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Find the volume of the solid generated by revolving the following region about the given axis
The region in the first quadrant bounded above by the curve y=x^2, below by the x-axis and on the right by the line x=1, about the line x=-4

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I'd suggest using shells for this one:

v = ∫[0,1] 2πrh dx
where r = x+4 and h = y = x^2

v = 2π∫[0,1](x+4)*x^2 dx
= 2π∫[0,1] x^3 + 4x^2 dx
= 2π (1/4 x^4 + 4/3 x^3) [0,1]
= 2π (1/4 + 4/3)
= 19π/6

It can be done with discs, but you have to make them washers:

v = ∫[0,1] π(R^2-r^2) dy
where R = 5, r=4+x = 4+√y
v = π∫[0,1] (25 - (4+√y)^2) dy
= π (9y - 16/3 y^3/2 - 1/2 y^2) [0,1]
= π (9 - 16/3 - 1/2)
= 19π/6

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