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#?

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#?

Samantha Adkison · Accepted Answer

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)#

Aurora Alvarado · Answer

undefined To 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 line $ x = 1 $, follow these steps:

1. Determine the axis of rotation. In this case, it's $ x = 1 $.

2. Draw a sketch of the region and the axis of rotation to visualize the problem.

3. Identify the bounds of integration. In this case, $ x $ goes from $-1$ to $0$.

4. Choose a representative horizontal rectangle parallel to the axis of rotation. Each rectangle has height $ e^{-x} $ and width $ dx $.

5. The radius of the shell is the distance from the rectangle's side to the axis of rotation, which is $ 1 - x $.

6. The volume element of each shell is $ 2\pi (1 - x) e^{-x} \, dx $.

7. Integrate the volume element over the given bounds of integration:

$$ V = \int_{-1}^{0} 2\pi (1 - x) e^{-x} \, dx $$

8. Evaluate the integral to find the volume of the solid generated by revolving the given region about $ x = 1 $.