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 and 2x+2y=5 rotated about the y=1/2?

1 Answer
Jan 1, 2017

=pi(2 ln 2+21/8)=12.60 cubic units, nearly.

Explanation:

graph{ ( xy-1)(x+y-5/2) = 0 [-5.37, 5.37, -2.684, 2.684]}

The rectangular hyperbola H; xy = 1 and the L: straight line

x+y=5/2 meet at (1/2. 2) and (2, 1/2).

The area is depicted by the graph.

The volume is

pi int ((y_L-1/2)^2-(y_H-1/2)^2 )dx, with x from 1/2 to 2

=pi int ((5/2-x-1/2)^2-(1/x-1/2)^2) dx,.with x from 1/2 to 2

=pi int ((4-4x+x^2)-(1/4-1/(4x)+1/x^2) )dx,

with x from 1/2 to 2

=pi int (1/(4x)-1/x^2+15/4-4x+x2) dx, with x from 1/2 to 2

=pi[(1/4)(ln 2 -ln(1/2))+(1/2-2)+15/4(2-1/2)-2(2^2-1/2^2)-15+1/3(2^3-1/2^3)]

=pi[1/2 ln 2-3/2+45/8-15/2 -15+21]

=pi(2 ln 2+21/8), cubic units