Calculus of a Single Variable, Chapter 7, 7.2, Section 7.2, Problem 22

Let's use the method of disc for evaluating the volume of the solid generated.
As per the method of discs V=intAdx or V=intAdy , where A stands for Area of a typical disc , A=pir^2
andr=f(x) or r=f(y) depending on the axis of revolution.
Given xy=3 , y=1 , y=4 , x=5
and the region is rotated about the line x=5
Consider a disc perpendicular to the line of revolution,
Then the radius of the disc will be (5-x)
Since xy=3, x=3/y
Radius of the disc = (5-3/y)
V=int_1^4pi(5-3/y)^2dy
V=piint_1^4(25-2(5)(3/y)+(3/y)^2)dy
V=piint_1^4(25-30/y+9/y^2)dy
V=pi[25y-30ln(y)+9(y^(-2+1)/(-2+1))]_1^4
V=pi[25y-30ln(y)-9/y]_1^4
V=pi{[25(4)-30ln(4)-9/4]-[25(1)-30ln(1)-9/1]}
V=pi(100-9/4-30ln(4)-25+9)
V=pi(84-9/4-30ln(4))
V=pi(327/4-30ln(4))
V~~126.17