## anonymous 3 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.

1. anonymous

2. KingGeorge

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

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. anonymous

5. anonymous

I will work on this right away

6. anonymous

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. anonymous

Hm. I was thinking parabolas, too. I think that might be just what I am looking for.

8. anonymous

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

9. anonymous

Let the very bottom line be the directrix and let the very center be the focus...

10. anonymous

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. anonymous

$x^2 = 4(1/2)y = 2y \implies y = x^2/2$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. anonymous

Thank you so much!

13. anonymous

No problem, it was a bit of a learning experience for me as well.

14. anonymous

Would you agree with the points I drew in and labeled ?

15. anonymous

Actually, let me think about it for a moment...

16. anonymous

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. anonymous

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. anonymous

be the*

19. anonymous

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. anonymous

Wait now that I think about it..

21. anonymous

Those numbers should be cut in half

22. anonymous

|dw:1356849810908:dw| I know it's not a circle but think of it this way.

23. anonymous

You're absolutely right. Back to the drawing board for me.

24. anonymous

|dw:1356849937491:dw|

25. anonymous

|dw:1356850006505:dw|

26. anonymous

In this case, it was just finding the right algebra.... lol

27. anonymous

$\sqrt{x^2 + x^2} + x = \sqrt{1^2+1^2} = \sqrt{2}$

28. anonymous

Right, sqrt(2) . Hm.

29. anonymous

Yeah, well, just solve for $$x$$...

30. anonymous

$\sqrt{x^2+x^2}+x = \sqrt{2x^2}+x = \sqrt{2} \cdot x+x$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. anonymous

@LogicalApple Hmm?

32. anonymous

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. anonymous

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. anonymous

Because the total length is 2, but we're only doing halfway... which is 1.

35. anonymous

Yeah scratch that

36. anonymous

I am in agreement that x = 2 - sqrt 2

37. anonymous

I will modify my equations

38. anonymous

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. anonymous

|dw:1356850703108:dw|

40. anonymous

4x as in 4 times

41. anonymous

I like that idea a lot actually

42. anonymous

The area of the square is easy. We just found half the length of its side. So just double that and square it.

43. anonymous

4(2 - sqrt 2)^2

44. anonymous

We also already found the equation of the parabola, and we know its limits.

45. anonymous

@LogicalApple Got an idea for finding the area under that parabola?

46. anonymous

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. anonymous

I don't know how you're getting that...

48. anonymous

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. anonymous

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. anonymous

We could just ignore that it a parabola altogether and just try using the distance formula.

51. anonymous

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. anonymous

Should be a plus in there.

53. anonymous

Now I think I'm getting a rational answer of 8/3 for the area of the region.

54. anonymous

$\sqrt{(y-1)^2} = \sqrt{x^2 + y^2} \\ (y-1)^2 = x^2 + y^2 \\ y^2 - 2y + 1 = x^2 + y^2 \\ -2y + 1 = x^2 \\ y = -\frac{x^2}{2} +\frac{1}{2}$We got a different equation because we had a different origin.

55. anonymous

But notice how our slope is pretty much the same.

56. anonymous

Just reflected because this time we took the concave down parabola at top.

57. anonymous

I'm following.. and I'm curious. I am going to review this entire post when it's all said and done.

58. anonymous

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. anonymous

$\int_{-(2-\sqrt{2})}^{2-\sqrt{2}} \left[ \frac{-x^2 + 1}{2} - (2-\sqrt{2}) \right]dx$

60. anonymous

Sorry, been having quite a few typos today.

61. anonymous

Don't sweat it

62. anonymous

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. anonymous

Right but I can just use symmetry again. Plus there is wolfram alpha

64. anonymous

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. anonymous

plugging that pellet in that is gonna be a pain in the retrice..

66. anonymous

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. anonymous

$\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} \frac{ \sqrt{2} - 1 }{ 4\sqrt{2} - 3} x^{2}dx -\int\limits_{-(2 - \sqrt{2)}}^{2 - \sqrt{2}} (2 - \sqrt{2}) dx = 4\sqrt{2} - 16$ 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.

68. anonymous

should be a dx in that second integral (sorry am still getting used to the equation creator)

69. anonymous

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. anonymous

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. anonymous

That looks right... makes me wonder where I went wrong.

72. anonymous

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. anonymous

@LogicalApple They are somehow wrong. @dumbcow 's method makes more sense.

74. anonymous

But didn't you calculate the distance from a point to the line y = 1 instead of to the line y = 2?

75. anonymous

Of course you did. Hm.

76. anonymous

Let me see what I obtain next.

77. anonymous

Wow it's really bothering me how I messed up. I can't figure out why my trig method was wrong.

78. anonymous

@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. anonymous

Ooooh! I figured out my error!

80. anonymous

|dw:1356855949526:dw| We found $$x$$! We wanted $$z$$ in the diagram!

81. anonymous

|dw:1356856038380:dw|

82. anonymous

$2z^2 = x^2 \\ \sqrt{2}\cdot z= x \\ z = x/\sqrt{2} = \frac{2-\sqrt{2}}{\sqrt{2}} = \frac{2}{\sqrt{2}} - 1 = \sqrt{2}-1$Lol... really worry about that @LogicalApple Thank @dumbcow for spotting it.

83. anonymous

sorry I mean^

84. anonymous

Modifying .. 1 sec

85. anonymous

86. anonymous

That looks correct.

87. anonymous

I will integrate now.. and I have a feeling it will be irrational this time ;(

88. anonymous

I now see why there was a discrepancy with my earlier parabolic equation.

89. anonymous

I knew from the very start there was very little chance of a rational result.

90. anonymous

My image is still wrong. One more modification ..

91. anonymous

92. anonymous

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

93. anonymous

Final answer is ... $\frac{ 4 }{ 3 }(4\sqrt{2} - 5)$

94. anonymous

For just the parabola?

95. anonymous

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

96. anonymous
97. anonymous

98. anonymous

$\int\limits_{-(\sqrt{2} - 1)}^{\sqrt{2} -1} (\frac{ 1 - x ^{2} }{ 2}) dx - \int\limits_{-(\sqrt{2} - 1)}^{\sqrt{2} -1} (\sqrt{2} - 1)dx = \frac{ 10 \sqrt{2}- 14 }{ 3 }$ 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.

99. anonymous

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.

100. anonymous

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....