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Find the volume of the solid formed when the region described is revolved about the x-axis using disk method the region under the curve y=sqrt[3]{x} on the interval 0le x le 8
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since there is only one curve to contend with; it makes life easier
the radius of each little circle we form is simply f(x) so integrate from 0 to 8; pi [cbrt(x)]^2 right?
\[\int_{0}^{8} pi(\sqrt[3]{x})^2dx\]
ya.
hey, I know the correct way. I do wrongly with integrate.Thanks.
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the trick here might be to convert the radical into its exponent form :)
I do it wrongly at 1st, when I integrade, I divide wrongly, so I get the wrong answer.
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