Q8. Evaluate ∭E z dV. E is the solid bounded by z=2-x^2-y^2 and z=1.

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.

Learn how to evaluate the triple integral ∭E z dV where E is the solid bounded by z=2-x^2-y^2 and z=1. Detailed step-by-step solution in cylindrical coordinates, suitable for advanced undergraduate level.

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