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#?

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)#