Double Bubble Gum

Relevant wiki: Volume of Revolution - Disc Method - Intermediate

Let $R$ be the radius of the sphere. Then the oval will have its minor and major axis equal to $R$ and $2R$ respectively.

By drawing an elliptical graph of the oval, we can then come up with the equation with center on the origin:

$\dfrac{x^2}{R^2} + \dfrac{y^2}{(2R)^2} = 1$

$x^2 = R^2 - \dfrac{y^2}{4}$

By revolving the graph around the y-axis, we can calculate volume $V$ of half the oval as:

$V = \int_0^{2R} (\pi x^2) \ \mathrm{dy} = \int_0^{2R} \pi (R^2 - \dfrac{y^2}{4}) \ \mathrm{dy} = \pi[ R^2 y - \dfrac{y^3}{12} |_0^{2R}] = \pi [2R^3 - \dfrac{8R^3}{12}] = \pi R^3(2 - \dfrac{2}{3}) = \dfrac{4}{3}\pi R^3$

Clearly, this is a formula for the sphere with radius $R$ .

In other words, half an oval has the same volume as one sphere.

Thus, the volume of $2$ spheres = the volume of $1$ oval.

Both options would give the same volume of gum.

This elliptical gum is just a sphere that has been stretched by a factor k =2 in one dimension. Thus the volume is scaled by 2, which equals the volume of two spheres.