Let's solve the problem step by step: ## **We need to:** Find the volume generated by rotating region \( R_1 \) about the line \( BC \). --- ### **Step 1: Identify Region \( R_1 \)** Region \( R_1 \) is bounded by: - \( y = 2x \) (bottom curve) - \( x = 0 \) (left, y-axis) - \( x = 1 \) (right) - \( y = 2 \) (top, which is the line \( BC \), since \( B(1,2) \) and \( C(0,2) \) ) So, \( R_1 \) is the region between \( y = 2x \) and \( y = 2 \) from \( x = 0 \) to \( x = 1 \). --- ### **Step 2: Set Up the Rotation** We are rotating **about the line \( BC \), which is \( y = 2 \)**. For any vertical slice at \( x \), the distance from \( y = 2x \) up to \( y = 2 \) is the thickness of the region. The radius of rotation for a point \( (x, y) \) is \( |2 - y| \). --- ### **Step 3: Use the Washer Method** For each \( x \) from 0 to 1, the region to be rotated is from \( y = 2x \) up to \( y = 2 \). - Inner radius: 0 (since the region starts at \( y = 2x \)) - Outer radius: \( 2 - 2x \) (distance from \( y = 2x \) up to \( y = 2 \)) The area of the washer at \( x \) is: \[ A(x) = \pi (2 - 2x)^2 \] But, because the region is bounded below by \( y = 2x \) and above by \( y = 2 \), and rotating about \( y = 2 \), the height is \( y = 2 - 2x \). But since the region is a vertical strip, the volume element is: \[ dV = \text{(circumference)} \times \text{height} \times dx \] Where: - Circumference of rotation: \( 2\pi \times \text{radius} \) - Height: \( y = 2 - 2x \) - Radius: Distance from \( y = 2x \) up to \( y = 2 \) is \( 2 - 2x \) So, \[ dV = 2\pi (\text{radius}) \cdot (\text{height}) dx = 2\pi (2 - 2x) \cdot dx \] Wait: This is the formula for rotating a strip around a **horizontal axis** above it (shell method). ### **Shell Method (about \( y = 2 \))** For a vertical strip at \( x \) from \( x = 0 \) to \( x = 1 \): - Height of the shell: \( y = 2x \) - Distance from the axis \( y = 2 \): \( 2 - y = 2 - 2x \) So, the volume is: \[ V = \int_{x=0}^{x=1} 2\pi (\text{radius}) \cdot (\text{height}) dx = \int_{x=0}^{x=1} 2\pi (2 - 2x)(2x) dx \] --- ### **Step 4: Compute the Integral** \[ V = 2\pi \int_{0}^{1} (2 - 2x)(2x) dx \] \[ = 2\pi \int_{0}^{1} (4x - 4x^2) dx \] \[ = 2\pi \left[ \int_{0}^{1} 4x\, dx - \int_{0}^{1} 4x^2\, dx \right] \] \[ = 2\pi \left[ 4 \int_{0}^{1} x\, dx - 4 \int_{0}^{1} x^2\, dx \right] \] \[ = 2\pi \left[ 4 \left(\frac{x^2}{2}\right)_{0}^{1} - 4 \left(\frac{x^3}{3}\right)_{0}^{1} \right] \] \[ = 2\pi \left[ 4 \left(\frac{1}{2} - 0\right) - 4 \left(\frac{1}{3} - 0\right) \right] \] \[ = 2\pi \left[ 2 - \frac{4}{3} \right] \] \[ = 2\pi \left[ \frac{6 - 4}{3} \right] \] \[ = 2\pi \left( \frac{2}{3} \right) \] \[ = \frac{4\pi}{3} \] --- ## **Final Answer** \[ \boxed{\frac{4\pi}{3}} \] This is the volume generated by rotating region \( \mathcal{R}_1 \) about the line \( BC \) (i.e., \( y = 2 \)).

Certainly! Here are two pure theoretical lines related to the problem of finding volume generated by rotation: 1. **Shell Method Formula:** *When revolving a region around a horizontal or vertical axis, the volume can be calculated using the shell method, which involves integrating the lateral surface areas of cylindrical shells formed by vertical or horizontal strips. For a vertical strip at position \( x \), rotated about a horizontal axis \( y = c \), the volume element is:* \[ dV = 2\pi \times (\text{radius}) \times (\text{height}) \, dx \] *where the radius is the distance from the strip to the axis of rotation, and the height is the length of the strip.* 2. **Integral Setup for Volume of Revolution:** *The volume \( V \) of a solid of revolution generated by rotating a region bounded by functions \( y = f(x) \) and \( y = g(x) \) over \( x \in [a, b] \), about a line \( y = c \), is given by:* \[ V = 2\pi \int_a^b (\text{distance from } y = c \text{ to the strip}) \times (\text{height of the strip}) \, dx \] *This integral accounts for the circumferences of the shells formed during rotation, integrating over the domain of the region.*

Certainly! Here are two pure theoretical statements related to volumes of revolution: 1. When a region is revolved around a line, the volume of the resulting solid can be calculated by summing the lateral surface areas of infinitesimal cylindrical shells formed by rotating thin strips of the region. 2. The volume generated by rotating a bounded region about a line is obtained by integrating the product of the circumference of the shells and their height over the domain of the region.