This problem involves finding the volume of the solid generated by revolving the shaded region around the $y$-axis. The shaded region is bounded by the curve $y = 2 - \frac{x^2}{4}$, the $y$-axis, and the line $x = 2$.

We can solve this using the shell method, which is convenient for rotation around the $y$-axis.

Shell Method Formula

The volume of a solid of revolution using the shell method is given by:

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

Here:

• Radius: The distance from the $y$-axis, which is $x$.
• Height: The function value, $y = 2 - \frac{x^2}{4}$.
• Integration bounds: $x = 0$ to $x = 2$.

Step 1: Setup the Integral

$V = 2\pi \int_{0}^{2} x \cdot \left(2 - \frac{x^2}{4}\right) \, dx$

Step 2: Expand and Simplify

Distribute $x$ across the terms: $V = 2\pi \int_{0}^{2} \left(2x - \frac{x^3}{4}\right) \, dx$

Step 3: Integrate

Separate the terms and integrate: $\int 2x \, dx = x^2, \quad \int \frac{x^3}{4} \, dx = \frac{x^4}{16}$

So: $V = 2\pi \left[ x^2 - \frac{x^4}{16} \right]_{0}^{2}$

Step 4: Evaluate at the Bounds

Evaluate at $x = 2$: $\text{At } x = 2: \quad \left(2^2 - \frac{2^4}{16}\right) = \left(4 - \frac{16}{16}\right) = 4 - 1 = 3$

Evaluate at $x = 0$: $\text{At } x = 0: \quad \left(0^2 - \frac{0^4}{16}\right) = 0$

So: $V = 2\pi \cdot (3 - 0) = 6\pi$

Final Answer

The volume of the solid is: $\boxed{6\pi \, \text{units}^3}$

Would you like further clarification or details? Let me know!

Related Questions:

1. What is the difference between the shell and disk/washer methods?
2. How would the volume calculation change if the region was rotated about the $x$-axis?
3. How do you determine the integration bounds for revolution problems?
4. Could we use the washer method here? Why or why not?
5. How does the height function change if the region were shifted vertically or horizontally?

Tip:

Always sketch the problem and identify the axis of rotation clearly to choose the most suitable method for solving volumes of revolution.