広島大学 先進理工系科学研究科 情報科学プログラム 2019年8月実施 専門科目I 問題2 Author samparker
Description D ( R ) = { ( x , y ) ∈ R 2 : 1 ≤ x 2 + y 2 ≤ ( R + 1 ) 2 , 0 ≤ y ≤ x } D(R) = \{(x,y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 \leq (R+1)^2, 0 \leq y \leq x\} D ( R ) = {( x , y ) ∈ R 2 : 1 ≤ x 2 + y 2 ≤ ( R + 1 ) 2 , 0 ≤ y ≤ x } R ≥ 0 R \geq 0 R ≥ 0
(1) 実数 α \alpha α G ( R ) = ∬ D ( R ) x α y d x d y G(R) = \iint_{D(R)} x^\alpha y \ dxdy G ( R ) = ∬ D ( R ) x α y d x d y
(2) α = − 3 \alpha = -3 α = − 3
lim R → + 0 G ( R ) − R 2 + R 2 4 R 3 \lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} R → + 0 lim R 3 G ( R ) − 2 R + 4 R 2 を求めよ。
(3) lim R → + ∞ G ( R ) \lim_{R \to +\infty} G(R) lim R → + ∞ G ( R ) α \alpha α
Let D ( R ) = { ( x , y ) ∈ R 2 : 1 ≤ x 2 + y 2 ≤ ( R + 1 ) 2 , 0 ≤ y ≤ x } D(R) = \{(x,y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 \leq (R+1)^2, 0 \leq y \leq x\} D ( R ) = {( x , y ) ∈ R 2 : 1 ≤ x 2 + y 2 ≤ ( R + 1 ) 2 , 0 ≤ y ≤ x } R ≥ 0 R \geq 0 R ≥ 0
(1) Calculate the integral G ( R ) = ∬ D ( R ) x α y d x d y G(R) = \iint_{D(R)} x^\alpha y \, dxdy G ( R ) = ∬ D ( R ) x α y d x d y α \alpha α
(2) Find the limit
lim R → + 0 G ( R ) − R 2 + R 2 4 R 3 \lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} R → + 0 lim R 3 G ( R ) − 2 R + 4 R 2 when α = − 3 \alpha = -3 α = − 3
(3) Determine the range of the real number α \alpha α lim R → + ∞ G ( R ) \lim_{R \to +\infty} G(R) lim R → + ∞ G ( R )
Kai (1)
Let
x = r cos θ , y = r sin θ x = r \cos \theta, y = r \sin \theta x = r cos θ , y = r sin θ Then the region D ( R ) D(R) D ( R )
1 ≤ r ≤ R + 1 , 0 ≤ θ ≤ π 4 1 \leq r \leq R+1, 0 \leq \theta \leq \frac{\pi}{4} 1 ≤ r ≤ R + 1 , 0 ≤ θ ≤ 4 π The integral becomes
∫ 0 π 4 ∫ 1 R + 1 r ( r cos θ ) α r sin θ d r d θ = ∫ 0 π 4 ∫ 1 R + 1 r α + 2 cos α θ sin θ d r d θ \begin{aligned}
\int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r (r \cos \theta)^{\alpha} r \sin \theta \ drd\theta &= \int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r^{\alpha+2} \cos^{\alpha}\theta \sin \theta \ drd\theta
\end{aligned} ∫ 0 4 π ∫ 1 R + 1 r ( r cos θ ) α r sin θ d r d θ = ∫ 0 4 π ∫ 1 R + 1 r α + 2 cos α θ sin θ d r d θ ∫ 1 R + 1 r α + 2 d r = r α + 3 α + 3 ∣ 1 R + 1 = ( R + 1 ) α + 3 − 1 α + 3 \int_{1}^{R+1} r^{\alpha +2} dr = \frac{r^{\alpha + 3}}{\alpha + 3} \bigg |_{1}^{R+1} = \frac{(R+1)^{\alpha+3} - 1}{\alpha+3} ∫ 1 R + 1 r α + 2 d r = α + 3 r α + 3 1 R + 1 = α + 3 ( R + 1 ) α + 3 − 1 ∫ 0 π 4 cos α θ sin θ d θ = − cos α + 1 θ α + 1 ∣ 0 π 4 = 1 − 2 − α + 1 2 α + 1 \int_{0}^{\frac{\pi}{4}} \cos^{\alpha}\theta \sin \theta \ d\theta = \frac{-\cos^{\alpha+1} \theta}{\alpha + 1} \bigg |_{0}^{\frac{\pi}{4}} = \frac{1 - 2^{-\frac{\alpha+1}{2}}}{\alpha + 1} ∫ 0 4 π cos α θ sin θ d θ = α + 1 − cos α + 1 θ 0 4 π = α + 1 1 − 2 − 2 α + 1 Hence
G ( R ) = ( ( R + 1 ) α + 3 − 1 ) ( 1 − 2 − α + 1 2 ) ( α + 3 ) ( α + 1 ) G(R) = \frac{((R+1)^{\alpha+3}-1)(1 - 2^{-\frac{\alpha+1}{2}})}{(\alpha + 3)(\alpha + 1)} G ( R ) = ( α + 3 ) ( α + 1 ) (( R + 1 ) α + 3 − 1 ) ( 1 − 2 − 2 α + 1 ) (2)
When α = − 3 \alpha = -3 α = − 3
G ( R ) = ∫ 0 π 4 ∫ 1 R + 1 r − 1 cos − 3 θ sin θ d r d θ = ln ( R + 1 ) ⋅ C G(R) = \int_0^{\frac{\pi}{4}} \int_{1}^{R+1} r^{-1} \cos^{-3}\theta \sin \theta \ drd\theta = \ln (R+1) \cdot C G ( R ) = ∫ 0 4 π ∫ 1 R + 1 r − 1 cos − 3 θ sin θ d r d θ = ln ( R + 1 ) ⋅ C where C = ∫ 0 π 4 cos − 3 θ sin θ d θ = 1 2 C = \int_0^{\frac{\pi}{4}} \cos^{-3}\theta \sin \theta \ d\theta = \frac{1}{2} C = ∫ 0 4 π cos − 3 θ sin θ d θ = 2 1
lim R → + 0 G ( R ) − R 2 + R 2 4 R 3 = lim R → + 0 1 2 ln ( R + 1 ) − R 2 + R 2 4 R 3 = lim R → + 0 1 2 ( R − R 2 2 + R 3 3 + O ( R 4 ) ) − R 2 + R 2 4 R 3 = 1 6 \begin{aligned}
\lim_{R \to +0} \frac{G(R) - \frac{R}{2} + \frac{R^2}{4}}{R^3} &=\lim_{R \to +0} \frac{\frac{1}{2}\ln (R+1) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \\
&= \lim_{R \to +0} \frac{\frac{1}{2}(R - \frac{R^2}{2} + \frac{R^3}{3} + O(R^4)) - \frac{R}{2} + \frac{R^2}{4}}{R^3} \\
&= \frac{1}{6}
\end{aligned} R → + 0 lim R 3 G ( R ) − 2 R + 4 R 2 = R → + 0 lim R 3 2 1 ln ( R + 1 ) − 2 R + 4 R 2 = R → + 0 lim R 3 2 1 ( R − 2 R 2 + 3 R 3 + O ( R 4 )) − 2 R + 4 R 2 = 6 1 (3)
To determine the range of α \alpha α lim R → + ∞ G ( R ) \lim_{R \to +\infty} G(R) lim R → + ∞ G ( R ) R R R ( R + 1 ) α + 3 (R+1)^{\alpha + 3} ( R + 1 ) α + 3
Since lim R → + ∞ ( R + 1 ) α + 3 \lim_{R \to +\infty} (R+1)^{\alpha + 3} lim R → + ∞ ( R + 1 ) α + 3 α ≤ − 3 \alpha \leq -3 α ≤ − 3 lim R → + ∞ G ( R ) \lim_{R \to +\infty} G(R) lim R → + ∞ G ( R ) α = − 3 \alpha = -3 α = − 3 α < − 3 \alpha < -3 α < − 3