# Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the curves x=0 , y=1 , x=y^15 , about the line y=1 Find the volume using integrals. Do it the way that you would do it in calculus 2

You should remember that the formula of volume of solid of revolution is given by:

`V = pi*int_a^b r^2(x) dx`

You should solve for x the equation `x = y^15 => y = root(15)x` , hence, the radius of the solid is `(1 - root(15)x)^2`

`V = pi*int_0^1 (1 - root(15)x)^2 dx...

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You should remember that the formula of volume of solid of revolution is given by:

`V = pi*int_a^b r^2(x) dx`

You should solve for x the equation `x = y^15 => y = root(15)x` , hence, the radius of the solid is `(1 - root(15)x)^2`

`V = pi*int_0^1 (1 - root(15)x)^2 dx `

Expanding the square yields:

`V = pi*int_0^1 (1 - 2root(15)x + x^(2/15))dx`

`V = pi*int_0^1 (1)dx -2pi*int_0^1 x^(1/15)dx +pi*int_0^1 x^(2/15)dx`

`V = pi*x|_0^1 - 2pi*(x^(16/15))/(16/15)|_0^1 + pi*(x^(17/15))/(17/15)|_0^1`

`V = pi*(1-0) - 2pi*15/16(1-0) + pi*15/17*(1-0)`

`V = pi - 30pi/16 + 15pi/17`

`V = (pi(272 - 510 + 240))/272 => V = 2pi/272 => V = pi/136`

Hence, evaluating the volume of the solid of revolution, under the given conditions, yields `V = pi/136` .

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