# How do you find the volume of the solid obtained by rotating the region bounded by the curves y = x^3, x=0, and x=1 rotated around the y=-2?

Apr 18, 2016

The region to be rotated about y=-2 is shown here shaded in red.
Consider an element of length y and width $\delta$x at a distance x from the origin. If this this is rotated about y=-2. It would form a circular disc of radius y+2. Its area would be $\pi {\left(y + 2\right)}^{2}$. Its volume would be $\pi {\left(y + 2\right)}^{2} \delta x$ From this subtract the volume of coaxial region $\pi {2}^{2} \delta x$

Required volume is that of the annular region $\pi {\left(y + 2\right)}^{2} \delta x - 4 \pi \delta x$

=$\pi \left({y}^{2} + 4 y\right) \delta x$.

The volume of the solid formed by the whole region would therefore be

${\int}_{x = 0}^{1} \pi \left({y}^{2} + 4 y\right) \mathrm{dx}$. Since y is =${x}^{3}$

Integral to solve would be ${\int}_{x = 0}^{1} \pi \left({x}^{6} + 4 {x}^{3}\right) \mathrm{dx}$

= $\pi {\left[{x}^{7} / 7 + 4 {x}^{4} / 4\right]}_{0}^{1}$
= $\frac{8 \pi}{7}$