calculus

Find the volume of the solid generated by revolving the region about the given line.
The region in the second quadrant bounded above by the curve y = 16 - x2, below by the x-axis, and on the right by the y-axis, about the line x = 1

I have gathered, that washer method is to be used - (-4,0) is the shaded area.

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  1. washers is a good way:
    v = ∫[0,16] π(R^2-r^2) dy
    where R=1+|x|, r=1
    v = π∫[0,16] (1+√(16-y))^2 - 1 dy
    = -π(4/3 (16-y)^(3/2) + 1/2 (16-y)^2) [0,16]
    = 640π/3

    Or, using shells,
    v = ∫[-4,0] 2πrh dx
    where r=1-x and h=y
    v = 2π∫[-4,0] (x+1)(16-x^2) dx
    = 2πx (x^3/4 - x^2/3 - 8x + 16)[-4,0]
    = 640π/3

    Hmm! shells is less complicated this time.

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  2. Where does the 4/3 and 3/2 come from?

    Thank you

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  3. ∫(1+√(16-y))^2 - 1 dy
    let u = 16-y
    du = -dy

    ∫(1+√u)^2 - 1 du
    ∫1 + 2√u + u - 1
    ∫2√u + u
    2(2/3)u^(3/2) + 1/2 u^2
    because
    ∫u^n = 1/(n+1) u^(n+1) where n=1/2

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