Key to "Find the area of the surface generated when..."

High School
Algebra
Pre-Algebra
Negative numbers and coordinate plane

cleritere39

Answered question

2021-11-13

Find the area of the surface generated when the given curve is revolved about the x-axis.

y = \frac{x^{3}}{6} + \frac{1}{2 x}, f or \frac{1}{2} \leq x \leq 1

Answer & Explanation

Aretha Frazier

Beginner2021-11-14Added 16 answers

We have,
$y = \frac{x^{3}}{6} + \frac{1}{2 x}; x \in [\frac{1}{2}, 1]$
Here, $f (x) = \frac{x^{3}}{6} + \frac{1}{2 x} \Rightarrow f^{'} (x) = \frac{x^{2}}{2} - \frac{1}{2 x^{2}}$
$1 + f^{'} {(x)}^{2} = 1 + \frac{x^{4}}{4} + \frac{1}{4 x^{4}} - \frac{1}{2} = {(\frac{x^{2}}{2} + \frac{1}{2 x^{2}})}^{2}$
$∴ S = 2 π \int_{\frac{1}{2}}^{1} (\frac{x^{3}}{6} + \frac{1}{2 x}) (\frac{x^{2}}{2} + \frac{1}{2 x^{2}}) d x$
$\Rightarrow S = 2 π \int_{\frac{1}{2}}^{1} [\frac{x^{5}}{12} + \frac{x}{12} + \frac{x}{4} + \frac{1}{4 x^{3}}] d x$
$\Rightarrow S = 2 π [\frac{263}{512}]$
Answer $S = 2 π [\frac{263}{512}]$

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