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#?

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Answer 1 · 2017-01-01T06:21:17

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

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