Question

Find the area of the surface that is generated by revolving the portion of

the curve y = x

3 between x = 0 and x = 1 about the x −axis.

Answer 1

$$V=\pi\displaystyle\int_{0}^{1}(x^3)^2dx=\pi\big[\dfrac{x^7}{7}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{\pi}{7} (units^3)$$

$V=\dfrac{\pi}{7}$ cubic units.