96. Let's calculate the area of the figure bounded by the lines

$$
y=x^{3}, \quad y=-1, \quad y=-3 \text{ and } x=0.
$$
Solution. Method 1. Perform a rotation around point $O$ by an angle of $-90^{\circ}$; in this case, the given figure (Fig. 45) will transform into an equal curvilinear trapezoid bounded by the lines

$$
y=(-x)^{\frac{1}{3}}, y=0, x=-3 \text { and } x=-1
$$

By formula (1), we have:

$$
S=\int_{-3}^{-1}(-x)^{\frac{1}{3}} d x=\frac{3}{4}(3 \sqrt[3]{3}-1)
$$

57

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-59.jpg?height=646&width=591&top_left_y=248&top_left_x=387)

Fig. 45

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-59.jpg?height=651&width=551&top_left_y=248&top_left_x=1112)

Fig. 46

Method 2. Perform a rotation around point $O$ by an angle of $90^{\circ}$; in this case, the given figure (Fig. 46) will transform into an equal curvilinear trapezoid bounded by the lines

$$
y=-x^{\frac{1}{3}}, y=0, x=1 \text { and } x=3
$$

By formula (2), we have:

$$
S=-\int_{1}^{3}\left(-x^{\frac{1}{3}}\right) d x=\frac{3}{4}(3 \sqrt[3]{3}-1)
$$

Method 3. Instead of calculating the area of the given figure, we will calculate the area of the figure symmetric to the given one relative to the line $y=-x$. This figure (Fig. 47) is bounded by the lines $y=x^{\frac{1}{3}}, y=0, x=1$ and $x=3$ and represents a curvilinear trapezoid. We will calculate its area using formula (1):

$$
S=\int_{1}^{3} x^{\frac{1}{3}} d x=\frac{3}{4}(3 \sqrt[3]{3}-1)
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-59.jpg?height=620&width=577&top_left_y=1709&top_left_x=405)

Fig. 47

![](https://cdn.mathpix.com/cropped/2024_05_21_b1a79769c6c63d19f37fg-59.jpg?height=614&width=597&top_left_y=1715&top_left_x=1072)

Fig. 48