Find the volume of the solid generated by revolving the region x=y^2+1 and the line x=3 about the line x=3 (use horizontal slicing)?

x = y² + 1

When x = 3 :
y² + 1 = 3
y² = 2
y = √2 or y = -√2

Volume of the solid generated
= ∫(-√2 to √2) π [3 - (y² + 1))]² dy
= ∫(-√2 to √2) π [2 - y²]² dy
= ∫(-√2 to √2) π [y⁴ - 4y² + 4] dy
= π [(1/5)y⁵ - (4/3)y³ + 4y] | (-√2 to √2)
= π [(1/5)(√2)⁵ - (4/3)(√2)³ + 4(√2)] - π [(1/5)(-√2)⁵ - (4/3)(-√2)³ + 4(-√2)]
= [(2/5)(√2)⁵ - (8/3)(√2)³ + 8(√2)]π
= [(8/5)(√2) - (16/3)(√2) + 8(√2)]π
= (40√2)π/3 sq. units