就 一位网友 棱台体积公式 微积分推导 具体过程 如下 供诸君参考

设
棱台
等高截面等积圆台
母线与对称轴所成角为θ
有
r/h=tanθ
即
r²/h²=tan²θ
即
πr²/h²=πtan²θ
即
S/h²=πtan²θ
(定值πtan²θ即这位网友所设常数k)
即
h=√S/(√πtanθ)
即
dh=1/(2√(πS)tanθ)dS
即
V
=
∫(h1,h2)Sdh
=
∫(S1,S2)S/(2√(πS)tanθ)dS
=
∫(S1,S2)√S/(2√πtanθ)dS
=
1/(2√πtanθ)∫(S1,S2)√SdS
=
1/(2√πtanθ)·2/3S^(3/2)|(S1,S2)
=
1/(3√πtanθ)(√S2³-√S1³)
=
(√S2-√S1)
(S1+S2+√(S1S2))
/
(3√πtanθ)
=
(h2-h1)
(S1+S2+√(S1S2))
/
3
=
H(S1+S2+√(S1S2))/3
ps.
抑或
(1)
有
S1/S2
=
h1²/h2²
即
S1/h1²=S2/h2²
设
S1/h1²=S2/h2²=k
有
S1=kh1²
S2=kh2²
即
S=kh
(2)
V
=
∫(h1,h2)Sdh
=
∫(h1,h2)πtan²θh²dh
=
πtan²θ∫(h1,h2)h²dh
=
πtan²θ/3·h³|(h1,h2)
=
πtan²θ(h2³-h1³)/3
=
πtan²θ(h2-h1)(h1²+h1h2+h2²)/3
=
πH(r1²+r1r2+r2²)/3
=
H(S1+√(S1S2)+S2)/3