To find the volume of the solid generated by rotating the curve $f(x) = 3x^3$ around the y-axis over the interval $[0,1]$, we can use the method of cylindrical shells.

The formula for the volume using cylindrical shells is:

$V = 2\pi \int_{a}^{b} x f(x) \, dx$

Where:

• $f(x) = 3x^3$
• $a = 0$, $b = 1$ (the limits of integration)

Step 1: Set up the integral

Substitute $f(x) = 3x^3$ into the formula:

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

Simplify the expression inside the integral:

$V = 2\pi \int_{0}^{1} 3x^4 \, dx$

Step 2: Solve the integral

Now, solve the integral:

$V = 2\pi \left[ \frac{3x^5}{5} \right]_{0}^{1}$

Evaluate this at the bounds $x = 1$ and $x = 0$:

$V = 2\pi \left( \frac{3(1)^5}{5} - \frac{3(0)^5}{5} \right)$

Simplify:

$V = 2\pi \times \frac{3}{5}$

$V = \frac{6\pi}{5}$

Final Answer:

The volume of the solid generated by rotating the curve $f(x) = 3x^3$ around the y-axis over the interval $[0, 1]$ is:

$V = \frac{6\pi}{5}$

Would you like more details on any of the steps, or have any questions? Here are some related questions:

1. How does the method of cylindrical shells work?
2. Can you explain the difference between cylindrical shells and the disk/washer method?
3. What if we rotated the region around the x-axis instead of the y-axis?
4. What if the interval was different, say [1, 2] instead of [0, 1]?
5. How would we find the surface area of the solid of revolution?

Tip: When rotating around different axes, make sure to choose the appropriate method (shell vs. disk/washer) based on the problem's symmetry.