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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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    Oh, icic, $dx$ can also be canceled out ...2012-04-30
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    @Jiew Well, it's rather an abuse of notation, but in general, it works.2012-04-30