愛知医科大学 2023年度 医学部一般選抜 数学 III
問題 ID: `MM-1f16746c14e5-Q3`
球面は x 2 + y 2 + z 2 = 1 x^2+y^2+z^2=1 x 2 + y 2 + z 2 = 1 P 0 = ( 3 2 , 0 , 1 2 ) \displaystyle
P_0=\left(\frac{\sqrt3}{2},0,\frac12\right) P 0 = ( 2 3 , 0 , 2 1 ) P 0 P_0 P 0 3 2 x + 1 2 z = 1. (1) \displaystyle
\frac{\sqrt3}{2}x+\frac12z=1. \tag{1} 2 3 x + 2 1 z = 1. ( 1 ) L L L P 0 P_0 P 0
1) 円板上の点のz座標の範囲
平面(1)上で高さ z = t z=t z = t x = 2 − t 3 . \displaystyle
x=\frac{2-t}{\sqrt3}. x = 3 2 − t . ( x , y , t ) (x,y,t) ( x , y , t ) ( x − 3 2 ) 2 + y 2 + ( t − 1 2 ) 2 ≦ 1. \displaystyle
\left(x-\frac{\sqrt3}{2}\right)^2+y^2+
\left(t-\frac12\right)^2\le1. ( x − 2 3 ) 2 + y 2 + ( t − 2 1 ) 2 ≦ 1. δ = t − 1 / 2 \delta=t-1/2 δ = t − 1/2 x − 3 / 2 = − δ / 3 x-\sqrt3/2=-\delta/\sqrt3 x − 3 /2 = − δ / 3 y y y δ 2 3 + δ 2 ≦ 1 すなわち ∣ δ ∣ ≦ 3 2 . \displaystyle
\frac{\delta^2}{3}+\delta^2\le1
\text{すなわち} |\delta|\le\frac{\sqrt3}{2}. 3 δ 2 + δ 2 ≦ 1 すなわち ∣ δ ∣ ≦ 2 3 . 1 − 3 2 ≦ t ≦ 1 + 3 2 . \displaystyle
\boxed{\frac{1-\sqrt3}{2}\le t\le\frac{1+\sqrt3}{2}}. 2 1 − 3 ≦ t ≦ 2 1 + 3 .
2) z軸に最も近い距離 d 1 d_1 d 1
高さ t t t x = ( 2 − t ) / 3 x=(2-t)/\sqrt3 x = ( 2 − t ) / 3 y 2 ≦ 1 − 4 3 ( t − 1 2 ) 2 . \displaystyle
y^2\le1-\frac43\left(t-\frac12\right)^2. y 2 ≦ 1 − 3 4 ( t − 2 1 ) 2 . x 2 + y 2 \sqrt{x^2+y^2} x 2 + y 2 y = 0 y=0 y = 0 d 1 = 2 − t 3 ( 1 − 3 2 ≦ t ≦ 1 + 3 2 ) . \displaystyle
\boxed{d_1=\frac{2-t}{\sqrt3}}
\quad\left(\frac{1-\sqrt3}{2}\le t\le\frac{1+\sqrt3}{2}\right). d 1 = 3 2 − t ( 2 1 − 3 ≦ t ≦ 2 1 + 3 ) .
3) 円板が通過してできる立体の体積
円 C C C r = 3 / 2 r=\sqrt3/2 r = 3 /2 P = ( r cos φ , r sin φ , 1 / 2 ) P=(r\cos\varphi,r\sin\varphi,1/2) P = ( r cos φ , r sin φ , 1/2 ) e r = ( cos φ , sin φ , 0 ) e_r=(\cos\varphi,\sin\varphi,0) e r = ( cos φ , sin φ , 0 ) e φ = ( − sin φ , cos φ , 0 ) e_\varphi=(-\sin\varphi,\cos\varphi,0) e φ = ( − sin φ , cos φ , 0 ) t t t X = q e r + s e φ + t ( 0 , 0 , 1 ) X=q e_r+s e_\varphi+t(0,0,1) X = q e r + s e φ + t ( 0 , 0 , 1 ) P ⋅ X = 1 P\cdot X=1 P ⋅ X = 1 q = 1 − t / 2 r = 2 − t 3 . \displaystyle
q=\frac{1-t/2}{r}=\frac{2-t}{\sqrt3}. q = r 1 − t /2 = 3 2 − t . ∣ X − P ∣ 2 = ( q − r ) 2 + s 2 + ( t − 1 / 2 ) 2 = 4 3 ( t − 1 2 ) 2 + s 2 . \displaystyle
|X-P|^2=(q-r)^2+s^2+(t-1/2)^2
=\frac43\left(t-\frac12\right)^2+s^2. ∣ X − P ∣ 2 = ( q − r ) 2 + s 2 + ( t − 1/2 ) 2 = 3 4 ( t − 2 1 ) 2 + s 2 . ∣ X − P ∣ 2 ≦ 1 |X-P|^2\le1 ∣ X − P ∣ 2 ≦ 1 ∣ s ∣ ≦ h ( t ) : = 1 − 4 3 ( t − 1 2 ) 2 . \displaystyle
|s|\le h(t):=\sqrt{1-\frac43\left(t-\frac12\right)^2}. ∣ s ∣ ≦ h ( t ) := 1 − 3 4 ( t − 2 1 ) 2 . φ \varphi φ t t t q = d 1 ( t ) q=d_1(t) q = d 1 ( t ) q 2 + h ( t ) 2 \sqrt{q^2+h(t)^2} q 2 + h ( t ) 2 π ( ( q 2 + h 2 ) − q 2 ) = π ( 1 − 4 3 ( t − 1 2 ) 2 ) . \displaystyle
\pi\bigl((q^2+h^2)-q^2\bigr)
=\pi\left(1-\frac43\left(t-\frac12\right)^2\right). π ( ( q 2 + h 2 ) − q 2 ) = π ( 1 − 3 4 ( t − 2 1 ) 2 ) . V = π ∫ ( 1 − 3 ) / 2 ( 1 + 3 ) / 2 ( 1 − 4 3 ( t − 1 2 ) 2 ) d t = π ∫ − 3 / 2 3 / 2 ( 1 − 4 3 δ 2 ) d δ = 2 3 3 π . \displaystyle
\begin{aligned}
V
&=\pi\int_{(1-\sqrt3)/2}^{(1+\sqrt3)/2}
\left(1-\frac43\left(t-\frac12\right)^2\right)dt\\
&=\pi\int_{-\sqrt3/2}^{\sqrt3/2}
\left(1-\frac43\delta^2\right)d\delta
=\boxed{\frac{2\sqrt3}{3}\pi}.
\end{aligned} V = π ∫ ( 1 − 3 ) /2 ( 1 + 3 ) /2 ( 1 − 3 4 ( t − 2 1 ) 2 ) d t = π ∫ − 3 /2 3 /2 ( 1 − 3 4 δ 2 ) d δ = 3 2 3 π .
**使用解法:** 球面の接平面、円板内距離条件、回転対称性を用いた水平断面の円環化、断面積の積分。
