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\( \leftarrow \) 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.)

Ask by Wang Reyes. in the United States
Mar 13,2025

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The lateral surface area of the cone is \( \frac{448\sqrt{74}\pi}{25} \).

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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: \[ \boxed{\frac{448\pi\sqrt{74}}{25}} \]

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