Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

Chapter 8 - Section 8.2 - Area of a Surface of Revolution - 8.2 Exercises - Page 555: 7

Answer

$$\frac{\pi}{27}(145 \sqrt{145}-1)$$

Work Step by Step

Given $$y= x^3, \ \ \ 0 \leq x \leq 2$$ Since the surface area of the surface of revolution formed by revolving the graph of $f (x)$ around the $x-$axis is given by \begin{align*} S_x&=\int_{a}^{b}2\pi f(x)\sqrt{1+[f'(x)]^2}dx\\ &=\int_{0}^{2}2\pi x^3\sqrt{1+9x^4}dx \end{align*} Let $u=1+9x^4\ \ \Rightarrow \ \ du =36x^3dx$, at $x=0\to u=1$, $x=2\to u=145$, then \begin{align*} S_x&=\frac{1}{18}\int_{1}^{145} \pi u^{1/2}du\\ &=\frac{\pi}{27} u^{3/2} \bigg|_{1}^{145}\\ &=\frac{\pi}{27}(145 \sqrt{145}-1)\ \ \text{unit}^2 \end{align*}

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