# How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y = 1/x^4, y = 0, x = 1, x = 4 revolved about the x=-4?

Dec 16, 2017

$\frac{57 \pi}{16}$

#### Explanation:

the formula for the shell method is ${\int}_{a}^{b} 2 \pi r h \mathrm{dx}$

$a$ and $b$ are the x-bounds, which are x=1 and x=4, so $a = 1$ and $b = 4$.

$r$ is the distance from a certain x-value in the interval $\left[1 , 4\right]$ and the axis of rotation, which is x=-4. $r = x - \left(- 4\right) = x + 4$

$h$ is the height of the cylinder at a certain x-value in the interval $\left[1 , 4\right]$, which is $\frac{1}{x} ^ 4 - 0 = \frac{1}{x} ^ 4$ (because $\frac{1}{x} ^ 4$ is always greater than $0$ and h must be positive).

plugging it all in: volume $= {\int}_{1}^{4} \left(2 \pi \left(x + 4\right) \left(\frac{1}{x} ^ 4\right)\right) \mathrm{dx}$
you should get: $\frac{57 \pi}{16}$