Find the volume of the solid of revolution formed by rotating the shaded area in Quadrant 1 between \( x=1 \) and \( x+2 y=7 \) about the \( y \)-axis. (graph not to scale)

To find the volume of the solid of revolution formed by rotating the shaded area in Quadrant 1 between \( x = 1 \) and the line \( x + 2y = 7 \) about the \( y \)-axis, we first need to express the line in terms of \( y \). Starting with the equation of the line: \[ x + 2y = 7 \] we can solve for \( y \): \[ 2y = 7 - x \quad \Rightarrow \quad y = \frac{7 - x}{2} \] Next, we need to determine the limits of integration. The line intersects the \( y \)-axis when \( x = 0 \): \[ 0 + 2y = 7 \quad \Rightarrow \quad 2y = 7 \quad \Rightarrow \quad y = \frac{7}{2} = 3.5 \] Thus, the line intersects the \( y \)-axis at \( (0, 3.5) \). Now, we need to find the volume of the solid formed by rotating the area between \( x = 1 \) and the line \( x + 2y = 7 \) about the \( y \)-axis. We will use the method of cylindrical shells. The volume \( V \) of the solid of revolution can be calculated using the formula: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] where \( f(x) \) is the height of the shell at position \( x \), and \( a \) and \( b \) are the limits of integration. In our case, the height of the shell is given by \( y = \frac{7 - x}{2} \), and the limits of integration are from \( x = 1 \) to \( x = 7 \) (the \( x \)-coordinate where the line intersects the \( y \)-axis). Thus, we have: \[ V = 2\pi \int_{1}^{5} x \cdot \frac{7 - x}{2} \, dx \] This simplifies to: \[ V = \pi \int_{1}^{5} x(7 - x) \, dx \] Now, we can expand the integrand: \[ x(7 - x) = 7x - x^2 \] So the volume integral becomes: \[ V = \pi \int_{1}^{5} (7x - x^2) \, dx \] Now we can compute the integral: \[ \int (7x - x^2) \, dx = \frac{7x^2}{2} - \frac{x^3}{3} \] Evaluating this from \( 1 \) to \( 5 \): \[ V = \pi \left[ \left( \frac{7(5^2)}{2} - \frac{5^3}{3} \right) - \left( \frac{7(1^2)}{2} - \frac{1^3}{3} \right) \right] \] Calculating the first part: \[ \frac{7(25)}{2} - \frac{125}{3} = \frac{175}{2} - \frac{125}{3} \] Finding a common denominator (which is 6): \[ \frac{175 \cdot 3}{6} - \frac{125 \cdot 2}{6} = \frac{525 - 250}{6} = \frac{275}{6} \] Now for the second part: \[ \frac{7(1)}{2} - \frac{1}{3} = \frac{7}{2} - \frac{1}{3} \] Finding a common denominator (which is 6): \[ \frac{7 \cdot 3}{6} - \frac{1 \cdot 2}{6} = \frac{21 - 2}{6} = \frac{19}{6} \] Now we can combine both parts: \[ V = \pi \left( \frac{275}{6} - \frac{19}{6} \right) = \pi \left( \frac{256}{6} \right) = \frac{128\pi}{3} \] Thus, the volume of the solid of revolution is: \[ \boxed{\frac{128\pi}{3}} \]