We can rrewrite the equation of the hyperbola:
y=x6.
We know, that the volume of the solid generated by a region under f(x) bounded by the x-axis and vertical lines x=a and x=b, which is revolved about the x-axis is
V=π∫ab[f(x)]2dx.
So we have:
V=π∫24(x6)2dx=π∫24x236dx=36π∫24x2dx=
=36π(x−1)∣24=36π(−41+21)=36π⋅41=
=9π≈28.27 cubic units.
Answer: =9π(≈28.27) cubic units.
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