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`

and`r=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`

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