To find the volume of the solid obtained by rotating the region $E$, which is bounded by the curve $x = y^2$, and the lines $y = 2$ and $x = 0$, about the x-axis, we will use the method of cylindrical shells.

When using the method of cylindrical shells, the formula for the volume of the solid is given by:

$V = 2\pi \int_a^b y \cdot f(y) \, dy$

where $y$ is the distance from the axis of rotation (x-axis in this case), and $f(y)$ is the function representing the radius of the shell (or the x-coordinate of the curve $x = y^2$).

For our problem:

• The region $E$ is bounded by $x = y^2$, $y = 2$, and $x = 0$.
• We are rotating around the x-axis.

Within these boundaries:

• The line $y = 2$ gives the upper limit for $y$.
• The line $x = 0$ corresponds to $x = y^2 = 0$, so $y = 0$ gives the lower limit for $y$.

Therefore, $y$ ranges from 0 to 2.

Now, the shell radius is $y$, and the shell height is given by the $x$ value from the curve: $x = y^2$.

We can now set up the integral for the volume:

$V = 2\pi \int_0^2 y (y^2) \, dy = 2\pi \int_0^2 y^3 \, dy$

Evaluate this integral:

$V = 2\pi \left[ \frac{y^4}{4} \right]_0^2$

$= 2\pi \left[ \frac{2^4}{4} - \frac{0^4}{4} \right]$

$= 2\pi \left[ \frac{16}{4} \right]$

$= 2\pi \cdot 4$

$= 8\pi$

Thus, the volume of the solid obtained by rotating the region $E$ about the x-axis is $8\pi$ cubic units.