Is the answer to this problem really an answer? Looking through some long-forgotten math texts of mine, I discovered this problem: > Suppose a spherical floating buoy has radius 1 m and density $\frac{1}{4}$ that of sea water. Given that the formula for the volume of a spherical cap is $V_{\text{cap}} = \frac{\pi{h}}{6}(3r^2 + h^2)$, to what depth does the buoy sink in water? After puzzling over this for a half hour I gave up and looked at the answer. To my surprise, it didn't actually seem directed at the question; it was: "When $x \approx 0.65$, or when the sphere is at a depth of approximately 0.65 metres." Can the problem be solved as stated? I suspect that the authors neglected to some information.

I recommend looking up the formula on Wikipedia, which makes clear that the $r^2$ in your formula is not the squared radius of the sphere. Rearranging gives the last expression in the article of $$V_{\text{cap}}=\pi \frac{h^2}{3}(3r_\text{radius}-h)$$ Here $h$ is the depth by which the sphere is submerged.

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The mass of the whole buoy is $\frac 4 3 \pi r_\text{radius}^3 * \frac 1 4 \rho_{\text{water}}$.

The mass of water displaced is $V_{\text{cap}} \rho_{\text{water}}$.

When are these equal? You should find an equation $h^3-3h^2+1=0$. This equation has three real roots, of which the physical one is the one in $[0,2]$, namely $0.652704\ldots$. (This has an exact expression involving complex numbers but I suspect you were meant to approximate the solution somehow.)