Let *R* be the region in the first quadrant bounded by the graphs of 

*y* = (4x)^(1/2), y=x^2/4

 Let *V* be the volume of the solid obtained by rotating *R* about the *y*-axis.

To find the volume of the solid obtained by rotating the region \( R \) about the \( y \)-axis, we'll use the **shell method**. The region \( R \) is bounded by the curves \( y = \sqrt{4x} \) and \( y = \frac{x^2}{4} \) in the first quadrant.

### Step 1: Determine the Points of Intersection
First, we need to find the points of intersection of the curves \( y = \sqrt{4x} \) and \( y = \frac{x^2}{4} \). We do this by setting the two equations equal to each other:

\[
\sqrt{4x} = \frac{x^2}{4}
\]

Square both sides to eliminate the square root:

\[
4x = \frac{x^4}{16}
\]

Multiply both sides by 16 to clear the fraction:

\[
64x = x^4
\]

Rearrange the equation:

\[
x^4 - 64x = 0
\]

Factor out \( x \):

\[
x(x^3 - 64) = 0
\]

This gives us:

\[
x = 0 \quad \text{or} \quad x^3 = 64
\]

Taking the cube root of 64:

\[
x = 0 \quad \text{or} \quad x = 4
\]

So, the curves intersect at \( x = 0 \) and \( x = 4 \).

### Step 2: Set Up the Volume Integral Using the Shell Method
The shell method formula for the volume \( V \) when rotating about the \( y \)-axis is:

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

Where:
- \( x \) is the radius of the shell.
- \( f(x) \) is the outer function, and \( g(x) \) is the inner function.
- \( [a, b] \) are the limits of integration.

In our case:
- The outer curve \( f(x) = \sqrt{4x} = 2\sqrt{x} \).
- The inner curve \( g(x) = \frac{x^2}{4} \).
- The limits of integration are from \( x = 0 \) to \( x = 4 \).

Thus, the volume \( V \) is:

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

### Step 3: Simplify and Integrate
Distribute the \( x \) inside the integral:

\[
V = 2\pi \int_{0}^{4} \left( 2x\sqrt{x} - \frac{x^3}{4} \right) \, dx
\]

This simplifies to:

\[
V = 2\pi \int_{0}^{4} \left( 2x^{3/2} - \frac{x^3}{4} \right) \, dx
\]

Now, integrate each term separately:

\[
\int x^{3/2} \, dx = \frac{2}{5}x^{5/2}, \quad \int x^3 \, dx = \frac{x^4}{4}
\]

So the integral becomes:

\[
V = 2\pi \left[ \frac{4}{5}x^{5/2} - \frac{x^4}{16} \right]_{0}^{4}
\]

### Step 4: Evaluate the Integral
Now, plug in the limits \( x = 4 \) and \( x = 0 \):

\[
V = 2\pi \left[ \frac{4}{5}(4)^{5/2} - \frac{(4)^4}{16} \right] - 2\pi \left[ \frac{4}{5}(0)^{5/2} - \frac{(0)^4}{16} \right]
\]

Calculating the powers:

\[
(4)^{5/2} = (2^2)^{5/2} = 2^5 = 32, \quad (4)^4 = 256
\]

Substitute back into the integral:

\[
V = 2\pi \left[ \frac{4 \times 32}{5} - 16 \right]
\]

Simplify:

\[
V = 2\pi \left[ \frac{128}{5} - 16 \right] = 2\pi \left[ \frac{128 - 80}{5} \right] = 2\pi \times \frac{48}{5} = \frac{96\pi}{5}
\]

So the volume \( V \) of the solid is:

\[
\boxed{V = \frac{96\pi}{5} \text{ cubic units}}
\]

Would you like more details on any of these steps, or do you have any other questions?

---

Here are five questions to further explore the concept:

1. How would the volume calculation change if the region was rotated about the x-axis instead?
2. What are the key differences between the disk method and the shell method?
3. Can you describe a situation where the washer method would be preferable to the shell method?
4. What would happen if we rotated the region about the line \( y = -1 \) instead of the y-axis?
5. How does the choice of axis of rotation affect the setup of the integral?

**Tip:** When using the shell method, always check the orientation of the region relative to the axis of rotation to correctly identify the radius and height of the shells.

Learn how to calculate the volume of a solid obtained by rotating a region bounded by y = sqrt(4x) and y = x^2/4 about the y-axis using the shell method. Detailed step-by-step solution with integral setup and evaluation. Suitable for advanced high school students and calculus enthusiasts.

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