Calculus

Find the area bounded by the y-axis and x = 4-y^2/3

asked by Rose
  1. I will assume you meant:
    x = 4 - y^(2/3)

    I assume you made a sketch
    let's take horizontal slices ....

    the y-intercept is (0,8) so the region is not closed.
    I will assume you want the region between the x-axis and and the y-axis

    area = ∫4 - y^(2/3) dy from 0 to 8
    = 4y - (3/5)y^(5/3) | from 0 to 8
    = (32 - (3/5)(8^(5/3)) - 0
    = 32 - (3/5)(32)
    = 64/5 or 12.8

    check my arithmetic

    posted by Reiny

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