# Question #3d4ae

Mar 12, 2017

$\frac{216 \pi}{5}$

#### Explanation:

The region to be revolved around x=7 is shown shaded in black in the attached diagram. For this region x would vary from x=0 to x=2. Now consider an element of length AB(thickness dx ), of this region at a distance x from from y-axis. Length AB would be $8 - {x}^{3}$ and its distance from x=7 would be 7-x.

Now if this element is rotated about x=7, the surface area of this cylindrical shell would be $2 \pi \left(7 - x\right) {x}^{3}$ and its volume would be $2 \pi \left(7 - x\right) \left(8 - {x}^{3}\right) \mathrm{dx}$

The volume of the solid so generated by revolving the whole shaded region would then be

${\int}_{0}^{2} 2 \pi \left(7 - x\right) \left(8 - {x}^{3}\right) \mathrm{dx}$

=$2 \pi {\int}_{0}^{2} \left(56 - 8 x - 7 {x}^{3} + {x}^{4}\right) \mathrm{dx}$

=$2 \pi {\left[56 x - 8 {x}^{2} / 2 - 7 {x}^{4} / 4 + {x}^{5} / 5\right]}_{0}^{2}$

=$2 \pi \left(112 - 16 - 28 + \frac{32}{5}\right) = 2 \pi \left(68 + \frac{32}{5}\right)$

$= 2 \pi \left(\frac{372}{5}\right) = \frac{744 \pi}{5}$