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Rejected Member
- Joined
- Feb 19, 2004
- Messages
- 85
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- HSC
- 2004
my maths tutor said its ok to use the dv = 2πx . f(x) . dx method but u should draw a cyclinder and write:
radius = ...
height = ...
width = ...
just to be sure u get all available marks
going from first principals for each specific question in an exam is very time consuming and should be avoided, if u want to go from first principals u should use a generic proof:
e.g. a "sandwich" limit
anulus area x smaller height < change in volume < anulus area x greater height
π[(x+Δx)<sup>2</sup>-x<sup>2</sup>] . f(x) < ΔV < π[(x+Δx)<sup>2</sup>-x<sup>2</sup>] . f(x+Δx)
π[2x + Δx] . f(x) < <sup>ΔV</sup>/<sub>Δx</sub> < π[2x + Δx] . f(x+Δx)
as Δx → 0, <sup>ΔV</sup>/<sub>Δx</sub> → 2πx . f(x)
ΔV → 2πx . f(x) . Δx
[note: inequality signs change for a decreasing function]
radius = ...
height = ...
width = ...
just to be sure u get all available marks
going from first principals for each specific question in an exam is very time consuming and should be avoided, if u want to go from first principals u should use a generic proof:
e.g. a "sandwich" limit
anulus area x smaller height < change in volume < anulus area x greater height
π[(x+Δx)<sup>2</sup>-x<sup>2</sup>] . f(x) < ΔV < π[(x+Δx)<sup>2</sup>-x<sup>2</sup>] . f(x+Δx)
π[2x + Δx] . f(x) < <sup>ΔV</sup>/<sub>Δx</sub> < π[2x + Δx] . f(x+Δx)
as Δx → 0, <sup>ΔV</sup>/<sub>Δx</sub> → 2πx . f(x)
ΔV → 2πx . f(x) . Δx
[note: inequality signs change for a decreasing function]