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## Homework Statement

Find the surface area traced out when the curve 8y^2 = x^2(1-x^2) is revolved around the x-axis.

## The Attempt at a Solution

x-axis means y = 0

When y = 0, x = 0, -1 or 1.

Since this curve is "the infinity symbol", the curve has symmetry at x = 0.

Isolating y,

[tex]8y^2 = x^2(1-x^2)[/tex]

[tex]y = \frac{\sqrt{x^2(1-x^2)}}{8}[/tex]

Surface area = [tex]2\pi \int_{C} y dr [/tex]

= [tex]4\pi \int^{1}_{0} y dr [/tex]

[tex] dr = {\sqrt{1+ ({\frac{\sqrt{2x(1-2x^2)}}{8})^2}} {dx}[/tex]

[tex] dr = {\frac{\sqrt{64+2x(1-2x^2)}}{8}} {dx}[/tex]

Surface area = [tex]{\frac{\pi}{8}} \int^{1}_{0} {\sqrt{x^2(1-x^2)(64+2x(1-2x^2))}{dx}[/tex]

And.......I have no idea how to proceed with the integral.

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