Example 8. Calculate the area bounded by the Bernoulli lemniscate $r^{2}=a^{2} \cos 2 \varphi$.
Solution. The curve is symmetric with respect to the coordinate axes (Fig. 5.12), so it is sufficient to determine one quarter of the desired area using formula (2.17):

$$
\frac{1}{4} S=\frac{1}{2} \int_{0}^{\frac{\pi}{4}} a^{2} \cos 2 \varphi d \varphi=\frac{a^{2}}{4} \int_{0}^{\frac{\pi}{4}} \cos 2 \varphi d(2 \varphi)=\left.\frac{a^{2}}{4} \sin 2 \varphi\right|_{0} ^{\frac{\pi}{4}}=\frac{a^{2}}{4}
$$

Thus, $S=a^{2}$. In particular, when $a=2$, we get $S=4$.

## Problems

Calculate the areas of the figures bounded by the lines:

1. $y=x^{2}-6 x+8, y=0$.
2. $x=4-y^{2}, x=0$.
3. $y=\ln x, x=e, y=0$.
4. $y^{3}=x^{2}, y=1$.
5. $y^{2}=2 p x, x^{2}=2 p y$.
6. $y=2-x^{2}, y^{3}=x^{2}$.

Find the areas of the figures bounded by the lines:
7. $x=a \cos ^{3} t, y=a \sin ^{3} t$ (astroid).
8. $x=a \cos t, y=b \sin t$ (ellipse).

9. One turn of the Archimedean spiral $r=a \varphi$.
10. $x^{2}+y^{4}=y^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-213.jpg?height=430&width=501&top_left_y=68&top_left_x=99)

Fig. 5.12
Calculate the areas bounded by the loop of the curve:
11. $y^{2}(2 a-x)=x(x-a)^{2}$ (strophoid).
12. $x^{3}+y^{3}-3 x y y=0$ (Cartesian leaf).

## Answers

1. $\frac{4}{3}$. 2. $\frac{32}{3}$. 3.1. 4. $\frac{4}{5}$. 5. $\frac{4}{3} p^{2}$.
2. $2 \frac{2}{15}$. 7. $\frac{3 \pi a^{2}}{8}$.
3. $\pi a b$.
4. $\frac{4}{3} \pi^{3} a^{2}$. 10. $\frac{4}{3}$. 11. $\frac{4-\pi}{2} a^{2}$. 12. $\frac{3 a^{2}}{2}$. Hint. By transitioning to polar coordinates $x=\rho \cos \varphi, y=\rho \sin \varphi$, we get:

$$
\rho=\frac{3 a \sin \varphi \cos \varphi}{\sin ^{3} \varphi+\cos ^{3} \varphi} ; S=\frac{9 a^{2}}{2} \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} \varphi \cos ^{2} \varphi}{\left(\sin ^{3} \varphi+\cos ^{3} \varphi\right)^{2}} d \varphi
$$

Since $\sin \varphi=\operatorname{tg} \varphi \cdot \cos \varphi$, the integrand can be reduced to the form $\frac{\operatorname{tg}^{2} \varphi d(\operatorname{tg} \varphi)}{\left(1+\operatorname{tg}^{3} \varphi\right)^{2}}$, from which the antiderivative is immediately found: $-\frac{1}{3} \cdot \frac{1}{1+\operatorname{tg}^{3} \varphi}$.

## § 5.3. Length of an Arc of a Curve

The length of an arc of a curve $y=f(x)$, where $a \leq x \leq b$, is calculated using the formula

$$
l=\int_{a}^{b} \sqrt{1+y^{\prime 2}} d x \text { or } l=\int_{a}^{b} \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x
$$

The length of an arc of a curve given parametrically

$$
x=\varphi_{1}(t), y=\varphi_{2}(t)\left(t_{1} \leq t \leq t_{2}\right)
$$

is determined by the formula

$$
l=\int_{t_{1}}^{t_{2}} \sqrt{x_{t}^{\prime 2}+y_{t}^{\prime 2}} d t \text { or } l=\int_{t_{1}}^{t_{2}} \sqrt{\varphi_{1}^{\prime 2}(t)+\varphi_{2}^{\prime 2}(t)} d t
$$

If the curve is given by the equation in polar coordinates $r=r(\varphi)$ $(\alpha \leq \varphi \leq \beta)$, then

$$
l=\int_{\alpha}^{\beta} \sqrt{r^{2}+r^{\prime 2}} d \varphi
$$