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 = 1 + x^2#, #y = 0#, #x = 0#, #x = 2# rotated about the line #x=4#?

1 Answer
Jason K.
Sep 28, 2017

#(76pi)/3#

Explanation:

This is a graph of the region that will be revolved around the vertical line #x =4# (not pictured).

graph{(y-1-x^2)(y)( sqrt(2-x) )(sqrt(x)) / (sqrt(2-x))/(sqrt(x))<=0 [0, 6, -1.51, 6.39]}

Recall the general form of the volume of a solid of revolution using the shells method when you are revolving about a vertical line:

#V = int_a^b 2pi * radius * f(x)" " dx #

The hardest conceptual part here is the radius. Since the axis of revolution is #x = 4#, the form of the radius is the expression #4 - x#.

(Why? Draw a segment from #x = 4# on the x-axis to #x = 1# on the x-axis. How long is that segment? It is #4 - 1 = 3#.

Thus:

#V = int_0^2 2pi * (4-x) * (1 + x^2) dx#
#= 2pi int_0^2 (4 + 4x^2 - x - x^3) dx#
#=2pi (4x + 4/3x^3 - 1/2x^2 - 1/4x^4)|_0^2#
#=2pi(4*2 + 4/3 * 2^3 - 1/2 * 2^2 - 1/4 * 2^4)#
#=2pi(8 + 32/3 - 2 - 4) = 2pi(2 + 32/3) = 2pi(38/3)=(76pi)/3#

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