# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=x^2, x = 2, x = 7, y = 0 revolved about the x=8?

Dec 15, 2017

$3565 \frac{\pi}{6}$

#### Explanation:

shell method: volume = ${\int}_{a}^{b} 2 \pi r h \mathrm{dx}$

In this problem, $a = 2$ and $b = 7$ (from the boundaries x=2 and x=7).

r is the distance from a certain x value in the interval [2,7] to the axis of rotation, which is x=8. Since r must be a positive distance, it is $8 - x$.

h is the height of each shell (parallel to axis of rotation). in this case it is the distance between the upper and lower bounds of the region, or ${x}^{2} - 0 = {x}^{2}$

plugging in: volume = ${\int}_{2}^{7} 2 \pi \left(8 - x\right) \left({x}^{2}\right) \mathrm{dx}$
$= 2 \pi {\int}_{2}^{7} \left(- {x}^{3} + 8 {x}^{2}\right) \mathrm{dx}$
$= 2 \pi \cdot \left[F \left(7\right) - F \left(2\right)\right]$, where $F \left(x\right) = - \frac{1}{4} {x}^{4} + \frac{8}{3} {x}^{3}$ (fundamental theorem of calculus).
$= 2 \pi \cdot \left[\left(- \frac{1}{4} {\left(7\right)}^{4} + \frac{8}{3} {\left(7\right)}^{3}\right) - \left(- \frac{1}{4} {\left(2\right)}^{4} + \frac{8}{3} {\left(2\right)}^{3}\right)\right]$

simplify: $= 2 \pi \cdot \left[- \frac{2401}{4} + \frac{2744}{3} + \frac{16}{4} - \frac{64}{3}\right]$
$= 2 \pi \cdot \left[- \frac{2385}{4} + \frac{2680}{3}\right]$
$= 2 \pi \cdot \left[\frac{3565}{12}\right]$
$= 3565 \frac{\pi}{6}$