Calculus

Let R be the region bounded by the curve x=9y-y^2 and the y- axis. Find the volume of the solid resulting from revolving R about the line y= -6.

I believe the integral limits are from y=0 to y-9
i set up
h(x) = 9y-y^2
r(x) = y+6

I am not sure if this is correct
Can anyone help me?
Thanks

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asked by Salome
  1. I answered my question.
    I had made an arithmetic error.

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    posted by Salome
  2. the shape you will get is half a torus.
    A torus is a donut or bagel shaped solid.
    We have cut ours in half (to make a tasty cream-cheese bagel)
    the y-intercepts:
    0 = y(9-y)
    so y = 0 or y = 9 , You had that

    so we need the radius of the outer and the inner rings:
    y^2 - 9y = -x
    y^2 - 9y + 20.25 = 20.25 -x , I completed the square
    (y - 4.5)^2 = 20.25-x
    y - 4.5 = ± √(20.25-x)
    y = 4.5 ± √(20.25-x)

    so the outer radius = 6 + 4.5 + √(20.25-x)
    = 10.5 + (20.25-x)^.5
    and the inner radius = 6 + 4.5 - √(20.25-x)
    = 10.5 - (20.25-x)^.5

    volume = π∫( (10.5 + (20.25-x)^.5 )^2 - ((10.5 - (20.25-x)^.5 )^2 dy from y = 0 to 9

    = π∫......
    I hope I did not mess up the algebra here, you might want to do this yourself. It is rather messy to type

    = π∫(42(20.25-x)^(1/2) dy from 0 to 9
    = π [-28(20.25-x)^(3/2) ] from 0 to 9
    = π( -28(20.25-9)^(3/2) -(-28(20.25-0)^(3/2) )
    = π ( - 1056.542... + 2551.5)
    = π(1494.9579)
    = appr 4696.55

    You better check my arithmetic in that last part, so easy to make an error

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    posted by Reiny

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