To evaluate the triple integral $\iiint_E z \, dV$, where $E$ is the solid bounded by $z = 2 - x^2 - y^2$ and $z = 1$, we will use cylindrical coordinates for simplicity.

In cylindrical coordinates $(r, \theta, z)$, the boundaries transform as follows:

• $x = r \cos \theta$
• $y = r \sin \theta$
• $z = z$
• The Jacobian determinant for the transformation to cylindrical coordinates is $r$.

The solid $E$ is bounded by $z = 1$ (bottom surface) and $z = 2 - r^2$ (top surface).

1. The bounds for $z$ are from $1$ to $2 - r^2$.
2. The bounds for $r$ are determined by the projection of the region onto the $xy$-plane. At $z = 1$, $1 = 2 - r^2$, so $r = \sqrt{1} = 1$. Thus, $r$ ranges from $0$ to $1$.
3. The bounds for $\theta$ are from $0$ to $2\pi$.

The integral in cylindrical coordinates becomes: $\iiint_E z \, dV = \int_0^{2\pi} \int_0^1 \int_1^{2-r^2} z \, r \, dz \, dr \, d\theta$

First, integrate with respect to $z$:

= \frac{(4 - 4r^2 + r^4)}{2} - \frac{1}{2} = 2 - 2r^2 + \frac{r^4}{2} - \frac{1}{2} = \frac{3}{2} - 2r^2 + \frac{r^4}{2}$$ Now, the integral becomes: $$\int_0^{2\pi} \int_0^1 \left( \frac{3}{2} - 2r^2 + \frac{r^4}{2} \right) r \, dr \, d\theta$$ Separate the integral: $$\int_0^{2\pi} d\theta \int_0^1 \left( \frac{3}{2}r - 2r^3 + \frac{r^5}{2} \right) dr$$ Evaluate the $$\theta$$ integral: $$\int_0^{2\pi} d\theta = 2\pi$$ Now, evaluate the $$r$$ integral: \[ \int_0^1 \left( \frac{3}{2}r - 2r^3 + \frac{r^5