Question

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`?

Answer 1

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 = 2piint xf(x)dx`.
where `x` is the radius and `f(x)` 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 [-1,0]`.

The function itself should be `f(x) = e^(-x)` since it is bounded by `y = 0`.

Therefore, you have:

`V = 2piint_(-1)^(0) (1-x)(e^(-x))dx`

`= 2piint_(-1)^(0) e^(-x)-xe^(-x)dx`

Let's see what `int xe^(-x)dx` is using Integration by Parts...

Let:
`u = x`
`du = dx`
`dv = e^(-x)dx`
`v = -e^(-x)`

`uv - intvdu`

`= -xe^(-x) + int e^(-x)dx`

`= -xe^(-x) -e^(-x)`

Overall we have:

`V = 2piint_(-1)^(0) e^(-x)dx - 2piint_(-1)^0 xe^(-x)dx`

`= 2pi [-e^(-x) - (-xe^(-x) -e^(-x))]|_(-1)^(0)`

`= 2pi [(-e^(0) - (0*e^(0) -e^(0))) - (-e^(1) - (e^(1) -e^(1)))]`

`= 2pi [(-1 - ( - 1)) - (-e)]`

`= 2pi [-1 + 1 + e]`

`color(blue)(= 2epi "u"^3)`