To solve this problem using the method of shells, let’s analyze it step by step. The diagram shows the region bounded by the curve $y = 1 + \frac{x^2}{4}$, the $x$-axis, and the vertical line $x = 2$. This region is rotated about the $y$-axis.

---

Shell Method Formula:

The volume $V$ of a solid of revolution using the shell method is given by: $V = 2\pi \int_a^b \left(\text{radius}\right) \cdot \left(\text{height}\right) \, dx$ Here:

• Radius is the distance from the $y$-axis to a point on the curve, which is $x$.
• Height is the value of the function, $y = 1 + \frac{x^2}{4}$.

---

Given:

• $a = 0$, $b = 2$, and $y = 1 + \frac{x^2}{4}$.

Substitute:

• Radius: $x$
• Height: $y = 1 + \frac{x^2}{4}$

$V = 2\pi \int_0^2 x \cdot \left(1 + \frac{x^2}{4}\right) dx$

---

Simplify the Integral:

Expand the expression: $V = 2\pi \int_0^2 \left( x \cdot 1 + x \cdot \frac{x^2}{4} \right) dx$ $V = 2\pi \int_0^2 \left( x + \frac{x^3}{4} \right) dx$

Split into two integrals: $V = 2\pi \left( \int_0^2 x \, dx + \int_0^2 \frac{x^3}{4} \, dx \right)$

---

Solve Each Integral:

1. For $\int_0^2 x \, dx$: $\int x \, dx = \frac{x^2}{2} \Big|_0^2 = \frac{2^2}{2} - \frac{0^2}{2} = 2$

2. For $\int_0^2 \frac{x^3}{4} \, dx$: $\int \frac{x^3}{4} \, dx = \frac{1}{4} \int x^3 \, dx = \frac{1}{4} \cdot \frac{x^4}{4} \Big|_0^2 = \frac{1}{16} \cdot (2^4 - 0^4) = \frac{1}{16} \cdot 16 = 1$

---

Combine Results:

$V = 2\pi \left( 2 + 1 \right) = 2\pi \cdot 3 = 6\pi$

---

Final Answer:

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

---

Would you like a detailed explanation of any step?

---

Related Questions:

1. How does the shell method differ from the disk/washer method?
2. What would change if the rotation were about the $x$-axis instead?
3. Can this problem be solved using polar coordinates? How?
4. What happens if the limits of integration are modified?
5. How does the shape of the curve influence the volume calculation?

---

Tip:

When using the shell method, always clearly identify the radius and height relative to the axis of rotation to set up the integral correctly.