The area (in Sq. units) of the region bounded by $y = -2$, $y = 2$, $x = y^3$ and $x = 0$ is equal to

We need to find the area of a region with these boundaries:

$y = 2$ (horizontal line at top)

$y = -2$ (horizontal line at bottom)

$x = y^3$ (a curve)

$x = 0$ (the y-axis)

Since $x$ is expressed in terms of $y$, we integrate with respect to $y$.

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Let's see what happens with the curve $x = y^3$:

When $y = 0$, we get $x = 0$

When $y = 2$, we get $x = 8$

When $y = -2$, we get $x = -8$

For negative $y$ values, $x$ becomes negative. The curve goes into the left side (negative $x$ region).

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We split the region into two parts:

From $y = -2$ to $y = 0$: The region lies between $x = y^3$ (left) and $x = 0$ (right)

Width at any point $= 0 - y^3 = -y^3$

From $y = 0$ to $y = 2$: The region lies between $x = 0$ (left) and $x = y^3$ (right)

Width at any point $= y^3 - 0 = y^3$

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Setting up the integration:

$\text{Total Area} = \int_{-2}^{0} (0 - y^3) \, dy + \int_{0}^{2} (y^3 - 0) \, dy$

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

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Calculating the first integral:

$\int_{-2}^{0} (-y^3) \, dy = \left[-\frac{y^4}{4}\right]_{-2}^{0}$

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

$= 0 - \left(-\frac{16}{4}\right)$

$= 0 + 4 = 4$

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Calculating the second integral:

$\int_{0}^{2} y^3 \, dy = \left[\frac{y^4}{4}\right]_{0}^{2}$

$= \frac{2^4}{4} - \frac{0^4}{4}$

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

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$\text{Total Area} = 4 + 4 = 8 \text{ square units}$