Find the volume of the solid obtained by rotating the region bounded by \[ y=8 x+64, y=0, x=0 \] about the \( y \)-axis. Volume \( = \)

To find the volume of the solid obtained by rotating the region bounded by the lines \( y = 8x + 64 \), \( y = 0 \), and \( x = 0 \) about the \( y \)-axis, we can use the method of cylindrical shells. First, we need to determine the points of intersection of the line \( y = 8x + 64 \) with the line \( y = 0 \): \[ 0 = 8x + 64 \implies 8x = -64 \implies x = -8 \] However, since we are also bounded by \( x = 0 \), we only consider the region where \( x \) is non-negative. Thus, we will find the volume of the region from \( x = 0 \) to the point where the line intersects the x-axis, which is at \( x = -8 \) (not relevant here since we are only considering \( x \geq 0 \)). Next, we express \( x \) in terms of \( y \) from the equation of the line: \[ y = 8x + 64 \implies 8x = y - 64 \implies x = \frac{y - 64}{8} \] The volume \( V \) of the solid of revolution about the \( y \)-axis can be calculated using the formula for cylindrical shells: \[ V = 2\pi \int_{y_1}^{y_2} x \cdot h(y) \, dy \] where \( h(y) \) is the height of the shell at a given \( y \), and \( x \) is the radius of the shell. In our case, the height \( h(y) \) is given by the \( x \)-coordinate expressed in terms of \( y \): \[ h(y) = x = \frac{y - 64}{8} \] The bounds for \( y \) are from \( y = 0 \) (the x-axis) to \( y = 64 \) (the y-intercept of the line). Thus, we have: \[ V = 2\pi \int_{0}^{64} \left(\frac{y - 64}{8}\right) y \, dy \] Now, we simplify the integrand: \[ V = 2\pi \int_{0}^{64} \frac{y^2 - 64y}{8} \, dy = \frac{2\pi}{8} \int_{0}^{64} (y^2 - 64y) \, dy = \frac{\pi}{4} \int_{0}^{64} (y^2 - 64y) \, dy \] Next, we compute the integral: \[ \int (y^2 - 64y) \, dy = \frac{y^3}{3} - 32y^2 \] Evaluating this from \( 0 \) to \( 64 \): \[ \left[ \frac{64^3}{3} - 32(64^2) \right] - \left[ \frac{0^3}{3} - 32(0^2) \right] \] Calculating \( 64^3 \) and \( 64^2 \): \[ 64^2 = 4096, \quad 64^3 = 64 \times 4096 = 262144 \] Now substituting back: \[ \frac{262144}{3} - 32 \times 4096 = \frac{262144}{3} - 131072 \] To combine these, we convert \( 131072 \) to a fraction: \[ 131072 = \frac{393216}{3} \] Thus, \[ \frac{262144}{3} - \frac{393216}{3} = \frac{262144 - 393216}{3} = \frac{-131072}{3} \] Now, substituting this back into the volume formula: \[ V = \frac{\pi}{4} \left( \frac{-131072}{3} \right) = \frac{-131072\pi}{12} = \frac{-32768\pi}{3} \] Since volume cannot be negative, we take the absolute value: \[ V = \frac{32768\pi}{3} \] Thus, the volume of the solid obtained by rotating the region about the \( y \)-axis is: \[ \boxed{\frac{32768\pi}{3}} \]