Volumes of revolution flashcards

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Volume of revolution
The volume of the solid swept out when a region is rotated through 2π radians (360°) about a line — in P4, the x-axis.
Disc (in a volume of revolution)
A thin slice perpendicular to the axis: radius y, thickness δx, volume ≈ πy² δx.
Washer
A disc with a smaller disc removed; its area is π(outer radius)² − π(inner radius)².
Composite solid
A solid made of simpler pieces (e.g. a curve part plus a cone); find each volume and add or subtract.
Capacity of a container
The volume of the space inside it; with lengths in cm it is in cm³ (1 litre = 1000 cm³).
Volume of revolution about the x-axis
V = π∫ₐᵇ y² dx. (No — learn it)
Volume for a curve given parametrically
V = π∫ y² (dx/dt) dt, with limits t₁ and t₂ (the t-values at x = a and x = b). (No — derive it)
Volume of the solid from the region between two graphs
V = π∫ₐᵇ (y_outer² − y_inner²) dx. (No — derive it)
Volume of a cone
⅓πr²h. (No — learn it)
sin²x and cos²x in terms of cos 2x
sin²x = ½(1 − cos 2x); cos²x = ½(1 + cos 2x). (No — learn it)
Integration by parts
∫u (dv/dx) dx = uv − ∫v (du/dx) dx. (Yes)
Region under y = rx/h from x = 0 to h, rotated about the x-axis
A cone of radius r and height h; π∫(rx/h)² dx = ⅓πr²h.
Upper half of x² + y² = r² rotated about the x-axis
A sphere: π∫₋ᵣʳ (r² − x²) dx = (4/3)πr³.
Region between a line (above) and a curve (below), rotated
A solid with a hollow core; each slice is a washer, so subtract the inner volume from the outer.
Part of the curve below the x-axis, rotated
Sweeps out a solid just as a part above would; y² ≥ 0, so no splitting at the roots is needed.
Region bounded by a curve, a straight line and the x-axis
Split at the meeting point: a curved piece (integrate) and a cone or frustum (formula).
Why is it π∫y² dx and not π(∫y dx)²?
Each disc has volume πy²δx; the squaring applies to each radius before the discs are added.
Why is π∫(y₁ − y₂)² dx wrong for the region between two graphs?
It gives a solid whose radius is the gap between the graphs, not the hollow solid. Use π∫(y₁² − y₂²) dx.
How do you change the limits in a parametric volume?
Find the values of t at which x equals each x-limit, and use those as the t-limits.
Why can a volume never be negative?
y² ≥ 0 for every x, so with a < b the integral π∫ₐᵇ y² dx is never negative.
Name two limitations of modelling a real bowl or vase as a solid of revolution
e.g. the walls have thickness, a real base usually has a foot or rim that the profile curve does not give, the real profile is not exactly the curve.
How do you find the mass of a solid of revolution?
Mass = density × volume, keeping the units consistent (e.g. g cm⁻³ × cm³ = g).
y = x², 0 ≤ x ≤ 1, rotated about the x-axis: volume?
π∫₀¹ x⁴ dx = π/5.
y = 1/x, 1 ≤ x ≤ 3, rotated: volume?
π∫₁³ x⁻² dx = π[−1/x]₁³ = 2π/3.
y = eˣ, 0 ≤ x ≤ ln 3, rotated: volume?
π∫e²ˣ dx = π[½e²ˣ] = π(9/2 − 1/2) = 4π.
y = sin x, 0 ≤ x ≤ π, rotated: volume?
π∫½(1 − cos 2x) dx = π[½x − ¼ sin 2x]₀^π = π²/2.
y = √x, 0 ≤ x ≤ 4, rotated: volume?
y² = x: π[½x²]₀⁴ = 8π.
Region between y = 2x and y = x² (0 ≤ x ≤ 2), rotated: volume?
Line outside: π∫₀² (4x² − x⁴) dx = π(32/3 − 32/5) = 64π/15.
x = t², y = t, 0 ≤ t ≤ 1, rotated: volume?
dx/dt = 2t: π∫₀¹ t² · 2t dt = π/2 (check: y² = x, π∫₀¹ x dx = π/2).
Cone from rotating y = ¾x, 0 ≤ x ≤ 4: volume?
Radius 3, height 4: ⅓π × 9 × 4 = 12π.
Dropping π part-way through
Write V = π∫ … at the start and keep the π to the final line.
Squaring (a + b) as a² + b²
Include the cross term: (a + b)² = a² + 2ab + b².
Keeping x-limits in a parametric integral
Once the integral is in t, every limit must be a t-value.
Giving a decimal when 'exact' is asked
Keep π, e, ln and surds, e.g. 125π/3 or π ln(9/4).
'Hence' after a partial-fraction or by-parts part
Use the earlier result directly in the volume integral; starting again loses the link.

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