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

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You need to use the disk method to evaluate the volume of solid generated by revolving the region bounded by the curve y = 1/sqrtx and the lines `x=1` and x=2, about `y = -1` , such that:

`V = pi*int_1^2 (R^2(x) - r^2(x))dx`

`R(x)` represents the outer radius

`R(x) = 1/sqrtx - (-1) => R(x) = 1/sqrtx + 1`

`r(x)` represents the inner radius

`r(x) = (0 - (-1)) = 1`

`V = pi*int_1^2 ((1/sqrtx + 1)^2 - 1^2)dx`

Expanding the square yields:

`V = pi*int_1^2(1/x + 1/sqrtx + 1 - 1) dx`

Reducing duplicate terms yields:

`V = pi*int_1^2(1/x + 1/sqrtx) dx`

Using the linearity of integral yields:

`V = pi*int_1^2 1/x dx + pi*int_1^2 1/sqrt x dx`

`V = pi*(ln x)|_1^2 + pi*(sqrtx/2)|_1^2`

Using the fundamental theorem of calculus yields:

`V = pi*(ln 2 - ln 1 + sqrt2/2 - sqrt1/2)`

`V = pi*(ln 2 + (sqrt2 - 1)/2)`

Hence, evaluating the volume of solid generated by revolving the region bounded by the curve `y = 1/sqrtx` and the lines `x=1` and `x=2` , about `y = -1` , using disk method, yields `V = pi*(ln 2 + (sqrt2 - 1)/2).`

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