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Find surface area generated by $x^{2/3}+y^{2/3}=1$ about $-1 \le x \le 1$

So I got $\frac{dy}{dx} = - \frac{y^{1/3}}{x^{1/3}}, \qquad \sqrt{1+f'(x)}=\frac{1}{x^{1/3}}$

The rest of the answer looks like $A = \color{blue}2 \int^{1}_{0} 2 \pi \frac{y}{x^{1/3}} dx $

How did the 2 appear?

$= 4\pi \int y^{2/3} \color{blue}{\frac{y^{1/3}}{x^{1/3}}} dx= -4\pi \int y^{2/3} dy$

How did they get this $\frac{y^{1/3}}{x^{1/3}}$ removed?

1 Answers 1

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You have stated that

$-\frac{{dy}}{{dx}} = \frac{{{y^{1/3}}}}{{{x^{1/3}}}}$

You have (note you're missing some limits of integration)

$\eqalign{ & 4\pi \int {{y^{2/3}}\frac{{{y^{1/3}}}}{{{x^{1/3}}}}dx} = \cr & - 4\pi \int {{y^{2/3}}\frac{{dy}}{{dx}}dx} = \cr & - 4\pi \int {{y^{2/3}}dy} \cr} $

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    @Jiew Well, it's rather an abuse of notation, but in general, it works.2012-04-30