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Circle Problem (Tex not working on other 1)

  1. Apr 3, 2006 #1
    Problem :

    A cube is inscribed in a sphere of diameter 1. What is the surface area of the cube?


    Solution (attempted!) :

    Let there be a circle of radius 0.5 with centre at (0, 0.5).

    The equation of this circle is therefore:

    [tex] x^2 + (y-0.5)^\frac{1}{2} = 0.5^2[/tex]

    [tex] x = (y-y^2)^\frac{1}{2}[/tex]

    Integrate x with respect to y gives the area to the left of the curve.

    So :

    Let [tex] x = (y-y^2)^\frac{3}{2} [/tex]

    Let [tex] u = (y-y^2)

    \Rightarrow x = u^\frac{3}{2} [/tex]

    And

    [tex]\frac{\delta x}{\delta y} = \frac{\delta x}{\delta u} \cdot \frac{\delta u}{\delta y}[/tex]

    [tex]\frac{\delta x}{\delta u} = \frac{3}{2} u^\frac{1}{2} = \frac{3}{2} (y-y^2)^\frac{1}{2}[/tex]

    [tex]\frac{\delta u}{\delta y} = 1 - 2y[/tex]

    [tex]\Rightarrow \frac{\delta x}{\delta y} = (\frac{3}{2} (y-y^2)^\frac{1}{2}) \cdot (1 - 2y)[/tex]

    [tex]\Rightarrow \frac{\delta x}{\delta y} = (y-y^2)^\frac{1}{2} \cdot (\frac{3}{2} - 3y)[/tex]

    So :

    [tex] \int_{a}^{b} (y-y^2)^\frac{1}{2} dy = \frac{(y-y^2)^\frac{3}{2}}{(\frac{3}{2} - 3y)} + C [/tex]

    [​IMG]

    Let B and A be defined by this drawing!

    B has co-ordinates (0.5,0.5)

    To find A co-ordinates. Consider the diagonal line. It is a diameter and thus has length 1.

    The edge of the square therefore has length [tex]\sqrt{\frac{1}{2}}[/tex]

    Therefore A has y co-ordinate [tex] 0.5 - \sqrt{\frac{1}{2}}[/tex]

    [tex] \Rightarrow \int_{0.5 - \sqrt{\frac{1}{2}}}^{\frac{1}{2}} (y-y^2)^\frac{1}{2} dy = \left[\frac{(y-y^2)^\frac{3}{2}}{(\frac{3}{2} - 3y)}\right]_{0.5 - \sqrt{\frac{1}{2}}}^{\frac{1}{2}}} = I[/tex]

    [tex]\Rightarrow 0.5^2\pi - 4(I - (\frac{1}{2})^2) = Surface area of 1 face [/tex]

    Then times by 6 to get SA of cube

    Is this correct?
     
    Last edited: Apr 3, 2006
  2. jcsd
  3. Apr 3, 2006 #2
    why is the circle at (0,0.5)? what is wrong with the origin? Also, is a sphere, not a circle, so the centre can be considered to be at (0,0,0). so, [tex] x^2 + y^2 + z^2 = 0.5[/tex], if a cube is inside it, then x = y = z where they touch, so [tex] 3r^2 = 0.5 [/tex], this is the distance along each axis (r), from the origin to the point of contact between the cube and the sphere, so the length of one side is double that, then the suface are is simply that squared and multiplied by 6 - Therefore, from all that I've done, if the method is right, the answer is 1.
     
    Last edited by a moderator: Apr 3, 2006
  4. Apr 3, 2006 #3
    Ye I did it that way too! (After I saw your post!!)

    But was wondering if my method beared water?

    6 I = 1 ?!!
     
    Last edited: Apr 3, 2006
  5. Apr 3, 2006 #4

    0rthodontist

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    Science Advisor

    You don't need any calculus. You know that a diagonal across the interior of the cube has length 1. So a line segment from the center to a cube corner has length 1/2. You know that any adjacent 2 of such line segments are perpendicular. You know that the third side of the triangle formed by those 2 adjacent line segments is a diagonal across a face of the cube. One more step and you have found the length of an edge of the cube, from which the cube's surface area is simple to compute.
     
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