Saturday
March 25, 2017

Post a New Question

Posted by on .

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

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question