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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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