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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]

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?

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# Homework Help: Circle Problem (Tex not working on other 1)

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