Find the volume of the solid obtained by revolving about the y-axis the region bounded by y=sqrt(x^2+1), y=0, x=0, and x=sqrt(8)

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The volume of the solid obtained by rotating about the y-axis the region bounded by `y = sqrt(1 +x^2)` , `y = 0` , `x = 0` and `x = sqrt 8` has to be determined.

The graph above shows the area that is rotated about the y-axis to create the solid.

The solid can be divided into discs of width `dx` and the radius of the discs is given by `y = sqrt(1+x^2)` where x takes on values from x = 0 to x = `sqrt 8` .

This gives the volume as: `int_0^(sqrt 8) pi*(1 + x^2) dx`

= `pi*int_0^(sqrt 8) 1 + x^2 dx`

= `pi*(x + x^3/3)_0^(sqrt 8)`

= `pi*(sqrt 8 + (8*sqrt 8)/3)`

= `pi*sqrt 8*(11/3)`

The required volume of the solid is `(pi*sqrt 3*11)/3`

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