# 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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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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