早稲田大学 創造理工学研究科 経営システム工学専攻 2015年8月実施 微积分 Author 思齐塾 , 祭音Myyura
Description 領域(domain) D D D x ≥ 0 x \geq 0 x ≥ 0 x 2 + y 2 ≤ 4 x^2 + y^2 \leq 4 x 2 + y 2 ≤ 4
I = ∬ D y 2 1 + x 2 + y 2 d x d y I = \iint_D \frac{y^2}{\sqrt{1 + x^2 + y^2}} dxdy I = ∬ D 1 + x 2 + y 2 y 2 d x d y 题目描述
定义区域
D = { ( x , y ) ∣ x ≥ 0 , x 2 + y 2 ≤ 4 } . D=\{(x,y)\mid x\geq0,\ x^2+y^2\leq4\}. D = {( x , y ) ∣ x ≥ 0 , x 2 + y 2 ≤ 4 } . 求二重积分
I = ∬ D y 2 1 + x 2 + y 2 d x d y I=\iint_D\frac{y^2}{\sqrt{1+x^2+y^2}}\,dx\,dy I = ∬ D 1 + x 2 + y 2 y 2 d x d y 的值。
Kai We transform the double integral to polar coordinates:
x = r cos θ , y = r sin θ x = r\cos\theta, y = r\sin\theta x = r cos θ , y = r sin θ x 2 + y 2 = r 2 x^2 + y^2 = r^2 x 2 + y 2 = r 2 d x d y = r d r d θ dxdy = r drd\theta d x d y = r d r d θ D D D x ≥ 0 x \geq 0 x ≥ 0 x 2 + y 2 ≤ 4 x^2 + y^2 \leq 4 x 2 + y 2 ≤ 4 0 ≤ r ≤ 2 0 \leq r \leq 2 0 ≤ r ≤ 2 − π 2 ≤ θ ≤ π 2 -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} − 2 π ≤ θ ≤ 2 π x ≥ 0 x \geq 0 x ≥ 0 − π 2 ≤ θ ≤ π 2 -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} − 2 π ≤ θ ≤ 2 π
I = ∫ − π 2 π 2 ∫ 0 2 ( r sin θ ) 2 1 + r 2 r d r d θ = ∫ − π 2 π 2 ∫ 0 2 r 3 sin 2 θ 1 + r 2 d r d θ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^2 \frac{(r\sin\theta)^2}{\sqrt{1 + r^2}} r dr d\theta = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_0^2 \frac{r^3\sin^2\theta}{\sqrt{1 + r^2}} dr d\theta I = ∫ − 2 π 2 π ∫ 0 2 1 + r 2 ( r sin θ ) 2 r d r d θ = ∫ − 2 π 2 π ∫ 0 2 1 + r 2 r 3 sin 2 θ d r d θ I = ∫ − π 2 π 2 sin 2 θ d θ ∫ 0 2 r 3 1 + r 2 d r I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2\theta d\theta \int_0^2 \frac{r^3}{\sqrt{1 + r^2}} dr I = ∫ − 2 π 2 π sin 2 θ d θ ∫ 0 2 1 + r 2 r 3 d r Let u = 1 + r 2 u = 1 + r^2 u = 1 + r 2 r 2 = u − 1 r^2 = u - 1 r 2 = u − 1 d u = 2 r d r du = 2rdr d u = 2 r d r r d r = 1 2 d u rdr = \frac{1}{2}du r d r = 2 1 d u r = 0 r = 0 r = 0 u = 1 u = 1 u = 1 r = 2 r = 2 r = 2 u = 5 u = 5 u = 5
I = ∫ − π 2 π 2 sin 2 θ d θ ∫ 1 5 r 2 u 1 2 d u = ∫ − π 2 π 2 sin 2 θ d θ ∫ 1 5 u − 1 2 u d u I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2\theta d\theta \int_1^5 \frac{r^2}{\sqrt{u}} \frac{1}{2} du = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2\theta d\theta \int_1^5 \frac{u - 1}{2\sqrt{u}} du I = ∫ − 2 π 2 π sin 2 θ d θ ∫ 1 5 u r 2 2 1 d u = ∫ − 2 π 2 π sin 2 θ d θ ∫ 1 5 2 u u − 1 d u I = ∫ − π 2 π 2 1 − cos ( 2 θ ) 2 d θ ∫ 1 5 u − 1 2 u d u = [ θ 2 − sin ( 2 θ ) 4 ] − π 2 π 2 ∫ 1 5 1 2 ( u 1 / 2 − u − 1 / 2 ) d u I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} d\theta \int_1^5 \frac{u - 1}{2\sqrt{u}} du = \left[\frac{\theta}{2} - \frac{\sin(2\theta)}{4}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \int_1^5 \frac{1}{2} (u^{1/2} - u^{-1/2}) du I = ∫ − 2 π 2 π 2 1 − cos ( 2 θ ) d θ ∫ 1 5 2 u u − 1 d u = [ 2 θ − 4 sin ( 2 θ ) ] − 2 π 2 π ∫ 1 5 2 1 ( u 1/2 − u − 1/2 ) d u I = [ π 4 − ( − π 4 ) ] 1 2 [ 2 3 u 3 / 2 − 2 u 1 / 2 ] 1 5 = π 2 1 2 [ 2 3 5 3 / 2 − 2 5 − 2 3 + 2 ] I = \left[\frac{\pi}{4} - (-\frac{\pi}{4})\right] \frac{1}{2} \left[\frac{2}{3}u^{3/2} - 2u^{1/2}\right]_1^5 = \frac{\pi}{2} \frac{1}{2} \left[\frac{2}{3}5^{3/2} - 2\sqrt{5} - \frac{2}{3} + 2\right] I = [ 4 π − ( − 4 π ) ] 2 1 [ 3 2 u 3/2 − 2 u 1/2 ] 1 5 = 2 π 2 1 [ 3 2 5 3/2 − 2 5 − 3 2 + 2 ] I = π 4 [ 10 5 3 − 2 5 + 4 3 ] = π 4 [ 4 5 3 + 4 3 ] = π 3 ( 5 + 1 ) I = \frac{\pi}{4} \left[\frac{10\sqrt{5}}{3} - 2\sqrt{5} + \frac{4}{3}\right] = \frac{\pi}{4} \left[\frac{4\sqrt{5}}{3} + \frac{4}{3}\right] = \frac{\pi}{3} (\sqrt{5} + 1) I = 4 π [ 3 10 5 − 2 5 + 3 4 ] = 4 π [ 3 4 5 + 3 4 ] = 3 π ( 5 + 1 ) Therefore, the value of the double integral is π 3 ( 5 + 1 ) \frac{\pi}{3}(\sqrt{5} + 1) 3 π ( 5 + 1 )