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 will use the method of cylindrical shells.

Step 1: Set up the integral using cylindrical shells

The formula for the volume of a solid formed by rotating a region around the y-axis using cylindrical shells is:

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

In this case:

• The radius of a shell at a point $x$ is simply $x$, since the distance from the y-axis to the point on the curve is $x$.
• The height of the shell at a point $x$ is the vertical distance between the curves $y = 4$ and $y = x^2$. This height is given by $4 - x^2$.

Step 2: Determine the limits of integration

The region is bounded by the curve $y = x^2$ and the line $y = 4$. To find the limits of integration, we solve for $x$ where the two curves intersect:

$x^2 = 4 \quad \Rightarrow \quad x = -2 \text{ and } x = 2$

Thus, the limits of integration are $x = -2$ to $x = 2$.

Step 3: Set up the integral

The volume $V$ is:

$V = 2\pi \int_{-2}^{2} x \cdot (4 - x^2) \, dx$

Step 4: Evaluate the integral

Now, we evaluate the integral:

$V = 2\pi \int_{-2}^{2} (4x - x^3) \, dx$

First, break the integral into two parts:

$V = 2\pi \left[ \int_{-2}^{2} 4x \, dx - \int_{-2}^{2} x^3 \, dx \right]$

1. The integral of $4x$:

$\int_{-2}^{2} 4x \, dx = 0 \quad \text{(since the integrand is an odd function, and the limits are symmetric around 0)}$

2. The integral of $x^3$:

\int_{-2}^{2} x^3 \, dx = 0 \quad \text{(since x^3 is also an odd function and the limits are symmetric)}

Thus, the volume is:

$V = 0$

It seems the integral is 0. Since both functions are symmetric we need a numerical approach