A ball of radius R floats fully submerged in water (the

A ball of radius $R$ floats fully submerged in water (the specific weight of both the ball and the water is $1$). How much work is required to pull it completely out of the water?

Step 1: Set up coordinates and identify the working range

Let the water surface be $z=0$ and let $y$ denote the height of the ball's center above the surface. The ball extends from $y-R$ to $y+R$.

The relevant transition begins when the top of the ball just touches the surface, i.e. $y+R=0$, so $y=-R$ (still entirely submerged, no part above water yet), and ends when the bottom of the ball just clears the surface, i.e. $y-R=0$, so $y=R$ (entirely out of the water). For any $y$ outside this range the ball is either fully submerged (no work, by the buoyancy argument above) or fully in air (no further work needed). So the work integral runs over $y\in[-R,R]$.

Step 2: Volume of the ball above the surface

When the center is at height $y$, the part of the ball poking above $z=0$ is a spherical cap of height $h=y+R$ (which indeed ranges from $0$ at $y=-R$ to $2R$ at $y=R$, the whole ball). The standard cap-volume formula $V_{\text{cap}}=\dfrac{\pi h^2}{3}(3R-h)$ gives

\[V_{\text{above}}(y)=\frac{\pi(y+R)^2}{3}\big(3R-(y+R)\big)=\frac{\pi(y+R)^2(2R-y)}{3}.\]

This correctly gives $V_{\text{above}}(-R)=0$, $V_{\text{above}}(0)=\tfrac23\pi R^3$ (half the ball, as expected when the center sits exactly at the surface), and $V_{\text{above}}(R)=\tfrac43\pi R^3$ (the full ball).

Step 3: Force needed at height $y$

With specific weight $\gamma=1$ for both ball and water, the unsupported weight above the surface is simply

\[F(y)=\gamma\,V_{\text{above}}(y)=\frac{\pi(y+R)^2(2R-y)}{3}.\]

Step 4: Expand and integrate over the full range

\[(y+R)^2(2R-y)=(R^2+2Ry+y^2)(2R-y)=2R^3+3R^2y-y^3.\]

\[W=\int_{-R}^{R}F(y)\,dy=\frac{\pi}{3}\int_{-R}^{R}\left(2R^3+3R^2y-y^3\right)dy.\]

The odd-power terms $3R^2y$ and $-y^3$ integrate to zero over the symmetric interval $[-R,R]$, leaving only the constant term:

\[W=\frac{\pi}{3}\int_{-R}^{R}2R^3\,dy=\frac{\pi}{3}\cdot2R^3\cdot2R=\frac{4\pi R^4}{3}.\]

\[\boxed{W=\frac{4\pi R^4}{3}}\]

Step 5: Sanity check via the ball's weight

The ball's own weight is $P=\gamma\cdot\tfrac43\pi R^3=\tfrac43\pi R^3$ (with $\gamma=1$). Comparing, $W=\tfrac43\pi R^4=P\cdot R$: the total work is exactly the ball's weight times its radius. This is a clean, checkable relation (equivalent to lifting the entire ball's weight through the distance from its center to its surface) and was confirmed independently by direct numerical integration of $F(y)$ over $[-R,R]$.

Step 6: Note on the integration range

Only integrating half the transition, from $y=0$ to $y=R$ (i.e. starting from the ball exactly half-submerged rather than from the true fully-submerged starting point $y=-R$), would omit the work already required to lift the ball from “just touching the surface” to “half out,” and gives a smaller, incorrect result; the full physical process starts at $y=-R$, where $V_{\text{above}}=0$, and the integral above uses that correct starting point throughout.