661. A ball lies at the bottom of a pool with a depth of $H=14 \partial$ m. Determine the work required to extract the ball from the water, if its radius $R=3 \partial$, and specific weight $\delta=2$.
Solution. When lifting the ball to the water surface, the force $P_{1}$, which performs work, is constant and equal to the difference between the weight of the ball and the weight of the displaced water:

$$
P_{1}=\frac{4}{3} \pi R^{3} \delta-\frac{4}{3} \pi R^{3}=\frac{4}{3} \pi R^{3}(\delta-1)
$$

Therefore, the work $Q_{1}$ required to lift the ball to the water surface is determined in an elementary manner as the product of the force $P_{1}$ and the height of the lift $H-2 R$:

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-212.jpg?height=187&width=367&top_left_y=975&top_left_x=823)

Fig. 122

$$
Q_{1}=P_{1}(H-2 R)=\frac{4}{3} \pi R^{3}(\delta-1)(H-2 R)
$$

When further lifting the ball, the force $p$, which performs work, will vary depending on the height $x$ of the part of the ball above water (Fig. 122):

$$
p(x)=P_{\text{ball}}-p_{\text{water}}
$$

where $P_{\text{ball}}$ is the weight of the ball, and $p_{\text{water}}$ is the weight of the water displaced by the submerged part of the ball, which is numerically equal to the volume of the spherical segment with height $h=2 R-x$:

$$
p_{\text{water}}=\pi h^{2}\left(R-\frac{h}{3}\right)=\frac{\pi}{3}(2 R-x)^{2}(R+x)=\frac{\pi}{3}\left(x^{3}-3 R x^{2}+4 R^{3}\right)
$$

Obviously, the work performed by the force $p(x)$ will be some function $q(x)$. Assuming that when lifting the ball a small additional height $dx$, the force $p(x)$ remains unchanged, we can find the approximate value of the work increment

$\Delta q \approx p(x) dx=\left(P_{\text{ball}}-p_{\text{water}}\right) dx=\frac{\pi}{3}\left[4 R^{3}(\delta-1)-x^{3}+3 R x^{2}\right] dx=d q$.

Integrating $d q$ from $x=0$ to $x=2 R$, we find the work $Q_{2}$ required to completely extract the ball from the water after it has been lifted to the water surface from the bottom of the pool:

$$
\begin{gathered}
Q_{2}=\frac{\pi}{3} \int_{0}^{2 R}\left[4 R^{3}(\delta-1)-x^{3}+3 R x^{2}\right] d x= \\
=\left.\frac{\pi}{3}\left[4 R^{3}(\delta-1) x-\frac{x^{4}}{4}+R x^{3}\right]\right|_{0} ^{2 R}=\frac{4}{3} \pi R^{4}(2 \delta-1)
\end{gathered}
$$

The total work required $Q=Q_{1}+Q_{2}=\frac{4}{3} \pi R^{3}[R+(\delta-1) H]=$ $=61.2 \pi(\kappa \Gamma \mu) \approx 600.4 \pi(\partial \text{zh})$.