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 need to set up the integral in cylindrical coordinates for simplicity.

Step-by-Step Solution:

1. Identify the bounds of the region $E$:

   The solid $E$ is bounded above by the surface $z = 2 - x^2 - y^2$ and below by the plane $z = 1$.

2. Convert to cylindrical coordinates:

   Recall that in cylindrical coordinates: $x = r \cos \theta, \quad y = r \sin \theta, \quad z = z$ and the volume element $dV = r \, dz \, dr \, d\theta$.

3. Determine the bounds for $r$, $\theta$, and $z$:

   • The region is rotationally symmetric about the z-axis, so $\theta$ ranges from $0$ to $2\pi$.
   • For $z$, it ranges from $z = 1$ (the lower bound) to $z = 2 - r^2$ (the upper bound given by the surface $z = 2 - x^2 - y^2$).
   • For $r$, it ranges from $0$ to $r = \sqrt{1}$, since $z = 1$ intersects $z = 2 - r^2$ at $r = 1$.

4. Set up the integral:

   $\iiint_E z \, dV = \int_0^{2\pi} \int_0^1 \int_1^{2-r^2} z \cdot r \, dz \, dr \, d\theta$

5. Evaluate the integral:

   • Integrate with respect to $z$: $\int_1^{2-r^2} z \, dz = \left[ \frac{z^2}{2} \right]_1^{2-r^2} = \frac{(2 - r^2)^2}{2} - \frac{1^2}{2} = \frac{(4 - 4r^2 + r^4)}{2} - \frac{1}{2} = 2 - 2r^2 + \frac{r^4}{2} - \frac{1}{2} = \frac{3}{2} - 2r^2 + \frac{r^4}{2}$

   • Integrate with respect to $r$: $\int_0^1 \left( \frac{3}{2} - 2r^2 + \frac{r^4}{2} \right) r \, dr = \int_0^1 \left( \frac{3r}{2} - 2r^3 + \frac{r^5}{2} \right) dr$ [ \int_0^1 \frac{3r}{2} , dr - \int_0^1 2r^3 , dr + \int_0^1 \frac{r^5}{2} , dr = \left[ \frac{3r^2}{4} \right]_0^1 - \left[ \frac{2r^4}{4} \right]_0^1 + \left[ \frac{r^6}{12} \right]_0^1 ] $= \frac{3}{4} - \frac{1}{2} + \frac{1}{12} = \frac{9}{12} - \frac{6}{12} + \frac{1}{12} = \frac{4}{12} = \frac{1}{3}$

   • Integrate with respect to $\theta$: $\int_0^{2\pi} \frac{1}{3} \, d\theta = \frac{1}{3} \cdot 2\pi = \frac{2\pi}{3}$

Final Answer:

$\iiint_E z \, dV = \frac{2\pi}{3}$

Would you like more details on any of these steps or have any questions?

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Here are 8 related questions to further your understanding:

1. How do cylindrical coordinates simplify the integration process for rotationally symmetric regions?
2. What are the limits of integration for $r$ if the solid were bounded by $z = 3 - x^2 - y^2$ and $z = 0$?
3. How would the bounds change if the lower bound were $z = 0$ instead of $z = 1$?
4. Can you explain the geometric interpretation of the bounds in cylindrical coordinates?
5. What if the upper bound were $z = 4 - x^2 - y^2$ and the lower bound were $z = 2$?
6. How would you set up the integral if $E$ were bounded by $z = 1 - x^2 - y^2$ and $z = -1$?
7. How does the Jacobian determinant $r$ in cylindrical coordinates affect the volume element $dV$?
8. Can you verify the result using Cartesian coordinates instead of cylindrical coordinates?

Tip: Always visualize the region of integration when dealing with triple integrals to better understand the bounds and the volume element.