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LogicalApple Group Title

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.

  • one year ago
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  1. LogicalApple Group Title
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    • one year ago
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  2. KingGeorge Group Title
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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.

    • one year ago
  3. KingGeorge Group Title
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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.

    • one year ago
  4. LogicalApple Group Title
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    Thank you for your time!

    • one year ago
  5. LogicalApple Group Title
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    I will work on this right away

    • one year ago
  6. wio Group Title
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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.

    • one year ago
  7. LogicalApple Group Title
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    Hm. I was thinking parabolas, too. I think that might be just what I am looking for.

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

    • one year ago
  9. wio Group Title
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    Let the very bottom line be the directrix and let the very center be the focus...

    • one year ago
  10. wio Group Title
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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

    • one year ago
  11. wio Group Title
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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.

    • one year ago
  12. LogicalApple Group Title
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    Thank you so much!

    • one year ago
  13. wio Group Title
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    No problem, it was a bit of a learning experience for me as well.

    • one year ago
  14. LogicalApple Group Title
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    Would you agree with the points I drew in and labeled ?

    • one year ago
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  15. wio Group Title
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    Actually, let me think about it for a moment...

    • one year ago
  16. wio Group Title
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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.

    • one year ago
  17. LogicalApple Group Title
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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

    • one year ago
  18. LogicalApple Group Title
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    be the*

    • one year ago
  19. LogicalApple Group Title
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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).

    • one year ago
  20. LogicalApple Group Title
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    Wait now that I think about it..

    • one year ago
  21. LogicalApple Group Title
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    Those numbers should be cut in half

    • one year ago
  22. wio Group Title
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    |dw:1356849810908:dw| I know it's not a circle but think of it this way.

    • one year ago
  23. LogicalApple Group Title
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    You're absolutely right. Back to the drawing board for me.

    • one year ago
  24. wio Group Title
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    |dw:1356849937491:dw|

    • one year ago
  25. wio Group Title
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    |dw:1356850006505:dw|

    • one year ago
  26. wio Group Title
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    In this case, it was just finding the right algebra.... lol

    • one year ago
  27. wio Group Title
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    \[ \sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2} \]

    • one year ago
  28. LogicalApple Group Title
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    Right, sqrt(2) . Hm.

    • one year ago
  29. wio Group Title
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    Yeah, well, just solve for \(x\)...

    • one year ago
  30. wio Group Title
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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.

    • one year ago
  31. wio Group Title
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    @LogicalApple Hmm?

    • one year ago
  32. LogicalApple Group Title
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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)?

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

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

    • one year ago
  35. LogicalApple Group Title
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    Yeah scratch that

    • one year ago
  36. LogicalApple Group Title
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    I am in agreement that x = 2 - sqrt 2

    • one year ago
  37. LogicalApple Group Title
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    I will modify my equations

    • one year ago
  38. wio Group Title
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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.

    • one year ago
  39. wio Group Title
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    |dw:1356850703108:dw|

    • one year ago
  40. wio Group Title
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    4x as in 4 times

    • one year ago
  41. LogicalApple Group Title
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    I like that idea a lot actually

    • one year ago
  42. wio Group Title
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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.

    • one year ago
  43. LogicalApple Group Title
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    4(2 - sqrt 2)^2

    • one year ago
  44. wio Group Title
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    We also already found the equation of the parabola, and we know its limits.

    • one year ago
  45. wio Group Title
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    @LogicalApple Got an idea for finding the area under that parabola?

    • one year ago
  46. LogicalApple Group Title
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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

    • one year ago
  47. wio Group Title
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    I don't know how you're getting that...

    • one year ago
  48. LogicalApple Group Title
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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

    • one year ago
  49. LogicalApple Group Title
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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)

    • one year ago
  50. wio Group Title
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    We could just ignore that it a parabola altogether and just try using the distance formula.

    • one year ago
  51. wio Group Title
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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 \[ \sqrt{(x-x)^2(y-1)^2} = \sqrt{x^2+y^2} \]

    • one year ago
  52. wio Group Title
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    Should be a plus in there.

    • one year ago
  53. LogicalApple Group Title
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    Now I think I'm getting a rational answer of 8/3 for the area of the region.

    • one year ago
  54. wio Group Title
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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.

    • one year ago
  55. wio Group Title
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    But notice how our slope is pretty much the same.

    • one year ago
  56. wio Group Title
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    Just reflected because this time we took the concave down parabola at top.

    • one year ago
  57. LogicalApple Group Title
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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.

    • one year ago
  58. wio Group Title
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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: \[ \int_{1(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx \]

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

    • one year ago
  60. wio Group Title
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    Sorry, been having quite a few typos today.

    • one year ago
  61. LogicalApple Group Title
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    Don't sweat it

    • one year ago
  62. wio Group Title
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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.

    • one year ago
  63. LogicalApple Group Title
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    Right but I can just use symmetry again. Plus there is wolfram alpha

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

    • one year ago
  65. wio Group Title
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    plugging that pellet in that is gonna be a pain in the retrice..

    • one year ago
  66. wio Group Title
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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.

    • one year ago
  67. LogicalApple Group Title
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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.

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

    • one year ago
  69. LogicalApple Group Title
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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..

    • one year ago
  70. dumbcow Group Title
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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 \[-\frac{1}{2}x^{2}+\frac{1}{2} = x\] \[x = \sqrt{2} - 1\]

    • one year ago
  71. wio Group Title
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    That looks right... makes me wonder where I went wrong.

    • one year ago
  72. LogicalApple Group Title
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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.

    • one year ago
  73. wio Group Title
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    @LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

    • one year ago
  74. LogicalApple Group Title
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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?

    • one year ago
  75. LogicalApple Group Title
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    Of course you did. Hm.

    • one year ago
  76. LogicalApple Group Title
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    Let me see what I obtain next.

    • one year ago
  77. wio Group Title
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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.

    • one year ago
  78. wio Group Title
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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.

    • one year ago
  79. wio Group Title
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    Ooooh! I figured out my error!

    • one year ago
  80. wio Group Title
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    |dw:1356855949526:dw| We found \(x\)! We wanted \(z\) in the diagram!

    • one year ago
  81. wio Group Title
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    |dw:1356856038380:dw|

    • one year ago
  82. wio Group Title
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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.

    • one year ago
  83. wio Group Title
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    sorry I mean^

    • one year ago
  84. LogicalApple Group Title
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    Modifying .. 1 sec

    • one year ago
  85. LogicalApple Group Title
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    I did worry about it!

    • one year ago
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  86. wio Group Title
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    That looks correct.

    • one year ago
  87. LogicalApple Group Title
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    I will integrate now.. and I have a feeling it will be irrational this time ;(

    • one year ago
  88. LogicalApple Group Title
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    I now see why there was a discrepancy with my earlier parabolic equation.

    • one year ago
  89. wio Group Title
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    I knew from the very start there was very little chance of a rational result.

    • one year ago
  90. LogicalApple Group Title
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    My image is still wrong. One more modification ..

    • one year ago
  91. LogicalApple Group Title
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    • one year ago
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  92. wio Group Title
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    That's pretty much it. It's an easy integral because it's just a 2 degree polynomial.

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

    • one year ago
  94. wio Group Title
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    For just the parabola?

    • one year ago
  95. LogicalApple Group Title
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    That is 4 times the parabola plus the area of the square.

    • one year ago
  96. wio Group Title
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    It lists your answer as an alternate form.

    • one year ago
  97. LogicalApple Group Title
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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.

    • one year ago
  98. wio Group Title
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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.

    • one year ago
  99. LogicalApple Group Title
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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....

    • one year ago
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