658. Find the pressure of water on the surface of a sphere with a diameter of 4 m, if its center is located at a depth of 3 m from the water surface.
Solution. Let us draw a vertical plane through the center of the sphere and choose a rectangular coordinate system $x O y$ on it, as shown in Fig. 119.

We will cut the sphere at a depth $h$ with a horizontal plane. Then the pressure of the water on the cut part of the sphere's surface will be some function $p(h)$.

When $h$ changes by an amount $d h$, the area $S$ of the cut part of the sphere's surface, as the area of a surface of revolution around the $O x$ axis, will change by an amount $\Delta s \approx 2 \pi y d l = d s$, where $d l$ is the differential of the arc of the circle, and the pressure $p(h)$ will change by an amount $\Delta p \approx 2 \pi h y d l = d p$.

![](https://cdn.mathpix.com/cropped/2024_05_22_8a069bcd8684ea7ea3dbg-210.jpg?height=356&width=320&top_left_y=515&top_left_x=869)

Fig. 119

Expressing $d p$ in terms of one variable $x$ and integrating from $x = -2$ to $x = 2$, we will find the pressure of the water on the entire surface of the sphere. From the equation of the circle $x^{2} + y^{2} = 4$, we find

$y' = -\frac{x}{y} \quad$ and then $\quad d l = \sqrt{1 + \left(y'\right)^{2}} d x = \sqrt{1 + \frac{x^{2}}{y^{2}}} d x = \frac{2}{y} d x ; \quad$ from the diagram we find $h = 3 + x$. Therefore,

$$
\begin{gathered}
P = 2 \pi \int_{-2}^{2} (3 + x) y \frac{2}{y} d x = 4 \pi \int_{-2}^{2} (3 + x) d x = \left.2 \pi (3 + x)^{2}\right|_{-2} ^{2} = \\
= 48 \pi(T) \approx 47088() \pi(H) \approx 0.471 \pi(M H)
\end{gathered}
$$

The pressure on the upper half of the sphere's surface will be obtained by integrating $d p$ from -2 to 0:

$$
P_{1} = \left.2 \pi (3 + x)^{2}\right|_{-2} ^{0} = 16 \pi(T) \approx 156960 \pi(H) \approx 0.157 \pi(M H)
$$

The pressure on the lower half of the sphere's surface will be

$$
P_{2} = \left.2 \pi (3 + x)^{2}\right|_{0} ^{2} = 32 \pi(T) \approx 313920 \pi(H) \approx 0.314 \pi(M H)
$$