99. Let's find the length $L$ of the arc of the curve: $x^{2}+y^{2}=R^{2}$.
Solution. This curve is a circle of radius $R$ centered at the origin (Fig. 53).

First, we find the length $l$ of the semicircle located in the first and second quadrants. Noting that its equation will be $y = \sqrt{R^2 - x^2}$, by formula (3) we have:

$$
\begin{gathered}
l = \int_{-R}^{R} \sqrt{1 + \left(\left(\sqrt{R^2 - x^2}\right)'\right)^2} \, dx = \int_{-R}^{R} \sqrt{1 + \left(-\frac{x}{\sqrt{R^2 - x^2}}\right)^2} \, dx = \\
= R \cdot \int_{-R}^{R} \frac{1}{\sqrt{R^2 - x^2}} \, dx = \left. R \arcsin \frac{x}{R} \right|_{-R}^{R} = \pi R.
\end{gathered}
$$

Therefore, $L = 2l = 2\pi R$.

Remark. In calculating the integral in the example, we used the fact that

$$
(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}
$$

## INTEGRAL AND SURFACE AREA

The integral can be used to derive a formula for calculating the surface area.

Suppose a figure is obtained by rotating the graph of the function $y = f(x)$ around the $Ox$ axis, where $f(x)$ has a continuous derivative $f'(x)$ on $[a; b]$ (Fig. 54). Then the area $S$ of this figure is calculated by the formula

$$
S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-63.jpg?height=709&width=862&top_left_y=228&top_left_x=583)

Fig. 54

Let's prove this. For this, we need the following

Lemma. The lateral surface area of a truncated cone is given by:

$$
S_{\text{lat}} = \pi (R + r) L,
$$

where $R$ and $r$ are the radii of the bases, and $L$ is the slant height.

Indeed, for a truncated cone (Fig. 55), we have:

$$
S_{\text{lat}} = S_{\text{lat} ASB} - S_{\text{lat} A_1SB_1} = \pi R (L + SB_1) - \pi r \cdot SB_1.
$$

But from the similarity of triangles $OSB$ and $O_1SB_1$,

$$
\frac{L + SB_1}{SB_1} = \frac{R}{r}, \text{ or } Lr + SB_1 \cdot r = SB_1 \cdot R, \text{ or } SB_1 = \frac{Lr}{R - r}.
$$

From equations (3) and (4), it follows that equation (2) holds.

Take $x \in [a; b]$ and consider the part of the surface of revolution located to the left of the point $x$. The area of this surface of revolution, denoted by $S(x)$, is defined on $[a; b]$. Let's compute the derivative of this function. To avoid complicating the calculations, we will provide a visual scheme for the case $\Delta x > 0$.

Then $\Delta S(x)$ can be approximately considered as the lateral surface area of a truncated cone with slant height $\Delta l(x)$ and radii of the bases $y + \Delta y$ and $y$, i.e., (see the lemma)

$\Delta S(x) \approx \pi ((y + \Delta y) + y) \Delta l(x)$, or $\alpha(\Delta x) = \Delta S(x) - \pi ((y + \Delta y) + y) \Delta l(x)$,
or

$$
\alpha(\Delta x) = \Delta S(x) - 2\pi y \Delta l(x) - \pi \Delta y \cdot \Delta l(x)
$$

where $\alpha(\Delta x)$ is an infinitesimal quantity as $\Delta x \to 0$ (see p. 60).

Taking the limit in equation (5), we get:

$$
\lim_{\Delta x \to 0} \alpha(\Delta x) = \lim_{\Delta x \to 0} (\Delta S(x) - 2\pi y \cdot \Delta l(x) - \pi \cdot \Delta y \cdot \Delta l(x)) = 0
$$

or, considering $\lim_{\Delta x \to 0} \Delta y = 0$ and $\lim_{\Delta x \to 0} \Delta l(x) = 0$:

$$
\begin{gathered}
\lim_{\Delta x \to 0} \Delta S(x) = 2\pi y \lim_{\Delta x \to 0} \Delta l(x), \text{ or } \lim_{\Delta x \to 0} \frac{\Delta S(x)}{\Delta x} = 2\pi y \cdot \lim_{\Delta x \to 0} \frac{\Delta l(x)}{\Delta x}, \text{ or } \\
S'(x) = 2\pi y \cdot l'(x)
\end{gathered}
$$

But according to formula (2) (p. 61),

$$
S'(x) = 2\pi y \sqrt{1 + \left(y'(x)\right)^2}
$$

Thus, $S(x)$ is an antiderivative of the function $2\pi y \sqrt{1 + \left(y'(x)\right)^2}$ on $[a; b]$. The area $S$ of the surface of revolution is obtained when $x = b$:

$$
S = S(b)
$$

Noting that $S(a) = 0$, we find $S = S(b) - S(a)$, or

$$
S = 2\pi \int_{a}^{b} y \sqrt{1 + \left(y'(x)\right)^2} \, dx
$$

where $y = f(x)$ and $y'(x) = f'(x)$.