\( \leftarrow \) Question 1 of 6 , Find the lateral (side) surface area of the cone generated by revolving the line segment \( y=\frac{7}{5} x, 0 \leq x \leq 8 \), about the \( x \)-axis. The lateral surface area of is \( \square \). (Type an exact answer, using \( \pi \) as needed.)

To find the lateral surface area of the cone generated by revolving the line segment \( y=\frac{7}{5} x \) about the \( x \)-axis from \( x = 0 \) to \( x = 8 \), we can use the formula for the lateral surface area of a surface of revolution, given by: \[ A = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, we find \( \frac{dy}{dx} \) for \( y=\frac{7}{5} x \): \[ \frac{dy}{dx} = \frac{7}{5} \] Now, calculate \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \): \[ \sqrt{1 + \left(\frac{7}{5}\right)^2} = \sqrt{1 + \frac{49}{25}} = \sqrt{\frac{74}{25}} = \frac{\sqrt{74}}{5} \] Now, substitute \( y \) and \( \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \) into the surface area formula: \[ A = \int_0^8 2\pi \left(\frac{7}{5}x\right) \left(\frac{\sqrt{74}}{5}\right) \, dx = \frac{14\pi\sqrt{74}}{25} \int_0^8 x \, dx \] Calculating the integral \( \int_0^8 x \, dx \): \[ \int_0^8 x \, dx = \left[\frac{x^2}{2}\right]_0^8 = \frac{64}{2} = 32 \] Now we substitute back into the equation for \( A \): \[ A = \frac{14\pi\sqrt{74}}{25} \cdot 32 = \frac{448\pi\sqrt{74}}{25} \] Thus, the lateral surface area of the cone is: \[ \boxed{\frac{448\pi\sqrt{74}}{25}} \]