point) 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 from \( x = 0 \) to \( x = -8 \) (which is not valid since \( x \) cannot be negative). Therefore, we need to find the intersection of the line with the \( y \)-axis, which occurs when \( x = 0 \): \[ y = 8(0) + 64 = 64 \] Thus, the region we are interested in is bounded by \( y = 8x + 64 \) from \( y = 0 \) to \( y = 64 \) and \( x = 0 \). Next, we express \( x \) in terms of \( y \): \[ y = 8x + 64 \implies 8x = y - 64 \implies x = \frac{y - 64}{8} \] Now, we can set up the volume integral using the method of cylindrical shells. The formula for the volume \( V \) when rotating around the \( y \)-axis is given by: \[ V = 2\pi \int_{a}^{b} x \cdot f(x) \, dx \] In our case, we will integrate with respect to \( y \) from \( y = 0 \) to \( y = 64 \): \[ V = 2\pi \int_{0}^{64} \left( \frac{y - 64}{8} \right) y \, dy \] Now, we simplify the integral: \[ V = 2\pi \int_{0}^{64} \left( \frac{y^2 - 64y}{8} \right) \, dy \] \[ = \frac{2\pi}{8} \int_{0}^{64} (y^2 - 64y) \, dy \] \[ = \frac{\pi}{4} \int_{0}^{64} (y^2 - 64y) \, dy \] Now we compute the integral: \[ \int (y^2 - 64y) \, dy = \frac{y^3}{3} - 32y^2 \] Evaluating 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: \[ = \frac{262144}{3} - 32 \times 4096 \] Calculating \( 32 \times 4096 \): \[ 32 \times 4096 = 131072 \] Now substituting: \[ = \frac{262144}{3} - 131072 \] To combine these, we convert \( 131072 \) to a fraction: \[ 131072 = \frac{393216}{3} \] Now we have: \[ = \frac{262144 - 393216}{3} = \frac{-131072}{3} \] Now substituting back into the volume formula: \[ V = \frac{\pi}{4} \cdot \frac{-131072}{3} = \frac{-131072\pi}{12} = \frac{-32768\pi}{3} \] 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: \[ \boxed{\frac{32768\pi}{3}} \]