To find the volume of the solid generated by rotating the region bounded by $y = x^3 - 2$, $y = 0$, and $x = 2$ about the y-axis using the method of disks, we will follow these steps:

Step 1: Express the equation in terms of $x$

The equation $y = x^3 - 2$ is already in terms of $x$. We can solve for $x$ as a function of $y$ since we are rotating about the y-axis. To do this:

$y = x^3 - 2 \implies x^3 = y + 2 \implies x = \sqrt[3]{y + 2}$

Step 2: Setup the integral

When using the method of disks (or washers) for rotation about the y-axis, the volume of the solid is given by the formula:

$V = \pi \int_{y_1}^{y_2} \left[ f(y)^2 \right] dy$

Where $f(y)$ is the radius of the solid at a given point $y$, and $y_1$ and $y_2$ are the bounds for $y$. The function $f(y)$ represents the distance from the y-axis to the curve, which in this case is $f(y) = \sqrt[3]{y + 2}$.

Step 3: Determine the limits of integration

• The region of interest is bounded by $x = 0$ (the y-axis) and $x = 2$, and the equation is given by $y = x^3 - 2$.
• At $x = 0$, $y = 0^3 - 2 = -2$.
• At $x = 2$, $y = 2^3 - 2 = 8 - 2 = 6$.

So, the limits of integration are from $y = -2$ to $y = 6$.

Step 4: Volume Integral

Now we can write the volume integral:

$V = \pi \int_{-2}^{6} \left( \sqrt[3]{y + 2} \right)^2 dy$

This simplifies to:

$V = \pi \int_{-2}^{6} (y + 2)^{2/3} \, dy$

Step 5: Solve the Integral

To solve the integral, we can use the power rule for integration:

$\int (y + 2)^{2/3} \, dy = \frac{3}{5} (y + 2)^{5/3} + C$

Now, apply the limits of integration:

$V = \pi \left[ \frac{3}{5} (y + 2)^{5/3} \right]_{-2}^{6}$

Substituting the limits:

$V = \pi \left[ \frac{3}{5} (6 + 2)^{5/3} - \frac{3}{5} (-2 + 2)^{5/3} \right]$ $V = \pi \left[ \frac{3}{5} (8)^{5/3} - \frac{3}{5} (0)^{5/3} \right]$ $V = \pi \left[ \frac{3}{5} \times 32 \right]$ $V = \pi \times \frac{96}{5}$

Thus, the volume of the solid is:

$V = \frac{96\pi}{5}$

Final Answer:

$V = \frac{96\pi}{5} \, \text{cubic units}.$

Would you like further explanation or details on any step?

---

Related questions to explore:

1. How would the volume change if the region were rotated about the x-axis?
2. What is the volume of the solid generated by rotating $y = x^2$ about the y-axis from $x = 0$ to $x = 3$?
3. What happens to the volume if the region is bounded by different curves or limits?
4. How does the disk method compare to the washer method in terms of application?
5. Can you use the shell method to solve for the volume of the same solid?

Tip: When performing integrals with fractional exponents, always double-check the algebra for any potential mistakes when simplifying powers, as they can be tricky.