Question: Find the area of the surface generated when the given curve is revolved about the given axis. \[…

Find the area of the surface generated when the given curve is revolved about the given axis.

\[ y = 22\sqrt{x}, \text{ for } 9 \leq x \leq 17; \text{ about the x-axis} \]

Solution

To find the area of the surface generated when the given curve is revolved about the x-axis, we use the formula for the surface area of revolution: \[ S = \int_a^b 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] Given the function \( y = 22\sqrt{x} \), we need to determine the derivative \( \frac{dy}{dx} \). First, let’s calculate \( \frac{dy}{dx} \): The function is \( y = 22x^{1/2} \). Differentiate with respect to \( x \): \[ \frac{dy}{dx} = 22 \cdot \frac{1}{2} x^{-1/2} = 11x^{-1/2} = \frac{11}{\sqrt{x}} \] Next, substitute \( y = 22\sqrt{x} \) and \( \frac{dy}{dx} = \frac{11}{\sqrt{x}} \) into the surface area formula: \[ S = \int_9^{17} 2\pi (22\sqrt{x}) \sqrt{1 + \left(\frac{11}{\sqrt{x}}\right)^2} \, dx \] Simplify the expression under the square root: \[ \sqrt{1 + \left(\frac{11}{\sqrt{x}}\right)^2} = \sqrt{1 + \frac{121}{x}} \] Now substitute back: \[ S = \int_9^{17} 44\pi \sqrt{x} \sqrt{1 + \frac{121}{x}} \, dx \] Simplify: \[ S = \int_9^{17} 44\pi \sqrt{x} \cdot \sqrt{\frac{x + 121}{x}} \, dx \] \[ S = \int_9^{17} 44\pi \frac{\sqrt{x + 121} \cdot \sqrt{x}}{\sqrt{x}} \, dx \] \[ S = \int_9^{17} 44\pi \sqrt{x + 121} \, dx \] Now evaluate the integral: Let \( u = x + 121 \), thus \( du = dx \), and when \( x = 9 \), \( u = 130 \), and when \( x = 17 \), \( u = 138 \). \[ S = \int_{130}^{138} 44\pi \sqrt{u} \, du \] Evaluate the integral: \[ S = 44\pi \left[ \frac{2}{3} u^{3/2} \right]{130}^{138} \] \[ S = \frac{88\pi}{3} \left[ u^{3/2} \right]{130}^{138} \] \[ S = \frac{88\pi}{3} \left[ 138^{3/2} - 130^{3/2} \right] \] This provides the surface area after calculation.