(1) ( i ) 0 ≤ θ ≤ π 6 \displaystyle (i)\ 0\le\theta\le\frac\pi6 ( i ) 0 ≤ θ ≤ 6 π x 0 = r + r sin θ ~ , θ ~ = π 6 − θ \displaystyle x_0=r+r\sin\tilde\theta,\ \tilde\theta=\frac\pi6-\theta x 0 = r + r sin θ ~ , θ ~ = 6 π − θ V ( θ ) \displaystyle V(\theta) V ( θ )
V ( θ ) = ∫ − r sin θ ~ r π ( r 2 − x 2 ) d x = π ( r 2 ( r + r sin θ ~ ) − 1 3 ( r 3 − ( − r sin θ ~ ) 3 ) ) = π 3 r 3 ( 3 ( 1 + sin θ ~ ) − ( 1 + sin 3 θ ~ ) ) = π 3 r 3 ( 2 + 3 sin θ ~ − sin 3 θ ~ ) = π 3 r 3 ( 2 + 3 sin ( π 6 − θ ) − sin 3 ( π 6 − θ ) ) \displaystyle \begin{aligned}V(\theta)&=\int_{-r\sin\tilde\theta}^r\pi(r^2-x^2)\,dx\\&=\pi\left(r^2(r+r\sin\tilde\theta)-\frac13(r^3-(-r\sin\tilde\theta)^3)\right)\\&=\frac\pi3r^3\left(3(1+\sin\tilde\theta)-(1+\sin^3\tilde\theta)\right)\\&=\frac\pi3r^3(2+3\sin\tilde\theta-\sin^3\tilde\theta)\\&=\frac\pi3r^3\left(2+3\sin\left(\frac\pi6-\theta\right)-\sin^3\left(\frac\pi6-\theta\right)\right)\end{aligned} V ( θ ) = ∫ − r s i n θ ~ r π ( r 2 − x 2 ) d x = π ( r 2 ( r + r sin θ ~ ) − 3 1 ( r 3 − ( − r sin θ ~ ) 3 ) ) = 3 π r 3 ( 3 ( 1 + sin θ ~ ) − ( 1 + sin 3 θ ~ ) ) = 3 π r 3 ( 2 + 3 sin θ ~ − sin 3 θ ~ ) = 3 π r 3 ( 2 + 3 sin ( 6 π − θ ) − sin 3 ( 6 π − θ ) )
( i i ) π 6 ≤ θ ≤ 2 π 3 \displaystyle (ii)\ \frac\pi6\le\theta\le\frac{2\pi}3 ( ii ) 6 π ≤ θ ≤ 3 2 π x 0 = r − r sin θ ~ ~ , θ ~ ~ = θ − π 6 \displaystyle x_0=r-r\sin\tilde{\tilde\theta},\ \tilde{\tilde\theta}=\theta-\frac\pi6 x 0 = r − r sin θ ~ ~ , θ ~ ~ = θ − 6 π V ( θ ) \displaystyle V(\theta) V ( θ )
V ( θ ) = ∫ r sin θ ~ ~ r π ( r 2 − x 2 ) d x = π ( r 2 ( r − r sin θ ~ ~ ) − 1 3 ( r 3 − r 3 sin 3 θ ~ ~ ) ) = π 3 r 3 ( 3 ( 1 − sin θ ~ ~ ) − ( 1 − sin 3 θ ~ ~ ) ) = π 3 r 3 ( 2 − 3 sin θ ~ ~ + sin 3 θ ~ ~ ) = π 3 r 3 ( 2 − 3 sin ( θ − π 6 ) + sin 3 ( θ − π 6 ) ) \displaystyle \begin{aligned}V(\theta)&=\int_{r\sin\tilde{\tilde\theta}}^r\pi(r^2-x^2)\,dx\\&=\pi\left(r^2(r-r\sin\tilde{\tilde\theta})-\frac13(r^3-r^3\sin^3\tilde{\tilde\theta})\right)\\&=\frac\pi3r^3\left(3(1-\sin\tilde{\tilde\theta})-(1-\sin^3\tilde{\tilde\theta})\right)\\&=\frac\pi3r^3(2-3\sin\tilde{\tilde\theta}+\sin^3\tilde{\tilde\theta})\\&=\frac\pi3r^3\left(2-3\sin\left(\theta-\frac\pi6\right)+\sin^3\left(\theta-\frac\pi6\right)\right)\end{aligned} V ( θ ) = ∫ r s i n θ ~ ~ r π ( r 2 − x 2 ) d x = π ( r 2 ( r − r sin θ ~ ~ ) − 3 1 ( r 3 − r 3 sin 3 θ ~ ~ ) ) = 3 π r 3 ( 3 ( 1 − sin θ ~ ~ ) − ( 1 − sin 3 θ ~ ~ ) ) = 3 π r 3 ( 2 − 3 sin θ ~ ~ + sin 3 θ ~ ~ ) = 3 π r 3 ( 2 − 3 sin ( θ − 6 π ) + sin 3 ( θ − 6 π ) ) 따라서 그릇에 담긴 물의 부피 V ( θ ) = π 3 r 3 ( 2 − 3 sin ( θ − π 6 ) + sin 3 ( θ − π 6 ) ) , 0 ≤ θ ≤ 2 π 3 \displaystyle V(\theta)=\frac\pi3r^3\left(2-3\sin\left(\theta-\frac\pi6\right)+\sin^3\left(\theta-\frac\pi6\right)\right),\ 0\le\theta\le\frac{2\pi}3 V ( θ ) = 3 π r 3 ( 2 − 3 sin ( θ − 6 π ) + sin 3 ( θ − 6 π ) ) , 0 ≤ θ ≤ 3 2 π 가 된다.
(2) 그릇의 기울어진 각도가 θ \displaystyle \theta θ W ( θ ) = 9 π 8 r 3 − V ( θ ) \displaystyle W(\theta)=\frac{9\pi}8r^3-V(\theta) W ( θ ) = 8 9 π r 3 − V ( θ ) θ \displaystyle \theta θ d d θ W ( θ ) = − d d θ π 3 r 3 ( 2 − 3 sin ( θ − π 6 ) + sin 3 ( θ − π 6 ) ) = − π r 3 ( − cos ( θ − π 6 ) + sin 2 ( θ − π 6 ) cos ( θ − π 6 ) ) = π r 3 cos 3 ( θ − π 6 ) \displaystyle \begin{aligned}\frac d{d\theta}W(\theta)&=-\frac d{d\theta}\frac\pi3r^3\left(2-3\sin\left(\theta-\frac\pi6\right)+\sin^3\left(\theta-\frac\pi6\right)\right)\\&=-\pi r^3\left(-\cos\left(\theta-\frac\pi6\right)+\sin^2\left(\theta-\frac\pi6\right)\cos\left(\theta-\frac\pi6\right)\right)\\&=\pi r^3\cos^3\left(\theta-\frac\pi6\right)\end{aligned} d θ d W ( θ ) = − d θ d 3 π r 3 ( 2 − 3 sin ( θ − 6 π ) + sin 3 ( θ − 6 π ) ) = − π r 3 ( − cos ( θ − 6 π ) + sin 2 ( θ − 6 π ) cos ( θ − 6 π ) ) = π r 3 cos 3 ( θ − 6 π ) 이고 0 ≤ cos ( θ − π 6 ) ≤ 1 \displaystyle 0\le\cos\left(\theta-\frac\pi6\right)\le1 0 ≤ cos ( θ − 6 π ) ≤ 1 θ = π 6 \displaystyle \theta=\frac\pi6 θ = 6 π d d θ W ( θ ) = π r 3 \displaystyle \frac d{d\theta}W(\theta)=\pi r^3 d θ d W ( θ ) = π r 3 한편, θ = π 6 \displaystyle \theta=\frac\pi6 θ = 6 π W ( π 6 ) = 9 π 8 r 3 − V ( π 6 ) = 9 π 8 r 3 − 2 π 3 r 3 = 11 π 24 r 3 \displaystyle W\left(\frac\pi6\right)=\frac{9\pi}8r^3-V\left(\frac\pi6\right)=\frac{9\pi}8r^3-\frac{2\pi}3r^3=\frac{11\pi}{24}r^3 W ( 6 π ) = 8 9 π r 3 − V ( 6 π ) = 8 9 π r 3 − 3 2 π r 3 = 24 11 π r 3
수리논술, 이게 다다.
