\[ SA = 2\pi \int_a^b f(x) \sqrt{1+f'(x)}dx \\ f(x) = x^3 \\ f'(x) = 3x^2 \\ SA = 2\pi \int_0^1 x^3 \sqrt{1 + 3x^2}dx \\ \text{Let}\space x = \frac{1}{\sqrt{3}}tan(\theta) \\ dx = \frac{1}{\sqrt{3}} sec^2(\theta)d\theta \\ SA = 2\pi \int_0^{\pi/4}\frac{1}{3^{3\2}}tan^3(\theta)\sqrt{1 + tan^2(\theta)}\frac{1}{\sqrt{3}} sec^2(\theta)d\theta \\ \text{N.B.}\space sec^2(\theta) = 1+ tan^2(\theta)\\ SA = \frac{2\pi}{9} \int_0^{\pi/4} tan^3(\theta)sec^3(\theta)d\theta \\ SA = \frac{2\pi}{9} \int_0^{\pi/4} tan^2(\theta)sec^2(\theta)tan(\theta)sec(\theta)d\theta \\ SA = \frac{2\pi}{9} \int_0^{\pi/4} (sec^2(\theta)-1)sec^2(\theta)tan(\theta)sec(\theta)d\theta \\ U = sec(\theta) \\ dU = sec(\theta)tan(\theta)d\theta \\ SA= \frac{2\pi}{9}\int_0^{\pi/4} (U^2 - 1)U^2dU \\ SA= \frac{2\pi}{9}\int_0^{\pi/4} U^4 -U^2dU \\ SA = \frac{2\pi}{9} [\frac{1}{5}sec^5(\theta) - \frac{1}{3}sec^3(\theta)]_0^{\pi/4} \\ SA = \frac{2\pi}{9} [(\frac{1}{5}(4\sqrt{2}) - \frac{1}{3}(2\sqrt{2})) - (\frac{1}{5}-\frac{1}{3})] \\ SA = 0.224725 \]