Discussion
The volume swept out by revolving \( y = f(x) \) about the \( x \)-axis between \( a \) and \( b \) is
Answer: C
A thin slice perpendicular to the axis is a disc of radius \( y \) and thickness \( dx \), so its volume is \( \pi y^{2}dx \) and
\[ V \;=\; \pi\int_{a}^{b} y^{2}\,dx. \]
The form \( 2\pi\int y\,dx \) is the surface of revolution's shape rather than the volume's, carrying one power of \( y \) because a surface has one dimension fewer
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