Answer
$$\dfrac{(2 \sqrt 2-1)\pi }{9}$$
Work Step by Step
We integrate in order to find the surface area:
$$S= \int_{c}^{d} (2\pi y) \sqrt {1+(y')^2}$$ $$\\ =2 \pi \int_{0}^{1} \dfrac{x^3}{3} \times \sqrt {1+x^4} dx \\=\dfrac{(\pi) \cdot (2)}{(6) \cdot (3)} \times [(1+x^4)^{(3/2)} ]_0^1\\=\dfrac{2 \pi }{18} \times [(1+x^4)^{(3/2)} ]_0^1 \\= \dfrac{ \pi(2 \sqrt 2-1)}{9}$$
