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If a sphere has a volume of 1, what would the minimum size of a box need to be to contain the sphere?

Imagine the below image represents a cross view of the blue sphere and yellow box. If you know the blue sphere has a volume of 1, how do you calculate what dimensions for the yellow box you would need?

box

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The volume of a sphere is: \begin{equation} V=\frac{4}{3}\pi r^3 \end{equation}

so

\begin{equation} r=\sqrt[3]\frac{3V}{4\pi} \end{equation}

The box must have side $l$ equal to $2r$:

\begin{equation} l=2\sqrt[3]\frac{3V}{4\pi} \end{equation}

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It's rather simple.

First of all notice that the diameter of the sphere has to be at least equal (or less, but we take the exactly quality to gain the minimum volume for the box) to the side of the box.

$$D = \ell$$

Since the volume of the box, supposed cubic, is $\ell^3$ you then have

$$V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (D/2)^3 = \frac{4}{3}\pi (D^3)/8 = \frac{1}{6}\pi D^3$$

In terms of the diameter. Which is also the side of the box $\ell$

From this you can get indeed $\ell$:

$$\frac{1}{6}\pi \ell^3 = V \to \boxed{\ell = \sqrt[3]{\frac{6V}{\pi}}}$$

The volume is unitary so

$$\ell = \sqrt[3]{\frac{6}{\pi}}$$

This is what the box side must be to allow the box to contain the sphere.

Notice that the volume of the box is nothing than $\ell^3$ that is

$$V_{\text{box}} = \ell^3 = \frac{6V_{\text{sphere}}}{\pi}$$

Which holds for a general sphere volume. In your case it's $1$.

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