Let's go step-by-step, analyzing each integral and matching it to the axis of revolution. --- ### **First, Analyze the Region \(R\)** Given: - \( y = \sqrt{2 - x} \) (right boundary, for \( x \leq 2 \)) - \( y = \sqrt{2 + x} \) (left boundary, for \( x \geq -2 \)) - \( y = 0 \) (bottom boundary) For \( y = \sqrt{2 - x} \), \( x = 2 - y^2 \). For \( y = \sqrt{2 + x} \), \( x = -2 + y^2 \). So, \( x \) ranges from \( -2 \) to \( 2 \), and \( y \) from \( 0 \) to \( \sqrt{2 + 2} = 2 \). --- ### **Integral Matching** #### **Question 1** \[ \pi \int_{-2}^{2} (|x| + 4\sqrt{2 - |x|} - 2) dx \] **Analysis:** - The presence of \(|x|\) and \(\sqrt{2 - |x|}\) suggests symmetry about \(x = 0\). - The integrand is more complex than a simple disk/washer, so it's likely a shell method. - The form \(|x|+4\sqrt{2-|x|}-2\) suggests a shell method about \(y = 2\). **Answer:** **y = 2** --- #### **Question 2** \[ \pi \int_{-2}^{2} (2 - |x|) dx \] **Analysis:** - The integrand \(2 - |x|\) is the vertical distance between the two curves for a given \(x\), which is the height of the region at \(x\). - Rotating about the \(x\)-axis (\(y = 0\)), this is a washer method. **Answer:** **y = 0** --- #### **Question 3** \[ \pi \int_{-2}^{2} (2 - |x| + 4\sqrt{2 - |x|})dx \] **Analysis:** - The integrand is similar to Question 1, but with a plus sign instead of minus. - This suggests a shell method, but about a different line. - The form fits rotation about \(y = -2\), as the distance from the shell to the axis increases by 2 units. **Answer:** **y = -2** --- #### **Question 4** \[ \pi \int_{0}^{\sqrt{2}} (16 - y^2) dy \] **Analysis:** - This is in terms of \(y\). \(x\) runs from \(-2\) to \(2\), so the length between \(x = -2 + y^2\) and \(x = 2 - y^2\) is \( (2 - y^2) - (-2 + y^2) = 4 - 2y^2 \). - But the integrand is \(16 - y^2\), which is likely for shells about \(x = 0\). **Answer:** **x = 0** --- ### **Final Answers** 1. **y = 2** 2. **y = 0** 3. **y = -2** 4. **x = 0** --- #### **Summary Table** | Question | Integral | Axis of Revolution | |----------|----------------------------------------------------|--------------------| | 1 | \( \pi \int_{-2}^{2} (|x| + 4\sqrt{2 - |x|} - 2) dx \) | y = 2 | | 2 | \( \pi \int_{-2}^{2} (2 - |x|) dx \) | y = 0 | | 3 | \( \pi \int_{-2}^{2} (2 - |x| + 4\sqrt{2 - |x|}) dx \) | y = -2 | | 4 | \( \pi \int_{0}^{\sqrt{2}} (16 - y^2) dy \) | x = 0 |

Let's analyze each integral carefully, understand the region \( R \), and match each integral to its axis of revolution step-by-step. --- ## Step 1: Understand the region \( R \) Given curves: - \( y = \sqrt{2 - x} \) - \( y = \sqrt{2 + x} \) - \( y = 0 \) **Domain of \(x\):** - Since \( y \ge 0 \), for the square roots to be real: - \( 2 - x \ge 0 \Rightarrow x \le 2 \) - \( 2 + x \ge 0 \Rightarrow x \ge -2 \) **Intersections:** - At \( x = -2 \), \( y = \sqrt{2 - (-2)} = \sqrt{4} = 2 \) - At \( x = 2 \), \( y = \sqrt{2 - 2} = 0 \) - Similarly, \( y = \sqrt{2 + x} \), at \( x = -2 \), \( y = 0 \); at \( x=2 \), \( y=2 \) **Vertical cross-section:** - For each \( x \in [-2, 2] \), the region is bounded above by the larger of the two curves, which are symmetric about \( x=0 \). **\( y \)-range:** - From \( y=0 \) up to \( y=2 \) **In \( xy \)-plane:** - The region \( R \) is between \( x=-2 \) and \( x=2 \), - For each \( x \), between \( y=0 \) and \( y=\sqrt{2 - x} \) or \( y=\sqrt{2 + x} \). --- ## Step 2: Express the region in terms of \( x \) and \( y \) - For a fixed \( y \), find the \( x \)-limits: From \( y = \sqrt{2 - x} \Rightarrow x = 2 - y^2 \) From \( y = \sqrt{2 + x} \Rightarrow x = y^2 - 2 \) - The region at height \( y \): \[ x \in [ y^2 - 2 , 2 - y^2 ] \] - Since \( y \in [ 0, 2 ] \), the width of the region at height \( y \) is: \[ \text{Width} = (2 - y^2) - (y^2 - 2) = 4 - 2 y^2 \] --- ## Step 3: Match integrals to axes of revolution ### **Question 4:** \[ V = \pi \int_{0}^{\sqrt{2}} (16 - y^2) dy \] - This integral is in \( y \), with \( y \) from 0 to \( \sqrt{2} \). - The integrand \( 16 - y^2 \) resembles a disc of radius \( \sqrt{16 - y^2} \), which suggests rotation about the \( x \)-axis (\( y=0 \)), forming disks. --- ### **Question 2:** \[ V = \pi \int_{-2}^{2} (2 - |x|) dx \] - The integral over \( x \) from \(-2\) to 2 of the linear function \( 2 - |x| \) is the area of a shape symmetric about \( x=0 \), which suggests a shell or washer method with rotation about the \( x \)-axis or \( y=0 \). - But since the integrand involves \( x \), and the integral is over \( x \), it matches rotation about the \( x \)-axis (\( y=0 \)). --- ### **Question 1:** \[ V = \pi \int_{-2}^{2} \left( |x| + 4 \sqrt{2 - |x|} - 2 \right) dx \] - The presence of \( |x| \) and \( \sqrt{2 - |x|} \) suggests the shape is symmetric about \( x=0 \). - The structure resembles a shell method around the line \( y=2 \) (since the \( 4\sqrt{2 - |x|} \) term indicates a radius related to \( y \)). --- ### **Question 3:** \[ V = \pi \int_{-2}^{2} \left( 2 - |x| + 4 \sqrt{2 - |x|} \right) dx \] - Similar to Question 1 but with a plus sign, indicating an axis of revolution at \( y=-2 \). --- ## Final matching: | Question | Integral Expression | Axis of Revolution | Reasoning | |------------|-------------------------|----------------------|-----------| | 1 | \( \pi \int_{-2}^{2} (|x| + 4 \sqrt{2 - |x|} - 2) dx \) | **\( y=2 \)** | Shell method around \( y=2 \), due to \( 4 \sqrt{2 - |x|} \) term and symmetry. | | 2 | \( \pi \int_{-2}^{2} (2 - |x|) dx \) | **\( y=0 \)** | Washer/disc method around \( y=0 \). The integral over \( x \) with \( 2 - |x| \) matches the width at each \( x \). | | 3 | \( \pi \int_{-2}^{2} (2 - |x| + 4 \sqrt{2 - |x|}) dx \) | **\( y=-2 \)** | Similar to Question 1 but shifted downward; the structure suggests rotation about \( y=-2 \). | | 4 | \( \pi \int_{0}^{\sqrt{2}} (16 - y^2) dy \) | **\( x=0 \)** | Rotation about the \( y \)-axis, because the integral is in \( y \), and the integrand suggests a disk radius \( \sqrt{16 - y^2} \). | --- ## **Final Answers:** ### **Question 1:** y = 2 ### **Question 2:** y = 0 ### **Question 3:** y = -2 ### **Question 4:** x = 0 --- ## **Summary:** - **Question 1:** Rotation about **\( y=2 \)** (top line of the region). - **Question 2:** Rotation about **\( y=0 \)** (the base). - **Question 3:** Rotation about **\( y=-2 \)** (bottom line). - **Question 4:** Rotation about **\( x=0 \)** (the y-axis). --- If you'd like further details on the derivation of each integral, let me know!

Let's analyze each question carefully, understand the region \( R \), and determine how the integrals relate to the axis of revolution step-by-step. --- ## Step 1: Understand the region \( R \) Given the curves: - \( y = \sqrt{2 - x} \) - \( y = \sqrt{2 + x} \) - \( y = 0 \) ### Domain in \( x \): - For \( y \ge 0 \), the square roots are real when: \[ 2 - x \ge 0 \Rightarrow x \le 2 \] \[ 2 + x \ge 0 \Rightarrow x \ge -2 \] Thus, \( x \in [-2, 2] \). ### Boundary points: - At \( x = -2 \), \( y = \sqrt{2 - (-2)} = \sqrt{4} = 2 \). - At \( x = 2 \), \( y = 0 \). - The curves are symmetric about \( x=0 \) with maximum height \( y=2 \). ### Cross-section: - For each \( x \in [-2, 2] \), the top boundary is given by the maximum of \( y = \sqrt{2 - x} \) and \( y = \sqrt{2 + x} \). Because of symmetry, the region is bounded between these two curves and the \( x \)-axis. ### Vertical cross-section: - For a fixed \( y \), the \( x \)-limits are found from: \[ y = \sqrt{2 - x} \Rightarrow x = 2 - y^2 \] \[ y = \sqrt{2 + x} \Rightarrow x = y^2 - 2 \] - The region at height \( y \) runs between: \[ x = y^2 - 2 \quad \text{and} \quad x = 2 - y^2 \] - The width at height \( y \): \[ \text{Width} = (2 - y^2) - (y^2 - 2) = 4 - 2 y^2 \] --- ## Step 2: Relationship of integrals to axes of revolution Now, analyze each integral and match it based on the shape and the axis about which the solid is generated. --- ### **Question 4:** \[ V = \pi \int_{0}^{\sqrt{2}} (16 - y^2) \, dy \] - The integrand \( 16 - y^2 \) suggests the radius squared of a disk: \[ r^2 = 16 - y^2 \] - Therefore, the volume is obtained by revolving around the \( x \)-axis (\( y=0 \)), creating disks with radius \( r = \sqrt{16 - y^2} \). - The limits \( y = 0 \) to \( y = \sqrt{2} \) match the height of the region (since maximum \( y \) is 2, and the integral is up to \( \sqrt{2} \)). **Conclusion:** **Integral 4** corresponds to **rotation about the \( x=0 \) line (the y-axis)**, because the integral is in \( y \), and the shape is a disk with radius depending on \( y \). --- ### **Question 2:** \[ V = \pi \int_{-2}^{2} (2 - |x|) \, dx \] - The integrand \( 2 - |x| \) is symmetric about \( x=0 \), representing the width of the region at position \( x \). - When revolving around the \( x \)-axis (\( y=0 \)), the cross-sectional disks at \( x \) have radius proportional to the height of the region at \( x \), which in this case is \( y = 2 - |x| \). - The integral over \( x \) from \(-2\) to \(2\) suggests a shell or washer method about the \( x \)-axis. **Conclusion:** **Integral 2** is for **rotation about \( y=0 \)** (the \( x \)-axis), because the integral is in \( x \), and the shape is symmetric. --- ### **Question 1:** \[ V = \pi \int_{-2}^{2} \left( |x| + 4 \sqrt{2 - |x|} - 2 \right) dx \] - The integrand combines \( |x| \) and \( \sqrt{2 - |x|} \) terms. - The structure hints at a shell method at a height \( y = 2 \), with the \( 4 \sqrt{2 - |x|} \) term resembling the radius from the shell to the axis at \( y=2 \). - The presence of \( |x| \) and the form of the integrand suggests this integral is for the volume generated by rotating around the line \( y=2 \). **Conclusion:** **Integral 1** corresponds to **rotation about \( y=2 \)**. --- ### **Question 3:** \[ V = \pi \int_{-2}^{2} \left( 2 - |x| + 4 \sqrt{2 - |x|} \right) dx \] - Similar to Question 1 but with the plus sign, indicating a shift downward. - The structure suggests rotation about \( y=-2 \). **Conclusion:** **Integral 3** corresponds to **rotation about \( y=-2 \)**. --- ## **Final matching:** | Question | Integral Expression | Axis of Revolution | Explanation | |------------|-------------------------|----------------------|--------------| | 1 | \( \pi \int_{-2}^{2} (|x| + 4 \sqrt{2 - |x|} - 2) dx \) | **\( y=2 \)** | Shell method at \( y=2 \) (top boundary). | | 2 | \( \pi \int_{-2}^{2} (2 - |x|) dx \) | **\( y=0 \)** | Rotating around \( y=0 \), the \( x \)-axis. | | 3 | \( \pi \int_{-2}^{2} (2 - |x| + 4 \sqrt{2 - |x|}) dx \) | **\( y=-2 \)** | Shell method at \( y=-2 \). | | 4 | \( \pi \int_{0}^{\sqrt{2}} (16 - y^2) dy \) | **\( x=0 \)** | Rotation about \( x=0 \) (the y-axis). | --- ## **Summary of final answers:** - **Question 1:** **\( y=2 \)** - **Question 2:** **\( y=0 \)** - **Question 3:** **\( y=-2 \)** - **Question 4:** **\( x=0 \)** --- ## **Note:** This detailed reasoning connects the shape, the integrals, and the axes of revolution. The key was understanding the shape's symmetry and the form of the integrals to identify the axes. If you need further clarification or derivations, I’m happy to help!