How do you find the volume of a solid of revolution using the disk method for $y= 3/(x+1)$ , $y=0$ , $x=0$ , $x=8$ revolved about the $x$ -axis?

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The question I am having trouble with is:

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the x-axis . (Using the disc method.)
$y= 3/(x+1)$ , $y=0$ , $x=0$ , $x=8$

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Answer 1 · 2018-04-08T15:50:39

Please see below.

Here is a graph of the region:

enter image source here

In order to use disks, a representative slice has been taken perpendicular to the axis of revolution. (In this case the axis is a horizontal line.)

For a disk we get representative volume

$pi r^2 " thickness"$

We have taken the slice at some value of $x$ and
the thickness is $dx$

In this case, $r =$ the $y$ -value on the curve, so
$r = 3/(x+1)$

The representative slice has volume: $pi(3/(x+1))^2dx$

The values of $x$ vary from $0$ to $8$ , so the volume of the solid is:

$V = int_0^8 pi(3/(x+1))^2dx = 9pi int_0^8 1/(x+1)^2 dx =8pi$