Calculus

Find the volume of the solid obtained by revolving the region x=(y-2)^2, the x-axis, the y-axis, about the x-axis.

So I understand that you're supposed to use disc method; however, idk where I evaluated wrong. My final answer is 136π/3, but according to Wolfram Alpha, the answer is 8π/3 and they used shell. Below is my work:

x=(0-2)^2
x=(-2)^2
x=4

x=(y-2)^2
x^(1/2)=y-2
x^(1/2)+2=y

π∫(x^(1/2)+2)^2 dx [0,4]
π∫(x+4(x)^(1/2)+4) dx [0,4]
π[(x^2/2)+(8/3)x^(3/2)+4x] [0,4]
=136π/3

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  1. using discs of thickness dx,
    v = ∫[0,4] πr^2 dx
    where r = y = 2-√x
    The parabola has two branches, and you took the upper branch 2+√x, but we want the curve bounded by the axes, which is 2-√x
    v = ∫[0,4] π(2-√x)^2 dx

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  2. Make your sketch.
    x = (y-2)^2
    ±√x = y - 2
    y = 2 + √x or y = 2 - √x
    the region that you are revolving has as its boundary y = 2 - √x
    and crosses the x-axis at (4,0)
    so we want:
    π∫ y^2 dx from x=0 to x=4
    = π∫ (2-√x)^2 dx from 0 to 4 = π∫ (2-x^(1/2))^2 dx from 0 to 4
    = π∫ (4 - 4x^(1/2) + x) dx from 0 to 4
    = π [4x - (8/3)x^(3/2) + (1/2)x^2 ] from 0 to 4
    = π( 8 -(8/3)(8) + 8 - 0 )
    = 8π/3 , as required

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