Posted by nicko on Sunday, January 23, 2011 at 10:00pm.
find the volume generated by the boundaries y=x^3+1 , x-axis , x=1 , x=2 revolved about y-axis
calculus - Reiny, Sunday, January 23, 2011 at 10:48pm
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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