Section 6.2: Problem 2 (1 point) Find the volume formed by rotating the region enclosed by: \[ x=9 y \text { and } y^{3}=x \text { with } y \geq 0 \] about the \( y \)-axis. Volume \( =-\frac{\pi}{2} \)

To find the volume of the region enclosed by \( x = 9y \) and \( y^3 = x \) for \( y \geq 0 \) when rotated about the \( y \)-axis, we'll use the **washer method**. Here's a step-by-step solution: 1. **Determine the Points of Intersection:** \[ 9y = y^3 \implies y^3 - 9y = 0 \implies y(y^2 - 9) = 0 \implies y = 0 \text{ or } y = \pm 3 \] Since \( y \geq 0 \), the points of intersection are \( y = 0 \) and \( y = 3 \). 2. **Set Up the Washer Method:** The volume \( V \) is given by: \[ V = \pi \int_{0}^{3} \left[ (x_{\text{outer}})^2 - (x_{\text{inner}})^2 \right] dy \] Here: - \( x_{\text{outer}} = 9y \) - \( x_{\text{inner}} = y^3 \) 3. **Substitute and Integrate:** \[ V = \pi \int_{0}^{3} \left[ (9y)^2 - (y^3)^2 \right] dy = \pi \int_{0}^{3} \left[ 81y^2 - y^6 \right] dy \] \[ V = \pi \left[ \frac{81y^3}{3} - \frac{y^7}{7} \right]_0^3 = \pi \left[ 27y^3 - \frac{y^7}{7} \right]_0^3 \] Evaluating from 0 to 3: \[ V = \pi \left[ 27(3)^3 - \frac{(3)^7}{7} \right] = \pi \left[ 729 - \frac{2187}{7} \right] \] \[ V = \pi \left[ \frac{5103}{7} - \frac{2187}{7} \right] = \pi \left[ \frac{2916}{7} \right] = \frac{2916\pi}{7} \] 4. **Final Answer:** The volume is \( \frac{2916\pi}{7} \). **Note:** The initial answer you provided was negative, which isn't possible for a volume. Volumes are always non-negative quantities. **Final Answer:** The volume is 2916 × π divided by seven. In symbols, Volume = 2916 π ⁄ 7