University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{ \pi (2\sqrt 2-1)}{9}$
The formula to determine the surface area is as follows: $S= \int_{m}^{n} 2 \pi y \sqrt {1+(\dfrac{dy}{dx})^2}$ Now, $S=\int_{0}^{1} 2 \pi \times \dfrac{y^3}{3} \times \sqrt {y^4+1} dx \\=(\dfrac{ \pi}{9})[ (y^4+1)^{3/2} ]_{0}^{1} \\=\dfrac{ \pi (2\sqrt 2-1)}{9}$