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NF杯2023

NF杯2023(A)

ユーザー解説 by shakayami

以下の重積分を求めればよい.

V=DdxdydzV=\iiint_D dxdydz ただし領域DDの定義は以下の通りである. D={(x,y,z)R3;x2+y2+z21,0xyz1}D=\lbrace (x,y,z)\in\mathbb{R}^3;x^2+y^2+z^2\leq 1,0\leq x\leq y\leq z\leq 1\rbrace

以下の変数変換をする.

x=rsinθcosϕ,y=rsinθsinϕ,z=rcosθx=r\sin{\theta}\cos{\phi},y=r\sin{\theta}\sin{\phi},z=r\cos{\theta}

このとき,ヤコビアンはr2sinθr^2\sin{\theta}であるので,

dxdydz=r2sinθdrdθdϕdxdydz=r^2\sin{\theta} dr d\theta d\phi

である.

ここで,変数変換後のパラメータの取りうる値の範囲を計算する.

θ,ϕ\theta,\phiの値にかかわらず0r10\leq r\leq 1である.

さらに,π/4ϕπ/2\pi/4 \leq \phi\leq \pi/2が成り立つ.

このときのθ\thetaの取りうる値の範囲は0tanθ1/sinϕ0\leq \tan{\theta}\leq 1/\sin{\phi}

より,0θtan1(1/sinϕ)0\leq \theta \leq \tan^{-1} (1/\sin{\phi})

よって,

V=01π/4π/20tan1(1/sinϕ)r2sinθdθdϕdrV=\int_{0}^{1} \int_{\pi/4 }^{\pi/2} \int_{0}^{\tan^{-1} (1/\sin{\phi})}r^2 \sin{\theta} d\theta d\phi dr

V=01r2drπ/4π/20tan1(1/sinϕ)sinθdθdϕV=\int_{0}^{1}r^2 dr\cdot \int_{\pi/4 }^{\pi/2} \int_{0}^{\tan^{-1} (1/\sin{\phi})} \sin{\theta} d\theta d\phi

V=13π/4π/2[cosθ]θ=0θ=tan1(1/sinϕ)dϕV=\dfrac{1}{3} \cdot \int_{\pi/4}^{\pi/2} \left[-\cos{\theta}\right]_{\theta=0}^{\theta=\tan^{-1} (1/\sin{\phi})} d\phi

V=13π/4π/21cos(tan1(1/sinϕ))dϕV=\dfrac{1}{3} \cdot \int_{\pi/4}^{\pi/2} 1-\cos (\tan^{-1} (1/\sin{\phi})) d\phi

ここで,実数ttについてcos(tan1(t))=1t2+1\cos(\tan^{-1}(t))=\dfrac{1}{\sqrt{t^2+1}}が成り立つため,

V=13π/4π/2111+(1/sinϕ)2dϕV=\dfrac{1}{3} \cdot \int_{\pi/4}^{\pi/2} 1-\dfrac{1}{\sqrt{1+(1/\sin{\phi})^2}} d\phi

V=13π/4π/21sinϕ1+sin2ϕdϕV=\dfrac{1}{3} \cdot \int_{\pi/4}^{\pi/2} 1-\dfrac{\sin{\phi}}{\sqrt{1+\sin^2{\phi}}} d\phi

V=13π413π/4π/2sinϕ2cos2ϕdϕV=\dfrac{1}{3} \cdot \dfrac{\pi}{4}-\dfrac{1}{3} \int_{\pi/4}^{\pi/2} \dfrac{\sin{\phi}}{\sqrt{2-\cos^2{\phi}}} d\phi

V=π121301/212u2duV=\dfrac{\pi}{12}-\dfrac{1}{3} \int_{0}^{1/\sqrt{2}} \dfrac{1}{\sqrt{2-u^2}} du

(ただしcosϕ=u\cos{\phi}=uと置換)

V=π121301/2222v2dvV=\dfrac{\pi}{12}-\dfrac{1}{3} \int_{0}^{1/2} \dfrac{\sqrt{2}}{\sqrt{2-2v^2}} dv

(ただしu=2vu=\sqrt{2}vと置換)

V=π121301/211v2dvV=\dfrac{\pi}{12}-\dfrac{1}{3} \int_{0}^{1/2} \dfrac{1}{\sqrt{1-v^2}} dv

V=π1213[sin1(v)]01/2V=\dfrac{\pi}{12}-\dfrac{1}{3} \left[\sin^{-1}(v)\right]_{0}^{1/2}

V=π1213(sin1(1/2)sin1(0))V=\dfrac{\pi}{12}-\dfrac{1}{3} (\sin^{-1}(1/2)-\sin^{-1}(0))

V=π1213π6V=\dfrac{\pi}{12}-\dfrac{1}{3} \cdot \dfrac{\pi}{6}

V=π12π18=π36V=\dfrac{\pi}{12}-\dfrac{\pi}{18}=\dfrac{\pi}{36}

よって体積は136π\dfrac{1}{36}\piと書けるので、解答するべき値は1+36=371+36=37である.