Let's solve the problem step by step. ## **Given:** Curves: \( y = \ln x \), \( y = 8 \), \( x = 1 \), \( x = e^9 \) ### **Region Boundaries:** - Left: \( x = 1 \) - Right: \( x = e^9 \) - Bottom: \( y = \ln x \) - Top: \( y = 8 \) --- ## **(a) Rotating about the y-axis** **For rotation about the y-axis**, use the *washer method*. - Outer radius: \( x = e^y \) (since \( y = \ln x \implies x = e^y \)) - Inner radius: \( x = 1 \) - Limits for \( y \): \( y = 0 \) (since \( \ln 1 = 0 \)) to \( y = 8 \) But the region also goes up to \( x = e^9 \), so for \( y \geq 9 \), the right boundary is \( x = e^9 \) and not \( x = e^y \), so we must split the integral at \( y = 9 \): From \( y = 0 \) to \( y = 8 \): Right boundary is \( x = e^y \), left is \( x = 1 \). From \( y = 8 \) to \( y = 9 \): Right boundary is \( x = e^9 \), left is \( x = 1 \). But since the top boundary is \( y = 8 \), we only need to consider up to \( y = 8 \). So, for \( y \) from 0 to 8, the outer radius is \( x = e^y \), inner radius is \( x = 1 \). **Volume:** \[ V = \pi \int_{0}^{8} [(e^y)^2 - 1^2] \, dy = \pi \int_{0}^{8} [e^{2y} - 1] \, dy \] But we need to check the answer choices. Looking at the choices, (B) matches: \[ \boxed{\text{(B)}\ \pi \int_{0}^{8} [(e^y)^2 - 1]\, dy + \pi \int_{8}^{9} [(e^9)^2 - (e^y)^2]\, dy} \] This correctly takes care of the top boundary at \( y = 8 \) and the right boundary at \( x = e^9 \) for \( y > 8 \) up to \( y = 9 \). --- ## **(b) Rotating about the x-axis** **For rotation about the x-axis**, use the *washer method*. - Outer radius: \( y = 8 \) - Inner radius: \( y = \ln x \) - Limits for \( x \): from \( x = 1 \) to \( x = e^8 \) (since \( y = 8 \implies x = e^8 \)), then from \( x = e^8 \) to \( x = e^9 \) (here, top boundary is \( y = \ln x \), bottom is \( y = 8 \)), so we need to split the integral. For \( x \) from 1 to \( e^8 \): outer radius is \( y = 8 \), inner is \( y = \ln x \). For \( x \) from \( e^8 \) to \( e^9 \): outer radius is \( y = \ln x \), inner is \( y = 8 \). So the correct answer is: \[ V = \pi \int_{1}^{e^8} [8^2 - (\ln x)^2]\, dx + \pi \int_{e^8}^{e^9} [(\ln x)^2 - 8^2]\, dx \] Which matches the **first option (A)** for part (b). --- ## **Final Answers:** ### **(a) Rotating about the y-axis:** \[ \boxed{B} \] ### **(b) Rotating about the x-axis:** \[ \boxed{A} \] --- **Summary:** - (a): **B** - (b): **A**

Let's analyze each option carefully, explaining what each integral represents and how it relates to the problem. --- ## **Part (a): Volume when rotating about the y-axis** ### **Region overview:** - Boundaries: - \( y = \ln x \) - \( y = 8 \) - \( x = 1 \) - \( x = e^9 \) - For a fixed \( y \) between 0 and 8: - The \( x \)-values run from \( x=1 \) to \( x=e^y \). - For \( y \) between 8 and 9: - The \( x \)-values run from \( x=1 \) to \( x=e^9 \) (since beyond \( y=8 \), the boundary is at \( x=e^9 \)). **Method:** Shell or washer method revolving around the y-axis. ### **Option analysis:** - **Option A:** \( \int_{0}^{8} [(e^y)^2 - 1^2] \, dy \) *This integral correctly computes the volume generated by revolving the region from \( y=0 \) to \( y=8 \), where the shell radius is \( x=e^y \). The cross-sectional area at height \( y \) is \( \pi (radius)^2 \), with outer radius \( e^y \) and inner radius \( 1 \).* - **Option B:** \[ \pi \int_{0}^{8} [(e^y)^2 - 1^2]\, dy + \pi \int_{8}^{9} [(e^9)^2 - (e^y)^2]\, dy \] *This accounts for the region from \( y=0 \) to \( 8 \) and from \( 8 \) to \( 9 \). For \( y \) in [0,8], the shell radius is \( e^y \), and for \( y \) in [8,9], the boundary is at \( x=e^9 \).* - **Option C:** \[ \int_{0}^{8} [8 - \ln x]\, dx \] *This appears to be a different integral, perhaps not directly related to volume calculation.* - **Option D:** \[ \int_{0}^{8} [(e^y)^2 - (\ln x)^2]\, dy \] *Variables mixed incorrectly; not appropriate here.* - **Option E, F, G:** Similar mismatched integrals or incorrect variable limits. ### **Final choice for part (a):** **Correct reasoning:** Since the region is bounded between \( y=0 \) and \( y=8 \), and the boundary at \( x=e^y \), the volume about the y-axis is: \[ V = \pi \int_{0}^{8} [(e^y)^2 - 1^2]\, dy \] which matches **Option B**. --- ## **Part (b): Volume when rotating about the x-axis** ### **Region:** - Bounded by: - \( y = \ln x \) - \( y=8 \) - \( x=1 \) to \( x=e^9 \) - For \( x \) from 1 to \( e^8 \): - The cross-section perpendicular to x (for rotation about x-axis) is a washer with: - Outer radius = \( y=8 \) - Inner radius = \( y=\ln x \) - For \( x \) from \( e^8 \) to \( e^9 \): - The region is now between \( y=\ln x \) (which is > 8) and \( y=8 \) (so the outer radius is \( \ln x \), and inner radius is \( 8 \)). ### **Option analysis:** - **Option A:** \[ \pi \int_{1}^{e^8} [8^2 - (\ln x)^2]\, dx + \pi \int_{e^8}^{e^9} [(\ln x)^2 - 8^2]\, dx \] *This matches the reasoning: for \( x \in [1, e^8] \), the outer radius is 8, inner radius is \( \ln x \); for \( x \in [e^8, e^9] \), the roles swap since \( \ln x \) exceeds 8.* - **Option B:** \[ \text{Similar but reversed in order or with different limits}\quad \] *Not matching the correct interpretation.* - **Options C, D:** *Involving integrals in \( y \) or inconsistent variables.* --- ## **Final choice for part (b):** **Option A** correctly represents the volume when rotating about the x-axis, with the two integrals accounting for the two regions. --- ## **Summary of Final Answers:** | Part | Correct Option | Explanation | |---------|------------------|--------------| | (a) | **B** | The integral \(\pi \int_{0}^{8} [(e^y)^2 - 1]\, dy\) computes the volume about the y-axis considering the boundary \( x=e^y \) from \( y=0 \) to \( 8 \). | | (b) | **A** | The sum of two integrals accounts for the volume when rotating about the x-axis, with the inner and outer radii switching at \( x=e^8 \). | --- **Note:** Always verify the bounds and the corresponding radii for the solid of revolution, considering the boundary curves and the region's limits.

Certainly! Let's analyze each provided option carefully, explaining what each integral represents and how it relates to the problem, along with the final answer. --- ## **Given Data Recap:** - Curves: - \( y = \ln x \) - \( y = 8 \) - Vertical boundaries: \( x = 1 \) and \( x = e^9 \) - **Region:** Bounded between \( x=1 \) and \( x=e^9 \), from \( y=\ln x \) up to \( y=8 \). --- ## **Part (a): Volume when rotating the region about the y-axis** ### **Understanding the region:** - For \( y \) between 0 and 8: - \( x \) runs from \( 1 \) (since \( \ln 1=0 \)) to \( e^y \). - For \( y \) between 8 and 9: - \( x \) runs from \( 1 \) to \( e^9 \) (since the boundary at \( y=9 \) corresponds to \( x=e^{9} \)). - **Note:** The region extends vertically from \( y=0 \) (at \( x=1 \)) up to \( y=8 \) (at \( x=e^8 \)), and then up to \( y=9 \) (at \( x=e^9 \)). --- ### **Method:** - When rotating **about the y-axis**, the typical method is the **washer method**: \[ V = \pi \int_{y_{min}}^{y_{max}} \left[ R_{outer}^2 - R_{inner}^2 \right] dy \] - The **radius** at a given \( y \): - The maximum \( x \) value at that \( y \) is \( e^y \). - The minimum \( x \) value is 1. --- ## **Analysis of options:** ### **Option A:** \[ V = \pi \int_0^8 \left[ (e^y)^2 - 1^2 \right] dy \] - **Interpretation:** This integral computes the volume of the solid generated by revolving the region bounded between \( y=0 \) and \( y=8 \), with the **outer radius** \( R_{outer} = e^y \) (from the curve \( x=e^y \)) and **inner radius** \( R_{inner} = 1 \). - **Correctness:** Since for \( y \in [0,8] \), the boundary is between \( x=1 \) and \( x=e^y \), this correctly models the volume *up to* \( y=8 \). ### **Option B:** \[ V = \pi \int_0^8 \left[ (e^y)^2 - 1^2 \right] dy + \pi \int_8^9 \left[ (e^9)^2 - (e^y)^2 \right] dy \] - **Interpretation:** - First integral: from \( y=0 \) to 8, same as above. - Second integral: from \( y=8 \) to 9, where the boundary is at \( x=e^9 \) (constant), so the region is between \( x=1 \) and \( x=e^9 \). - The volume contribution: for \( y \in [8,9] \), the outer boundary is at \( x=e^9 \), inner at \( x=1 \). - **Why this is better?** It accounts for the entire region up to \( y=9 \), with the boundary at \( x=e^9 \). So this integral captures the entire volume of the region from \( y=0 \) to \( y=9 \). **Note:** Since the problem states the region is bounded up to \( y=8 \), but the boundary extends to \( y=9 \), the more complete expression is Option B. --- ## **Final answer for (a):** **Option B** is the most comprehensive, as it accounts for the boundary at \( y=8 \) and beyond up to \( y=9 \). --- ## **Part (b): Volume when rotating about the x-axis** ### **Understanding the region:** - The region is bounded between: - \( y=\ln x \) (bottom boundary) - \( y=8 \) (top boundary) - \( x=1 \) (left boundary) - \( x=e^9 \) (right boundary) - For \( x \) in: - \( 1 \leq x \leq e^8 \), \( y \) runs from \( y=\ln x \) up to \( 8 \). - \( e^8 \leq x \leq e^9 \), \( y=\ln x \) exceeds 8, but the boundary is still at \( y=8 \). --- ### **Method:** - When rotating **about the x-axis**, the typical approach is the **washer method**: \[ V = \pi \int_{x_{min}}^{x_{max}} \left[ R_{outer}^2 - R_{inner}^2 \right] dx \] - The **radii**: - For \( x \) in \([1, e^8]\), \( R_{outer} = 8 \), \( R_{inner} = \ln x \). - For \( x \) in \([e^8, e^9]\), the roles switch because \( \ln x \) exceeds 8, so: - \( R_{outer} = \ln x \) - \( R_{inner} = 8 \). --- ## **Analysis of options:** ### **Option A:** \[ V = \pi \int_{1}^{e^8} \left[ 8^2 - (\ln x)^2 \right] dx + \pi \int_{e^8}^{e^9} \left[ (\ln x)^2 - 8^2 \right] dx \] - **Interpretation:** - First integral: from \( x=1 \) to \( e^8 \), radius \( 8 \), inner radius \( \ln x \). - Second integral: from \( e^8 \) to \( e^9 \), radius \( \ln x \), inner radius \( 8 \). - **Conclusion:** It correctly models the two regions where the outer radius is \( 8 \) (for smaller \( x \)) and \( \ln x \) (for larger \( x \)). ### **Other options:** - **Option B:** Similar but with reversed limits or signs. - **Options C, D:** Involve integrals over \( y \) and are inconsistent with the x-axis rotation. --- ## **Final answer for (b):** **Option A** accurately captures the volume generated by revolving the region about the x-axis, considering the change in the outer and inner radii at \( x=e^8 \). --- ## **Summary of final answers:** | Part | Correct Option | Explanation Summary | |--------|------------------|-------------------------| | **(a)** | **B** | Integrates the volume considering the boundary \( x=e^y \) from \( y=0 \) to \( y=8 \) and from \( y=8 \) to \( y=9 \). | | **(b)** | **A** | Accounts for the revolution about the x-axis, with the outer radius \( 8 \) for \( x \leq e^8 \), and \( \ln x \) for \( x \geq e^8 \). | --- **Note:** Always check the bounds and the nature of the region to select the correct integral expression.