To find the volume of the solid formed by rotating the region bounded by $y = x^2$ and $y = 4$ around the $y$-axis, we use the method of cylindrical shells or washers. Let's analyze the problem step-by-step.

Step 1: Understand the Region and Rotation

The region is bounded by:

1. $y = x^2$: A parabola opening upwards.
2. $y = 4$: A horizontal line.

The intersection points of these curves can be found by solving $x^2 = 4$, giving $x = \pm 2$. So the region lies between $x = -2$ and $x = 2$.

Step 2: Use the Shell Method

For rotation around the $y$-axis, the shell method is appropriate. The volume of the solid is given by: $V = 2\pi \int_{x=-2}^{x=2} \text{(radius)} \cdot \text{(height)} \, dx$

• Radius: The distance from the $y$-axis, $|x| = x$ (since $x$ is positive or negative).
• Height: The vertical difference between $y = 4$ and $y = x^2$, which is $4 - x^2$.

Substitute these into the formula: $V = 2\pi \int_{-2}^2 x(4 - x^2) \, dx$

Step 3: Simplify the Integral

Expand the integrand: $x(4 - x^2) = 4x - x^3$ Thus: $V = 2\pi \int_{-2}^2 (4x - x^3) \, dx$

Since $4x$ is odd and $x^3$ is odd, their integral over $[-2, 2]$ simplifies: $\int_{-2}^2 4x \, dx = 0 \quad \text{and} \quad \int_{-2}^2 x^3 \, dx = 0$ So the volume is: $V = 2\pi \cdot 0 = 0$

This matches option a.