# calculus

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find the volume generated by the boundaries y=x^3+1 , x-axis , x=1 , x=2 revolved about y-axis

• calculus -

I hope you made a sketch

the line x=1 and x=2 intersect with the curve at (1,2) and (2,9)

so first of all i will calculate a solid cyliner about the y-axis with a radius of 2 and a height of 9
= π(2^2)(9) = 36π

I will then hollow out a part with radius 1 and up to a height of 2
that is π(1^2)(2) = 2π

now I will hollow-out the curved part from a height of 2 to a height of 9 with a radius of
x = (y-1)^(1/3)

taking horizontal slices
Volume cut out
= π[integral] x^2 dy from 2 to 9
= π[integral] (y-1)^(2/3) dy
= π (5/3)(y-1)^(5/3 from 2 to 9
= π( (3/5)(8^(5/3)) - (3/5)(1^(5/3))
= π(93/5) = 18.6π

so final volume = 36π - 2π - 18.6π = 15.4π

check my arithmetic

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