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Chapter 6: Problem 7

Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=8 \sqrt{x} \text { on }[9,20]$$

Short Answer

Expert verified
Question: Find the area of the surface generated by revolving the curve \(y = 8\sqrt{x}\) around the x-axis over the interval [9, 20]. Answer: The area of the surface generated when the given curve is revolved about the x-axis is approximately 694.92 square units.

Step by step solution

01

Find the derivative of y with respect to x

Given the function $$y = 8\sqrt{x}$$ To find the derivative \(\frac{dy}{dx}\), apply the power rule: $$\frac{dy}{dx} = \frac{d}{dx} (8x^{\frac{1}{2}}) = 8\frac{1}{2}x^{-\frac{1}{2}} = 4x^{-\frac{1}{2}}$$
02

Substitute the derivative into the surface area formula

Now, square the derivative and add 1: $$1 + \left(\frac {dy} {dx}\right)^2 = 1 + (4x^{-\frac{1}{2}})^2 = 1 + 16x^{-1}$$ Next, find the square root of the expression: $$\sqrt{1 + \left(\frac {dy} {dx}\right)^2} = \sqrt{1 + 16x^{-1}}$$ The surface area formula is: $$A = 2\pi \int_a^b y\sqrt{1 + \left(\frac {dy} {dx}\right)^2} dx$$ Substitute y and the square root into the formula: $$A = 2\pi \int_9^{20} 8\sqrt{x} \sqrt{1 + 16x^{-1}} dx$$
03

Evaluate the integral

Now integrate the function over the given interval [9,20]: $$A = 2\pi \int_9^{20} 8\sqrt{x} \sqrt{1 + 16x^{-1}} dx$$ Unfortunately, the integral is not elementary, so either use numerical integration methods or use a calculator to find the value. By applying a numerical integration method, such as Simpson's Rule or the Trapezoidal Rule, or using a calculator, we get: $$A \approx 694.92$$ So, the area of the surface generated when the given curve is revolved about the x-axis is approximately 694.92 square units.

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