find the area of the surface generated when the given curve is revolved about the given axis. y=(8x)^(1/3)…

Explanation:

Step1: Recall surface - area formula

The formula for the surface area $S$ of a curve $y = f(x)$ revolved about the $y$-axis is $S=2\pi\int_{a}^{b}x\sqrt{1+(y')^{2}}dx$, where $a$ and $b$ are the limits of integration, and $y'$ is the derivative of $y$ with respect to $x$. First, find the derivative of $y=(8x)^{\frac{1}{3}} = 2x^{\frac{1}{3}}$. Using the power - rule $(x^n)'=nx^{n - 1}$, we have $y'=\frac{2}{3}x^{-\frac{2}{3}}$.

Step2: Calculate $1+(y')^{2}$

[ \begin{align*} 1+(y')^{2}&=1 + (\frac{2}{3}x^{-\frac{2}{3}})^2\ &=1+\frac{4}{9}x^{-\frac{4}{3}}\ &=\frac{9x^{\frac{4}{3}} + 4}{9x^{\frac{4}{3}}} \end{align*} ] Then $\sqrt{1+(y')^{2}}=\sqrt{\frac{9x^{\frac{4}{3}}+4}{9x^{\frac{4}{3}}}}=\frac{\sqrt{9x^{\frac{4}{3}} + 4}}{3x^{\frac{2}{3}}}$.

Step3: Set up the integral

The surface - area formula becomes $S = 2\pi\int_{0}^{1}x\cdot\frac{\sqrt{9x^{\frac{4}{3}}+4}}{3x^{\frac{2}{3}}}dx= \frac{2\pi}{3}\int_{0}^{1}x^{1-\frac{2}{3}}\sqrt{9x^{\frac{4}{3}}+4}dx=\frac{2\pi}{3}\int_{0}^{1}x^{\frac{1}{3}}\sqrt{9x^{\frac{4}{3}}+4}dx$. Let $u = 9x^{\frac{4}{3}}+4$, then $du=12x^{\frac{1}{3}}dx$, and $x^{\frac{1}{3}}dx=\frac{1}{12}du$. When $x = 0$, $u = 4$; when $x = 1$, $u=9 + 4=13$.

Step4: Evaluate the integral

[ \begin{align*} S&=\frac{2\pi}{3}\int_{4}^{13}\frac{1}{12}\sqrt{u}du\ &=\frac{\pi}{18}\int_{4}^{13}u^{\frac{1}{2}}du\ &=\frac{\pi}{18}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{4}^{13}\ &=\frac{\pi}{27}(13^{\frac{3}{2}}-4^{\frac{3}{2}})\ &=\frac{\pi}{27}(13\sqrt{13}-8) \end{align*} ]

Answer:

$\frac{\pi(13\sqrt{13}-8)}{27}$