# How compute by any methods the volume of the solid obtained by revolving around y axis the region bounded by x^2+(y-1)^2=1?

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You need to evaluate the volume of the solid obtained by revolving the region bounded by the curve `x^2 + (y - 1)^2 = 1` , around y axis, using disks method, such that:

`V = pi int_a^b f^2(y)dy`

You need to express x in terms of y in the given equation `x^2 + (y - 1)^2 = 1` , such that:

`x^2 + (y-1)^2 = 1 => x^2 = 1 - (y-1)^2`

`x = sqrt(1 - (y-1)^2)`

Hence, evaluating the requested volume, yields:

`V = pi*int_0^2 (sqrt(1 - (y-1)^2))^2 dy`

`V = pi*int_0^2 (1 - (y-1)^2) dy`

Expanding the square, yields:

`V = pi*int_0^2 (1 - y^2 + 2y - 1) dy`

Reducing duplicate terms yields:

`V = pi*int_0^2 (2y - y^2) dy`

Using the property of linearity of integrals, yields:

`V = pi*int_0^2 2ydy - pi*int_0^2 y^2 dy`

`V = pi*(2y^2/2 - y^3/3)|_0^2`

`V = pi*(y^2 - y^3/3)|_0^2`

Using the fundamental theorem of calculus yields:

`V = pi*(2^2 - 2^3/3 - 0^2 + 0^3/3)`

`V = pi*(4 - 8/3) => V = (4pi)/3`

**Hence, evaluating the volume of the solid obtained by revolving the region bounded by the curve `x^2 + (y - 1)^2 = 1` , around y axis, using disks method, yields **`V = (4pi)/3.`

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