## University Calculus: Early Transcendentals (3rd Edition)

$2 \pi$
We need to compute the surface area. $S= \int_{m}^{n} 2 \pi y \sqrt {1+(y')^2} \\=(2 \pi) \times \int_{1/2}^{3/2} \sqrt {2x-x^2} \times \sqrt {\dfrac{1}{2x-x^2}} \space dx\\=(2 \pi) \int_{1/2}^{3/2} x \space dx \\=2 \pi$