3佐賀大学 2020年 数学 第3問解答・解説

$0<t<2$ のとき、各積分範囲で半径の2乗は0以上である。円形の断面積を積分すると、 $$\displaystyle V_1(t)=\pi\int_0^t [x(3-x)]^2dx =\pi\left(3t^3-\frac32t^4+\frac15t^5\right).$$ $$\displaystyle V_2(t)=4\pi\int_t^2 x^2(x-2)^2dx =\pi\left(\frac{64}{15}-\frac{16}{3}t^3+4t^4-\frac45t^5\right).$$ したがって $$\displaystyle V(t)=\pi\left(\frac{64}{15}-\frac73t^3+\frac52t^4-\frac35t^5\right), \quad V'(t)=-\pi t^2(3t-7)(t-1).$$ $0<t<1$ では $V'<0$、$1<t<2$ では $V'>0$ である。したがって $0<t<2$ の範囲では減少した後に増加し、唯一の極値は $t=1$ での極小値 $V(1)=\boxed{23\pi/6}$ である。この範囲で最大値はない。

Tags. Unit: definite integrals, volume of revolution, differentiation/monotonicity. Method: disk integral, polynomial antiderivative, derivative factorization and sign chart. Theme: optimize the sum of two adjacent solids.

Quality comparison and disposition Independent derivation was saved before consulting the prior answer. The prior final result agrees: $\displaystyle V_1=\pi\left(3t^3-\frac32t^4+\frac15t^5\right)$; V2=pi(64/15-16t^3/3+4t^4-4t^5/5); V decreases on (0,1), increases on (1,2), minimum 23pi/6 at t=1; no maximum.. All three formulas, derivative signs, and the minimum agree. The original accepted result files and their evidence metadata remain untouched. Prior actual-result evidence was not independently validated, so actual model status stays UNVERIFIED.　 Disposition: mathematics PASS; source eligible; display matched; adopt this independent high-school solution.