The problem asks to:

1. Sketch the region $R$, defined by the equations $y = x^2$, $y = 4$, and the $y$-axis (already done in your sketch).
2. Calculate the volume of the solid formed when the region $R$ is rotated about the $x$-axis using the method of cylindrical shells.

Volume Calculation Using Cylindrical Shells

The formula for the volume using cylindrical shells is:

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

Here:

• Radius = $y$, since the rotation is about the $x$-axis.
• Height = $x$, based on the relation $y = x^2$ (or $x = \sqrt{y}$).
• Limits of integration: From $y = 0$ to $y = 4$.

Substitute: $V = \int_{0}^{4} 2\pi (y)(\sqrt{y}) \, dy$

$V = 2\pi \int_{0}^{4} y^{3/2} \, dy$

Solving the Integral

1. Integrate $y^{3/2}$: $\int y^{3/2} \, dy = \frac{2}{5} y^{5/2}$

2. Apply limits $y = 0$ to $y = 4$: $V = 2\pi \left[ \frac{2}{5} y^{5/2} \right]_{0}^{4}$

   At $y = 4$: $\frac{2}{5} (4^{5/2}) = \frac{2}{5} (32) = \frac{64}{5}$

   At $y = 0$: $0$

   Total volume: $V = 2\pi \cdot \frac{64}{5} = \frac{128\pi}{5}$

Final Answer

The volume is: $V = \frac{128\pi}{5} \, \text{units}^3$

Would you like me to explain any part of the derivation or the method further?

Questions for deeper understanding:

1. What is the principle behind the method of cylindrical shells?
2. How do you decide the integration limits in problems like this?
3. Could this problem be solved with a different method, like washers?
4. Why is the radius $y$ in this specific scenario?
5. How would the formula change if the region rotated around the $y$-axis?

Tip: Always verify the rotation axis to avoid confusion between radii and heights!