To find the volume of the solid generated by rotating the region enclosed by $y = 2x^3$, $y = 0$, $x = 0$, and $x = 1$ around the y-axis, we use the shell method.

Step 1: Shell Method Formula

The shell method states that the volume is given by:

$V = \int_{a}^{b} 2\pi (\text{radius}) (\text{height}) \,dx$

Here:

• Radius = $x$ (distance from the y-axis)
• Height = $y = 2x^3$
• Limits: $x = 0$ to $x = 1$

Step 2: Set Up the Integral

$V = \int_{0}^{1} 2\pi x (2x^3) \,dx$

$V = 2\pi \int_{0}^{1} 2x^4 \,dx$

$V = 4\pi \int_{0}^{1} x^4 \,dx$

Step 3: Evaluate the Integral

$\int x^4 dx = \frac{x^5}{5}$

$V = 4\pi \left[ \frac{x^5}{5} \right]_{0}^{1}$

$V = 4\pi \left( \frac{1}{5} - 0 \right)$

$V = \frac{4\pi}{5}$

Final Answer:

$\frac{4\pi}{5}$

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Related Problems:

1. Find the volume if the same region is rotated around the x-axis.
2. Compute the volume if the region is rotated around the line $x = -1$.
3. Use the disk/washer method to confirm the volume if rotated around the x-axis.
4. Modify the function to $y = 3x^2$ and compute the volume around the y-axis.
5. Calculate the surface area of the solid formed when the region is rotated.

Tip:

The shell method is useful when integrating with respect to x for rotation about the y-axis and when integrating with respect to y for rotation about the x-axis.