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Find the areas of the surfaces generated by revolvin the curves about the indicated axes.

x = (1/3)y^(3/2) - y^(1/2), 1≦y≦3; revolved about y-axis

so i use the general formula "S = Integral 2π (radius)(dS)"

and the radiu in this case is x which is (1/3)y^(3/2) - y^(1/2) but yea..

i just keep getting negative value when i subsitute everything in the equation...

thanks.

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# Can we get a negative value for areas of surfaces of revolution?

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