Use the method of cylindrical shells to find the volume V generated by rotating the region bounded by the given curves about the specified axis.

Question

y = x²,    x = y²;    about y = −5

Answer 1

For this Solid of Revolution, my intuition wants me to use the Disks method, but Cylindrical Shells are requested, so here we go.

Setup:

We are rotating (vertically around y = -5) horizontal x-strips by integrating on ‘y’ between 2 curves, between the points of intersection of these 2 curves: y = x² and x = y², i.e. y = √x.

Intersections occurs at x = y = 0, and x = y = 1.

In the range [0, 1], we know that y = x² ≤ y = √x [Why ?], so x = f(y) = √y ≥ x = g(y) = y², so our horizontal strip lengths, we will use √y - y² = h(y) = f(y) - g(y)

For a horizontal cylinder at yᵢ between x values, we have horizontal “height" of cylinder hᵢ = f(yᵢ) - g(yᵢ), the base of the cylindrical shell will be the area of a shell between yᵢ and yᵢ + Δy would be approximately 2πyᵢΔy, _if_ the revolution were about y = 0… but it is not. We revolve around y = -5.

Therefore Cylindrical Shell Vᵢ = [2π(yᵢ + 5)Δy]h(yᵢ) = 2π(yᵢ + 5)[f(yᵢ) - g(yᵢ)]Δy ==>

V = ∫₀¹ 2π(y + 5)[f(y) - g(y)]dy = ∫₀¹ 2π(y + 5)(√y - y²)dy # That was the fun part, now we get to work !

V = ∫₀¹ 2π(y + 5)[f(y) - g(y)]dy = ∫₀¹ 2π(y + 5)(√y - y²)dy = 2π ∫₀¹(y√y - y³ + 5√y - 5y²)dy ==>

V = 2π ∫₀¹ (y³/² - y³ + 5y¹/² - 5y²) dy = 2π ( [y⁵/² / (5/2)] - y⁴/4 + [5y³/² / (3/2)] - 5y³/3 ) | ₀¹

V = 2π (2/5 - 1/4 + 5(2/3) - 5/3) = (2π/60)(24 - 15 + 200 - 100) ==> V = 109π / 30

Answer 2

x = √y; √y = y²; y = 0 or y =1. Volume v = 2π∫₀ ¹(y + 5)(√y - y²)dy = 2π(2/5y^5/2 - y⁴/4 + 10/3y^3/2- 5y³/3)₀¹ =

2π(2/5 - 1/4 + 10/3 - 5/3) = 109π/30