Example 2.49. Compute the area of the figure bounded by the curve $y^{2}=x(x-2)^{2}$.
Solution. Due to the symmetry of the figure (Fig. 2.4), we can calculate half of the area and then double the result.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-080.jpg?height=349&width=457&top_left_y=129&top_left_x=143)

Fig. 2.2.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-080.jpg?height=326&width=457&top_left_y=559&top_left_x=144)

Fig. 2.4.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-080.jpg?height=324&width=461&top_left_y=149&top_left_x=625)

Fig. 2.3.

![](https://cdn.mathpix.com/cropped/2024_05_22_0f84e17422d7c0df8d68g-080.jpg?height=309&width=487&top_left_y=573&top_left_x=612)

Fig. 2.5.

Let's find the limits of integration, when \( y=0 \Rightarrow x_{1}=0, x_{2}=2 \).

\[
\begin{gathered}
\left|\int_{0}^{2} \sqrt{x}(x-2) d x\right|=\left|\int_{0}^{2}\left(x^{3 / 2}-2 x^{1 / 2}\right) d x\right|= \\
\left.=\left|\left[\frac{2}{5} x^{5 / 2}-\frac{4}{3} x^{3 / 2}\right]\right|_{0}^{2}|=|-\frac{2^{9 / 2}}{15} \right\rvert\,
\end{gathered}
\]

The minus sign under the modulus indicates that the area of the lower half has been found. The entire area

\[
S=2 \cdot \frac{2^{9 / 2}}{15}=\frac{2^{11 / 2}}{15}
\]