Let's break down the problem step by step: **Given:** - Region \( R \) in the first quadrant bounded by: - \( y = x^2 \) - \( y = 6 - x \) - \( x = 0 \) - We are to **set up the integral** (do not solve, just set up) to find the **volume** generated when \( R \) is rotated about the **y-axis**. --- ### **Step 1: Find the bounds of integration (intersection points)** Set \( y = x^2 \) and \( y = 6 - x \): \[ x^2 = 6 - x \implies x^2 + x - 6 = 0 \] \[ (x + 3)(x - 2) = 0 \] So, \( x = -3 \) or \( x = 2 \). Since we are in the **first quadrant** (\( x \geq 0 \)), use \( x = 0 \) and \( x = 2 \). --- ### **Step 2: Express in terms of \( x \) (method of shells)** For rotation about the **y-axis**, use the **shell method**: - Height of shell: \( (6-x) - (x^2) \) - Radius of shell: \( x \) - Thickness: \( dx \) Volume of shell: \( 2\pi x \left[(6-x) - x^2 \right] dx \) --- ### **Step 3: Set up the integral** Bounds: \( x = 0 \) to \( x = 2 \) \[ V = \int_{x=0}^{x=2} 2\pi x \left[(6-x) - x^2\right] dx \] --- ### **Final Answer (integral setup):** \[ \boxed{ V = \int_{0}^{2} 2\pi x \left[ (6-x) - x^2 \right] dx } \] This is the required integral to find the volume when the region \( R \) is rotated about the y-axis.