Area of region bounded by the curves $x = y^3$, $x = 0$ between $y = -1$ and $y = 2$ is:

The area is bounded by the curve $x = y^3$, the y-axis ($x = 0$), between $y = -1$ and $y = 2$.

Since the equation is given as $x$ in terms of $y$, integrate with respect to $y$.

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The curve $x = y^3$ at different y-values:

At $y = -1$: $x = (-1)^3 = -1$ (left of y-axis)

At $y = 0$: $x = 0^3 = 0$ (on the y-axis)

At $y = 2$: $x = 2^3 = 8$ (right of y-axis)

The curve is to the left of the y-axis from $y = -1$ to $y = 0$, and to the right from $y = 0$ to $y = 2$.

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For integration with respect to $y$:

Area $= \int (x_{\text{right}} - x_{\text{left}}) \, dy$

From $y = -1$ to $y = 0$:

Right boundary: $x = 0$, Left boundary: $x = y^3$

$A_1 = \int_{-1}^{0} (0 - y^3) \, dy$

$A_1 = \int_{-1}^{0} -y^3 \, dy$

From $y = 0$ to $y = 2$:

Right boundary: $x = y^3$, Left boundary: $x = 0$

$A_2 = \int_{0}^{2} (y^3 - 0) \, dy$

$A_2 = \int_{0}^{2} y^3 \, dy$

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$A_1 = \int_{-1}^{0} -y^3 \, dy$

$A_1 = \left[-\frac{y^4}{4}\right]_{-1}^{0}$

$A_1 = -\frac{(0)^4}{4} - \left(-\frac{(-1)^4}{4}\right)$

$A_1 = 0 + \frac{1}{4}$

$A_1 = \frac{1}{4}$

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$A_2 = \int_{0}^{2} y^3 \, dy$

$A_2 = \left[\frac{y^4}{4}\right]_{0}^{2}$

$A_2 = \frac{(2)^4}{4} - \frac{(0)^4}{4}$

$A_2 = \frac{16}{4} - 0$

$A_2 = 4$

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Total Area $= A_1 + A_2$

Total Area $= \frac{1}{4} + 4$

Total Area $= \frac{1}{4} + \frac{16}{4}$

Total Area $= \frac{17}{4}$ square units