} - 16\] The above integral is based on the attached picture (below) It determines the area of one of the parabolic regions bounded by the inner square. Multiply this result by 4 and add the area of the square, we obtain 8/3 as the area of the region.

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  68. LogicalApple
    • 2 years ago
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    should be a dx in that second integral (sorry am still getting used to the equation creator)

  69. LogicalApple
    • 2 years ago
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    Oh wait there is already one there. Nvm. Anyway, thanks a lot to @wio for your patience. This was an unusually challenging problem but you found a way to simplify the heck out of it! And thanks to KingGeorge for visualizing it and getting me on the right track. OpenStudy is pretty cool..

  70. dumbcow
    • 2 years ago
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    after looking this over...i agree with the methods except i believe the intersection points are incorrect the limits should be +-(sqrt2 -1) Not +-(2-sqrt2) when \[-\frac{1}{2}x^{2}+\frac{1}{2} = x\] \[x = \sqrt{2} - 1\]

  71. wio
    • 2 years ago
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    That looks right... makes me wonder where I went wrong.

  72. LogicalApple
    • 2 years ago
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    I think your geometry and algebra are correct in concluding the corner points of the region do lay at +- (2 - sqrt 2). I will double check.

  73. wio
    • 2 years ago
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    @LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

  74. LogicalApple
    • 2 years ago
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    But didn't you calculate the distance from a point to the line y = 1 instead of to the line y = 2?

  75. LogicalApple
    • 2 years ago
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    Of course you did. Hm.

  76. LogicalApple
    • 2 years ago
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    Let me see what I obtain next.

  77. wio
    • 2 years ago
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    Wow it's really bothering me how I messed up. I can't figure out why my trig method was wrong.

  78. wio
    • 2 years ago
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    @dumbcow 's method makes sense... you want the point where our parabola intersects y = x... But my method seemed correct as well... gotta figure out the fallacy I made.

  79. wio
    • 2 years ago
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    Ooooh! I figured out my error!

  80. wio
    • 2 years ago
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    |dw:1356855949526:dw| We found \(x\)! We wanted \(z\) in the diagram!

  81. wio
    • 2 years ago
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    |dw:1356856038380:dw|

  82. wio
    • 2 years ago
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    \[ 2z^2 = x^2 \\ \sqrt{2}\cdot z= x \\ z = x/\sqrt{2} = \frac{2-\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}} - 1 = \sqrt{2}-1 \]Lol... really worry about that @LogicalApple Thank @dumbcow for spotting it.

  83. wio
    • 2 years ago
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    sorry I mean^

  84. LogicalApple
    • 2 years ago
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    Modifying .. 1 sec

  85. LogicalApple
    • 2 years ago
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    I did worry about it!

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  86. wio
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    That looks correct.

  87. LogicalApple
    • 2 years ago
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    I will integrate now.. and I have a feeling it will be irrational this time ;(

  88. LogicalApple
    • 2 years ago
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    I now see why there was a discrepancy with my earlier parabolic equation.

  89. wio
    • 2 years ago
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    I knew from the very start there was very little chance of a rational result.

  90. LogicalApple
    • 2 years ago
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    My image is still wrong. One more modification ..

  91. LogicalApple
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  92. wio
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    That's pretty much it. It's an easy integral because it's just a 2 degree polynomial.

  93. LogicalApple
    • 2 years ago
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    Final answer is ... \[\frac{ 4 }{ 3 }(4\sqrt{2} - 5) \]

  94. wio
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    For just the parabola?

  95. LogicalApple
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    That is 4 times the parabola plus the area of the square.

  96. wio
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    It lists your answer as an alternate form.

  97. LogicalApple
    • 2 years ago
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    \[\int\limits_{-(\sqrt{2} - 1)}^{\sqrt{2} -1} (\frac{ 1 - x ^{2} }{ 2}) dx - \int\limits_{-(\sqrt{2} - 1)}^{\sqrt{2} -1} (\sqrt{2} - 1)dx = \frac{ 10 \sqrt{2}- 14 }{ 3 }\] Oh yeah, Wolfram is great! The above should be the area of just one parabolic region. Multiplying this by 4 and then adding the area of the square yields the previous result.

  98. wio
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=%5Cint+from+-%28sqrt%282%29+-1%29+to+%28sqrt%282%29-1%29+++%5B+%281-x%5E2%29%2F2+-+%28sqrt%282%29+-+1%29%5Ddx Wolfram agreeing with that too. Good algebra.

  99. LogicalApple
    • 2 years ago
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    I feel like something was accomplished tonight. Then again, I wonder what the one guy felt like spending 7 years solving Fermat's last theorem....

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is replying to Can someone tell me what button the professor is hitting...

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