九州大学 工学府 機械系専攻 2019年度 数学 問3 Author 祭音Myyura (refined with GPT 6 Astra)
Description C C C ( 0 , r ) (0,r) ( 0 , r ) X Y XY X Y r r r C C C X X X θ [ r a d ] \theta\,[\mathrm{rad}] θ [ rad ] P P P C C C C C C P P P ( 0 , 0 ) (0,0) ( 0 , 0 )
P P P ( x , y ) (x,y) ( x , y ) r r r θ \theta θ C C C P P P X X X
次に、X Y Z XYZ X Y Z S S S ( 0 , r , 0 ) (0,r,0) ( 0 , r , 0 ) 2 r 2r 2 r C ′ C' C ′ θ = 1 \theta=1 θ = 1 C C C Z Z Z 2 r 2r 2 r
S S S C ′ C' C ′ X Y XY X Y S S S C ′ C' C ′
题目描述
半径为 r r r C C C ( 0 , r ) (0,r) ( 0 , r ) X X X θ \theta θ P P P P P P P P P X X X ( 0 , r , 0 ) (0,r,0) ( 0 , r , 0 ) 2 r 2r 2 r S S S θ = 1 \theta=1 θ = 1 C C C Z Z Z 2 r 2r 2 r C ′ C' C ′ X Y XY X Y
Kai (1)
円の中心は ( r θ , r ) (r\theta,r) ( r θ , r ) P P P ( − r sin θ , − r cos θ ) (-r\sin\theta,-r\cos\theta) ( − r sin θ , − r cos θ )
x = r ( θ − sin θ ) , y = r ( 1 − cos θ ) . \boxed{x=r(\theta-\sin\theta),\qquad y=r(1-\cos\theta)}. x = r ( θ − sin θ ) , y = r ( 1 − cos θ ) . (2)
d x = r ( 1 − cos θ ) d θ dx=r(1-\cos\theta)d\theta d x = r ( 1 − cos θ ) d θ
A = ∫ 0 2 π r 2 ( 1 − cos θ ) 2 d θ = 3 π r 2 . A=\int_0^{2\pi}r^2(1-\cos\theta)^2d\theta
=\boxed{3\pi r^2}. A = ∫ 0 2 π r 2 ( 1 − cos θ ) 2 d θ = 3 π r 2 . (3)
θ = 1 \theta=1 θ = 1 ( r , r ) (r,r) ( r , r )
S : x 2 + ( y − r ) 2 ≤ 4 r 2 , C ′ : ( x − r ) 2 + ( y − r ) 2 ≤ r 2 . S:\ x^2+(y-r)^2\le4r^2,\qquad
C':\ (x-r)^2+(y-r)^2\le r^2. S : x 2 + ( y − r ) 2 ≤ 4 r 2 , C ′ : ( x − r ) 2 + ( y − r ) 2 ≤ r 2 . 半径 2 r 2r 2 r r r r ( 2 r , r ) (2r,r) ( 2 r , r )
(4)
v = y − r v=y-r v = y − r x = ρ cos φ x=\rho\cos\varphi x = ρ cos φ v = ρ sin φ v=\rho\sin\varphi v = ρ sin φ
0 ≤ ρ ≤ 2 r cos φ , − π 2 ≤ φ ≤ π 2 . 0\le\rho\le2r\cos\varphi,\qquad
-\frac\pi2\le\varphi\le\frac\pi2. 0 ≤ ρ ≤ 2 r cos φ , − 2 π ≤ φ ≤ 2 π . この底面の各点で、共通部分は 0 ≤ z ≤ 4 r 2 − ρ 2 0\le z\le\sqrt{4r^2-\rho^2} 0 ≤ z ≤ 4 r 2 − ρ 2
V = ∫ − π / 2 π / 2 ∫ 0 2 r cos φ ρ 4 r 2 − ρ 2 d ρ d φ = 8 r 3 3 ∫ − π / 2 π / 2 ( 1 − ∣ sin φ ∣ 3 ) d φ = 8 r 3 3 ( π − 2 ⋅ 2 3 ) = 8 r 3 9 ( 3 π − 4 ) . \begin{aligned}
V&=\int_{-\pi/2}^{\pi/2}\int_0^{2r\cos\varphi}
\rho\sqrt{4r^2-\rho^2}\,d\rho\,d\varphi\\
&=\frac{8r^3}{3}\int_{-\pi/2}^{\pi/2}
\left(1-|\sin\varphi|^3\right)d\varphi\\
&=\frac{8r^3}{3}\left(\pi-2\cdot\frac23\right)
=\boxed{\frac{8r^3}{9}(3\pi-4)}.
\end{aligned} V = ∫ − π /2 π /2 ∫ 0 2 r c o s φ ρ 4 r 2 − ρ 2 d ρ d φ = 3 8 r 3 ∫ − π /2 π /2 ( 1 − ∣ sin φ ∣ 3 ) d φ = 3 8 r 3 ( π − 2 ⋅ 3 2 ) = 9 8 r 3 ( 3 π − 4 ) . 