2.6. Find the area of the figure bounded by the given line:

a) $x=2 \cos t-\cos 2 t, y=2 \sin t-\sin 2 t$ (cardioid);
b) $x=a \cos ^{3} t, y=a \sin ^{3} t, a>0$ (astroid);

c) $y^{2}=x^{2}+x^{3}$.
Solution. The areas of figures $\Omega$ bounded by closed piecewise-smooth lines $L$ without self-intersections can be calculated using formula (2.65):

$$
\mu(\Omega)=\oint_{L}-y d x+x d y
$$

where the contour $L$ is traversed counterclockwise.

a) In the case of a cardioid $t \in [0, 2\pi]$, $d x = (-2 \sin t + 2 \sin 2t) d t$, $d y = (2 \cos t - 2 \cos 2t) d t$. The desired area is

$$
\begin{aligned}
& \oint_{L}-y d x + x d y = \int_{0}^{2\pi}[(2 \cos t - 2 \cos 2t)(2 \cos t - \cos 2t) + \\
& + (2 \sin t - 2 \sin 2t)(2 \sin t - \sin 2t)] d t = \int_{0}^{2\pi}(6 - 6 \cos t) d t = 12 \pi.
\end{aligned}
$$

b) In the case of an astroid $t \in [0, 2\pi]$, $d x = -3a \cos^2 t \sin t d t$, $d y = 3a \sin^2 t \cos t d t$. The area bounded by the astroid is

$$
\begin{aligned}
& \oint -y d x + x d y = \int_{0}^{2\pi} \left(3a^2 \cos^2 t \sin^4 t + 3a^2 \cos t \sin^2 t\right) d t = \\
& = 3a^2 \int_{0}^{2\pi} \cos^2 t \sin^2 t d t = \frac{3a^2}{4} \int_{0}^{2\pi} \sin^2 2t d t = \\
& = \frac{3a^2}{8} \int_{0}^{2\pi} (1 - \cos 4t) d t = \frac{3\pi a^2}{4}
\end{aligned}
$$

c) First, we find the parametric representation of the curve $y^2 = x^2 + x^3$. For this, let $y = tx$, where $t$ is a parameter. After substitution and simplification by $x^2$, we get:

$$
\begin{aligned}
& x = t^2 - 1 \\
& y = tx = t^3 - t
\end{aligned}
$$

The curve is shown in Fig. 2.8. The closed loop is traversed counterclockwise as the parameter $t$ increases from -1 to +1. We calculate the area of the region bounded by this loop.

![](https://cdn.mathpix.com/cropped/2024_05_22_05b7516b505cf6654012g-126.jpg?height=293&width=330&top_left_y=1400&top_left_x=807)

Fig. 2.8

$$
\begin{aligned}
& \mu(\Omega) = \oint_{L} -y d x + x d y = \int_{-1}^{1} \left(2(t^3 - t)t + (t^2 - 1)(3t^2 - 1)\right) d t = \\
& = \int_{-1}^{1} (t^2 - 1)^2 d t = 2 \int_{0}^{1} (t^4 - 2t^2 + 1)^2 d t = 2 \left(\frac{1}{5} - \frac{2}{3} + 1\right) = \frac{16}{15}
\end{aligned}
$$