To solve this problem, we will use the shell method for revolving the shaded region around the y-axis.

Shell Method Formula:

The volume $V$ of the solid generated by revolving a region around the y-axis using the shell method is given by: $V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx$ Where:

• $f(x)$ is the height of the shell at each value of $x$,
• $x$ is the radius of the shell,
• $[a, b]$ is the interval over which the region is defined.

Step-by-step Solution:

1. Set up the integral

• The region is defined from $x = 0$ to $x = \sqrt{3}$.
• The function representing the height is given by $y = 6 - 3x^2$.

Thus, the volume is: $V = 2\pi \int_{0}^{\sqrt{3}} x \cdot (6 - 3x^2) \, dx$

2. Expand the integrand

$V = 2\pi \int_{0}^{\sqrt{3}} (6x - 3x^3) \, dx$

3. Integrate

Now, integrate term by term: $V = 2\pi \left[ \frac{6x^2}{2} - \frac{3x^4}{4} \right]_{0}^{\sqrt{3}}$ Simplifying the coefficients: $V = 2\pi \left[ 3x^2 - \frac{3x^4}{4} \right]_{0}^{\sqrt{3}}$

4. Evaluate at the bounds

First, evaluate at $x = \sqrt{3}$: $3(\sqrt{3})^2 - \frac{3(\sqrt{3})^4}{4} = 3(3) - \frac{3(9)}{4} = 9 - \frac{27}{4} = \frac{36}{4} - \frac{27}{4} = \frac{9}{4}$

At $x = 0$, the expression evaluates to 0: $3(0)^2 - \frac{3(0)^4}{4} = 0$

Thus, the volume is: $V = 2\pi \left( \frac{9}{4} \right) = \frac{9\pi}{2}$

Final Answer:

The volume is $\boxed{\frac{9\pi}{2}}$.

Would you like a further breakdown of any step?

Here are 5 related questions:

1. What is the difference between the disk method and the shell method for finding volumes?
2. How do you decide whether to revolve around the x-axis or y-axis?
3. What are other real-world applications of the shell method?
4. Can the shell method be used for solids with holes? How does it work?
5. How does changing the limits of integration affect the volume calculation?

Tip: Always check the bounds of integration and the axis of revolution to choose the correct method for finding the volume.