高専の数学解説

第8章 8.10

(1)

0≤x≤10≤y≤1−x0 \le x \le 1 \quad 0 \le y \le \sqrt{1-x}

∫01∫01−y2(1−x2) dx dy=∫01[x−13x3]01−y2dy=∫011−y2−13(1−y2)3 dy\int_0^1\int_0^{1-y^2}(1-x^2)\,dx\,dy=\int_0^1\left[x-\frac{1}{3}x^3\right]_0^{1-y^2}dy=\int_0^1 1-y^2-\frac{1}{3}(1-y^2)^3\,dy

=∫011−y2−13+y2−y4+13y6 dy=[23y−15y5+121y7]01=1835=\int_0^1 1-y^2-\frac{1}{3}+y^2-y^4+\frac{1}{3}y^6\,dy=\left[\frac{2}{3}y-\frac{1}{5}y^5+\frac{1}{21}y^7\right]_0^1=\frac{18}{35}

(2)

2y=xx+y=2a2y=x \quad x+y=2a \quad これから y=23a0≤y≤23a\quad y=\frac{2}{3}a \quad 0\le y\le\frac{2}{3}a

∫023a∫2y2a−y1axy dx dy=∫023a[12ax2y]2y2a−ydy=∫023a12a{y(2a−y)2−4y3}dy\int_0^{\frac{2}{3}a}\int_{2y}^{2a-y}\frac{1}{a}xy\,dx\,dy=\int_0^{\frac{2}{3}a}\left[\frac{1}{2a}x^2y\right]_{2y}^{2a-y}dy=\int_0^{\frac{2}{3}a}\frac{1}{2a}\{y(2a-y)^2-4y^3\}dy

=∫023a12a(4a2y−4ay2+y3−4y3)dy=12a[2a2y2−43ay3−34y4]023a=1481a3=\int_0^{\frac{2}{3}a}\frac{1}{2a}(4a^2y-4ay^2+y^3-4y^3)dy=\frac{1}{2a}\left[2a^2y^2-\frac{4}{3}ay^3-\frac{3}{4}y^4\right]_0^{\frac{2}{3}a}=\frac{14}{81}a^3

(3)

1≤x≤31−1−(x−2)2≤y≤1+1−(x−2)21\le x\le3 \quad 1-\sqrt{1-(x-2)^2}\le y\le1+\sqrt{1-(x-2)^2} \quad これから

∫13∫1−1−(x−2)21+1−(x−2)2xy dy dx=∫13[12xy2]1−1−(x−2)21+1−(x−2)2dx\int_1^3\int_{1-\sqrt{1-(x-2)^2}}^{1+\sqrt{1-(x-2)^2}}xy\,dy\,dx=\int_1^3\left[\frac{1}{2}xy^2\right]_{1-\sqrt{1-(x-2)^2}}^{1+\sqrt{1-(x-2)^2}}dx

=∫1312x(1+21−(x−2)2+1−(x−2)2)−12x(1−21−(x−2)2+1−(x−2)2) dx=\int_1^3 \frac{1}{2}x(1+2\sqrt{1-(x-2)^2}+1-(x-2)^2)-\frac{1}{2}x(1-2\sqrt{1-(x-2)^2}+1-(x-2)^2)\,dx

=∫132x1−(x−2)2 dx=\int_1^3 2x\sqrt{1-(x-2)^2}\,dx

=[2x⋅12{(x−2)1−(x−2)2+sin⁡−1(x−2)}]13−∫13(x−2)1−(x−2)2+sin⁡−1(x−2) dx=\left[2x\cdot\frac{1}{2}\{(x-2)\sqrt{1-(x-2)^2}+\sin^{-1}(x-2)\}\right]_1^3-\int_1^3(x-2)\sqrt{1-(x-2)^2}+\sin^{-1}(x-2)\,dx

x−2=tx-2=t とおくと dx=dtx:1→3t:−1→1dx=dt \quad x:1\to3 \quad t:-1\to1 \quad これから

=2(3⋅12⋅π2+12⋅π2)−∫−11t1−t2+sin⁡−1t dt=2\left(3\cdot\frac{1}{2}\cdot\frac{\pi}{2}+\frac{1}{2}\cdot\frac{\pi}{2}\right)-\int_{-1}^1 t\sqrt{1-t^2}+\sin^{-1}t\,dt

=2π−[−13(1−t2)3]−11−[tsin⁡−1t]−11+∫−11t1−t2dt=2π−(π2−π2)−[1−t2]−11=2π=2\pi-\left[-\frac{1}{3}\sqrt{(1-t^2)^3}\right]_{-1}^1-[t\sin^{-1}t]_{-1}^1+\int_{-1}^1\frac{t}{\sqrt{1-t^2}}dt=2\pi-\left(\frac{\pi}{2}-\frac{\pi}{2}\right)-\left[\sqrt{1-t^2}\right]_{-1}^1=2\pi

(4)

z=4axy2=ax−x2y=ax−x20≤x≤a0≤y≤ax−x2z = \sqrt{4ax} \qquad y^2 = ax - x^2 \qquad y = \sqrt{ax-x^2} \qquad 0 \le x \le a \qquad 0 \le y \le \sqrt{ax-x^2}

4∫0a∫0ax−x22ax dy dx=8∫0aax(ax−x2) dx=8a∫0axa−x dx4\int_0^a \int_0^{\sqrt{ax-x^2}} 2\sqrt{ax}\, dy\, dx = 8\int_0^a \sqrt{ax(ax-x^2)}\, dx = 8\sqrt{a}\int_0^a x\sqrt{a-x}\, dx

=8a{[−23x(a−x)3]0a+23∫0a(a−x)3 dx}=8a⋅23[−25(a−x)5]0a=3215a3= 8\sqrt{a}\left\{\left[-\frac{2}{3}x\sqrt{(a-x)^3}\right]_0^a + \frac{2}{3}\int_0^a \sqrt{(a-x)^3}\, dx\right\} = 8\sqrt{a}\cdot\frac{2}{3}\left[-\frac{2}{5}\sqrt{(a-x)^5}\right]_0^a = \frac{32}{15}a^3

解説: r31bn1z