Solving the Cube and Sphere Volume Ratio Problem

This problem involves comparing the volumes of a cube and a sphere when their surface areas are equal. We need to use the formulas for surface area and volume for both shapes to find the relationship between their dimensions and then calculate the required ratio of the squared volumes.

Understanding the Problem Setup

We are given two conditions:

• The surface area of the cube is equal to the surface area of the sphere.
• \(x\) is the volume of the cube.
• \(y\) is the volume of the sphere.

We need to find the value of \(x^2 : y^2\).

Essential Geometric Formulas

Let's list the key formulas for a cube and a sphere. Assume the cube has side length \(a\) and the sphere has radius \(r\).

From the problem statement, we have:

• Volume of cube, \(x = a^3\)
• Volume of sphere, \(y = \frac{4}{3}\pi r^3\)
• Surface area of cube = Surface area of sphere

Using the Surface Area Equality

The condition that the surface areas are equal allows us to relate the side length of the cube (\(a\)) and the radius of the sphere (\(r\)).

Set the surface area formulas equal:

\(6a^2 = 4\pi r^2\)

We can solve this equation to express \(a^2\) in terms of \(r^2\):

\(a^2 = \frac{4\pi r^2}{6}\)

\(a^2 = \frac{2\pi r^2}{3}\)

Taking the square root of both sides gives the relationship between \(a\) and \(r\):

\(a = \sqrt{\frac{2\pi r^2}{3}} = r\sqrt{\frac{2\pi}{3}}\)

Calculating the Volumes \(x\) and \(y\)

Now we need to express the volumes \(x\) and \(y\) using this relationship. \(y\) is already in terms of \(r\).

Volume of cube, \(x = a^3\). Substitute the expression for \(a\) in terms of \(r\):

\(x = \left(r\sqrt{\frac{2\pi}{3}}\right)^3\)

\(x = r^3 \left(\sqrt{\frac{2\pi}{3}}\right)^3\)

\(x = r^3 \left(\frac{2\pi}{3}\right)^{3/2}\)

Volume of sphere, \(y\), is already in terms of \(r\):

\(y = \frac{4}{3}\pi r^3\)

Finding the Ratio \(x^2 : y^2\)

We need to find the ratio of \(x^2\) to \(y^2\). Let's calculate \(x^2\) and \(y^2\) first.

\(x^2 = \left(r^3 \left(\frac{2\pi}{3}\right)^{3/2}\right)^2\)

Using the power rule \((u^v)^w = u^{v \times w}\):

\(x^2 = r^{3 \times 2} \left(\frac{2\pi}{3}\right)^{(3/2) \times 2}\)

\(x^2 = r^6 \left(\frac{2\pi}{3}\right)^3\)

\(x^2 = r^6 \frac{(2\pi)^3}{3^3}\)

\(x^2 = r^6 \frac{8\pi^3}{27}\)

Now calculate \(y^2\):

\(y^2 = \left(\frac{4}{3}\pi r^3\right)^2\)

\(y^2 = \left(\frac{4}{3}\right)^2 \pi^2 (r^3)^2\)

\(y^2 = \frac{16}{9} \pi^2 r^6\)

Now find the ratio \(x^2 : y^2\), which is \(\frac{x^2}{y^2}\):

\(\frac{x^2}{y^2} = \frac{r^6 \frac{8\pi^3}{27}}{\frac{16}{9} \pi^2 r^6}\)

Cancel out the \(r^6\) term as long as \(r \ne 0\), which it must be for a sphere:

\(\frac{x^2}{y^2} = \frac{\frac{8\pi^3}{27}}{\frac{16\pi^2}{9}}\)

To divide fractions, multiply by the reciprocal of the denominator:

\(\frac{x^2}{y^2} = \frac{8\pi^3}{27} \times \frac{9}{16\pi^2}\)

Simplify the numerical and \(\pi\) terms:

\(\frac{x^2}{y^2} = \left(\frac{8}{16}\right) \times \left(\frac{9}{27}\right) \times \left(\frac{\pi^3}{\pi^2}\right)\)

\(\frac{x^2}{y^2} = \left(\frac{1}{2}\right) \times \left(\frac{1}{3}\right) \times (\pi)\)

\(\frac{x^2}{y^2} = \frac{\pi}{6}\)

So the ratio \(x^2 : y^2\) is \(\pi : 6\).

Conclusion

Given that the surface area of a cube equals that of a sphere, the ratio of the square of the cube's volume (\(x^2\)) to the square of the sphere's volume (\(y^2\)) is \(\pi : 6\).

Revision Table: Cube and Sphere Formulas

Additional Information: Relating Geometric Properties

This problem is a good example of how properties like surface area and volume are interconnected through the dimensions of geometric shapes. When two shapes have an equal property (like surface area here), it establishes a relationship between their dimensions (side \(a\) and radius \(r\)). This relationship can then be used to compare other properties, such as their volumes or squared volumes as done in this problem.

The constant \(\pi\) naturally appears when dealing with spheres (or other curved shapes like circles and cylinders) and often shows up in ratios when comparing them to shapes with flat surfaces, like cubes.

Understanding how to manipulate algebraic expressions involving exponents and variables (like \(r\) and \(a\)) is crucial for solving such geometry problems.