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?

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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.