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.

\[ y = \frac{1}{(8x)^3}, \text{ for } 0 \leq x \leq 8; \text{ about the y-axis} \]

Solution

To find the area of the surface generated when the curve is revolved about the y-axis, we use the formula for the surface area of revolution given by: \[ A = 2 \pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] First, identify the function and the interval: Given \( y = (8x)^{\frac{1}{3}} \) for \( 0 \leq x \leq 8 \). Find the derivative \(\frac{dy}{dx}\): \[ y = (8x)^{\frac{1}{3}} \] Let \( u = 8x \), then \(\frac{du}{dx} = 8\). By the chain rule: \[ \frac{dy}{dx} = \frac{1}{3}(8x)^{-\frac{2}{3}} \cdot 8 = \frac{8}{3} (8x)^{-\frac{2}{3}} \] Now, find the expression under the square root: \[ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \left(\frac{8}{3} (8x)^{-\frac{2}{3}}\right)^2 \] \[ = 1 + \frac{64}{9} (8x)^{-\frac{4}{3}} \] Now, compute the integral: \[ A = 2 \pi \int_{0}^{8} x \sqrt{1 + \frac{64}{9} (8x)^{-\frac{4}{3}}} \, dx \] Substitute \( u = (8x)^{\frac{1}{3}} \), so \( x = \frac{u^3}{8} \) and \( dx = \frac{3}{8}u^2 du \). The limits change from \( x: 0 \to 8 \) to \( u: 0 \to 4 \). \[ \begin{align*} A &= 2 \pi \int_{0}^{4} \frac{u^3}{8} \sqrt{1 + \frac{64}{9} u^{-4}} \cdot \frac{3}{8}u^2 \, du \\ &= \frac{3 \pi}{32} \int_{0}^{4} u^5 \sqrt{1 + \frac{64}{9} u^{-4}} \, du \\ \end{align*} \] This integral requires further simplification or numerical methods for evaluation, as it involves a non-elementary integral. The detailed computation or approximation needs advanced calculus tools.