# find the volume of the solid generated by revolving the region bounded by `y=1/(sqrt(x))` for `1<=x<=2` about the line y=-1

*print*Print*list*Cite

### 1 Answer

We want to use the "slicing" (aka "disks") method. Visualize a whole bunch of vertical cuts into the shape, which make (approximately) a bunch of cylinders.

The volume of a cylinder is `pi r^2 t` where t is the thickness of the cylinder

Our radius will be the distance from 0 to `(1)/(sqrt(x))` plus the distance from -1 to 0.

Thus the radius of one of our cylinders is `1+(1)/(sqrt(x))` , or `1+x^(-1/2)`

The thickness of a cylinder is `Delta x` , depending on how thin we sliced the shape into disks.

Thus the volume of one disk is:

`pi (1+x^(-1/2))^2 Delta x=pi (1+ 2x^(-1/2)+x) Delta x`

When we add up all the disks, and let the thickness of a slice go to 0, we get an integral:

`int_1^2 pi (1+2x^(-1/2)+x) dx`

`=pi [ x + 4 x^(1/2) + (1/2)x^2 ]_1^2`

`=pi[(2+4 sqrt(2) + 2)-(1+4+(1/2))]`

`=pi (4 sqrt(2) - (3/2) )`