The volume of the solid obtained by rotating the region bounded by y=x^2−2x and y=x about the line y=9, has the form a/bπ, where a and b are positive coprime integers. What is the value of a+b?

Answer 1

The curves intersect at (0,0) and (3,3)

Using discs (washers), the volume is

v = ∫[0,3] π(R^2-r^2) dx
where r=9-x and R = 9-(x^2-2x), so

v = π∫[0,3] (9-(x^2-2x))^2 - (9-x)^2 dx
= π(x^5/5 - x^4 - 5x^2 + 27x^2) [0,3]
= 378/5 π

Using shells, we need to separate the region into two parts, because the parabola has 2 values of x for each value of y.

The two branches of the parabola are
x = 1±√(1+y)

The volume is thus

∫[-1,0] 2πrh dy
where r=9-y and h = (1+√(1+y)) - (1-√(1+y))
+∫[0,3] 2πrh dy
where r=9-y and h = 1+√(1+y)-y

2π∫[-1,0] (9-y)(2√(1+y)) dy
= 8/15 π(47-3y)(1+y)√(1+y)
= 376/15 π

2π∫[0,3] (9-y)(1+√(1+y)-y) dy
= 2/15 π (1+y)(5y^2 - 80y + 215 - 2(3y-47)√(1+y)) [0,3]
= 758/15 π

(376/15 + 758/15)π = 378/5 π