https://file.aiquickdraw.com/mathbot/file/17264144859863fzpia17.jpg

We are tasked with using the shell method to find the volume of the solid generated by revolving the shaded region around the y-axis.

### Step 1: Understanding the problem
- The region is bounded by:
  - \( y = 2x^2 + 1 \) (the curve)
  - \( x = \frac{\sqrt{2}}{2} \) (the vertical line)
  - \( y = 2 \) (horizontal axis)
- The region is rotated about the **y-axis**.

### Step 2: Applying the Shell Method
The formula for the volume of a solid of revolution using the shell method is:

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

Where:
- \( x \) is the radius of the shell (distance from the y-axis).
- \( f(x) \) is the height of the shell (which corresponds to the curve function).
- \( [a, b] \) is the interval over which the region is bounded.

### Step 3: Setting up the integral
From the graph and the equations:
- The limits of integration are \( x = 0 \) to \( x = \frac{\sqrt{2}}{2} \).
- The height of the shell at any point is given by \( f(x) = 2x^2 + 1 \).

Thus, the volume integral becomes:

\[
V = 2\pi \int_{0}^{\frac{\sqrt{2}}{2}} x (2x^2 + 1) \, dx
\]

### Step 4: Solving the integral
We now expand the integrand and integrate:

\[
V = 2\pi \int_{0}^{\frac{\sqrt{2}}{2}} (2x^3 + x) \, dx
\]

Separate the terms:

\[
V = 2\pi \left( \int_{0}^{\frac{\sqrt{2}}{2}} 2x^3 \, dx + \int_{0}^{\frac{\sqrt{2}}{2}} x \, dx \right)
\]

- The first integral is \( \int x^3 \, dx = \frac{x^4}{4} \).
- The second integral is \( \int x \, dx = \frac{x^2}{2} \).

Substitute the limits:

\[
V = 2\pi \left( 2 \cdot \frac{\left(\frac{\sqrt{2}}{2}\right)^4}{4} + \frac{\left(\frac{\sqrt{2}}{2}\right)^2}{2} \right)
\]

### Step 5: Simplification
- \( \left( \frac{\sqrt{2}}{2} \right)^4 = \frac{1}{4} \).
- \( \left( \frac{\sqrt{2}}{2} \right)^2 = \frac{1}{2} \).

Substitute these values into the equation:

\[
V = 2\pi \left( 2 \cdot \frac{1}{16} + \frac{1}{4} \right)
\]

\[
V = 2\pi \left( \frac{2}{16} + \frac{4}{16} \right)
\]

\[
V = 2\pi \cdot \frac{6}{16} = 2\pi \cdot \frac{3}{8}
\]

\[
V = \frac{3\pi}{4}
\]

### Final Answer:
The volume is \( \frac{3\pi}{4} \).

Would you like more details on any step? Here are 5 related questions:

1. How does the shell method differ from the disk/washer method?
2. Can we use the disk method for this problem, and if so, how would it change?
3. Why is the radius of the shell \(x\) and not something else?
4. How do we decide the limits of integration when using the shell method?
5. Can you explain more about the bounds of this region and how they were set?

**Tip:** Always sketch the region when applying volume methods to visualize the setup more clearly.

Use the shell method to find the volume of the solid generated by revolving the shaded region about the y-axis.

Discover how to calculate the volume of a solid of revolution by revolving a shaded region about the y-axis using the shell method. This tutorial covers the setup of integral calculus problems and detailed steps including the calculation of the integral of polynomial functions.

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