# How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region  y = e^ (-x), bounded by: y = 0, x = -1, x = 0 rotated about the x=1?

Aug 21, 2015

This looks like:

graph{(y - e^(-x))(y)(x + 1)(sqrt(0.25 - (x + 0.5)^2))/(sqrt(0.25 - (x + 0.5)^2)) <= 0 [-3.29, 5.48, -0.855, 3.52]}

The Shell Method suggests using the formula $V = 2 \pi \int x f \left(x\right) \mathrm{dx}$.
where $x$ is the radius and $f \left(x\right)$ is your function.

Rotating it about $x = 1$ gives the revolved shape a radius of $1 - x$ instead of $x$ because the radius extends from the axis of rotation, $x = 1$, to each value of $x$ in the interval $x \in \left[- 1 , 0\right]$.

The function itself should be $f \left(x\right) = {e}^{- x}$ since it is bounded by $y = 0$.

Therefore, you have:

$V = 2 \pi {\int}_{- 1}^{0} \left(1 - x\right) \left({e}^{- x}\right) \mathrm{dx}$

$= 2 \pi {\int}_{- 1}^{0} {e}^{- x} - x {e}^{- x} \mathrm{dx}$

Let's see what $\int x {e}^{- x} \mathrm{dx}$ is using Integration by Parts...

Let:
$u = x$
$\mathrm{du} = \mathrm{dx}$
$\mathrm{dv} = {e}^{- x} \mathrm{dx}$
$v = - {e}^{- x}$

$u v - \int v \mathrm{du}$

$= - x {e}^{- x} + \int {e}^{- x} \mathrm{dx}$

$= - x {e}^{- x} - {e}^{- x}$

Overall we have:

$V = 2 \pi {\int}_{- 1}^{0} {e}^{- x} \mathrm{dx} - 2 \pi {\int}_{- 1}^{0} x {e}^{- x} \mathrm{dx}$

$= 2 \pi \left[- {e}^{- x} - \left(- x {e}^{- x} - {e}^{- x}\right)\right] {|}_{- 1}^{0}$

$= 2 \pi \left[\left(- {e}^{0} - \left(0 \cdot {e}^{0} - {e}^{0}\right)\right) - \left(- {e}^{1} - \left({e}^{1} - {e}^{1}\right)\right)\right]$

$= 2 \pi \left[\left(- 1 - \left(- 1\right)\right) - \left(- e\right)\right]$

$= 2 \pi \left[- 1 + 1 + e\right]$

$\textcolor{b l u e}{= 2 e \pi {\text{u}}^{3}}$