Problem 10 - Calculus Early Transcendentals

Question

Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=\sqrt{4 x+6} \text { on }[0,5]$$

Step-by-Step Solution

Answer

The area of the surface generated when the curve \(y=\sqrt{4x+6}\) on \([0,5]\) is revolved about the x-axis is approximately 224.69.

Step 1: Find the derivative of the function

Differentiate the function \(y=\sqrt{4x+6}\) with respect to \(x\) using the chain rule. The chain rule for a function \(y(u(x))\) is given by: $$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx}$$ In this case, let \(y=\sqrt{u}\) and \(u=4x+6\). The derivative of these functions with respect to their respective variables is: $$\frac{dy}{du} = \frac{1}{2\sqrt{u}} \text{ and } \frac{du}{dx} = 4$$ Now, find the derivative of the function \(\frac{dy}{dx}\) using the chain rule: $$\frac{dy}{dx} = \frac{dy}{du}\times \frac{du}{dx} = \frac{1}{2\sqrt{u}} \times 4 = \frac{2}{\sqrt{4x+6}}$$

Step 2: Set up the integral

Use the surface area of revolution formula with the given derivative and the given interval \([0, 5]\): $$A = 2\pi \int_0^5 \sqrt{4x+6} \sqrt{1+(\frac{2}{\sqrt{4x+6}})^2} dx$$

Step 3: Simplify the integral

First, simplify the expression inside the square root: $$1+(\frac{2}{\sqrt{4x+6}})^2 = 1 + \frac{4}{4x+6}$$ Now, multiply the two square root terms together: $$\sqrt{4x+6}\sqrt{1+\frac{4}{4x+6}} = \sqrt{(4x+6)(1+\frac{4}{4x+6})}$$ Simplify the expression inside the square root: $$\sqrt{(4x+6)(\frac{4x+6}{4x+6}+\frac{4}{4x+6})} = \sqrt{(4x+6)(\frac{4x+10}{4x+6})}$$ Now the integral becomes: $$A = 2\pi \int_0^5 \sqrt{(4x+6)(\frac{4x+10}{4x+6})} dx$$

Step 4: Solve the integral

Since the terms inside the square root cancel each other, the integral simplifies to: $$A = 2\pi \int_0^5 \sqrt{\frac{4x+10}{1}} dx = 2\pi \int_0^5 \sqrt{4x+10} dx$$ Now, we can use u-substitution to solve the integral. Let \(u = 4x + 10\). Then, \(\frac{du}{dx} = 4\), and \(dx = \frac{du}{4}\). When \(x = 0\), \(u = 10\). When \(x = 5\), \(u = 30\). Our integral becomes: $$A = 2\pi \int_{10}^{30} \sqrt{u} \frac{du}{4} = \frac{\pi}{2} \int_{10}^{30} \sqrt{u} du$$ Next, integrate \(\sqrt{u}\) with respect to \(u\): $$\frac{\pi}{2}[ \frac{2}{3}u^{\frac{3}{2}}]_{10}^{30}$$ Finally, evaluate the integral with limits 10 and 30: $$A = \frac{\pi}{2}\left[\frac{2}{3}(30^{\frac{3}{2}}) - \frac{2}{3}(10^{\frac{3}{2}})\right]$$ $$A = \pi\left[\frac{1}{3}(30^{3/2} - 10^{3/2})\right] \approx 224.69$$ The area of the surface generated when the curve \(y=\sqrt{4x+6}\) on \([0,5]\) is revolved about the x-axis is approximately 224.69.