高専の数学解説

第8章 8.23

(1)

x2+y2≤1−y≤z≤2yx^2+y^2\le1 \qquad -y\le z\le2y

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと上関係式から y≥0y\ge0 となり 0≤r≤10\le r\le1 0≤θ≤π0\le\theta\le\pi

∬D3y dx dy=∫0π∫013rsin⁡θ⋅r dr dθ=∫0π[3⋅13r3sin⁡θ]01dθ=∫0πsin⁡θ dθ=[−cos⁡θ]0π=2\iint_D 3y\,dx\,dy=\int_0^\pi\int_0^1 3r\sin\theta\cdot r\,dr\,d\theta=\int_0^\pi\left[3\cdot\frac13r^3\sin\theta\right]_0^1d\theta=\int_0^\pi\sin\theta\,d\theta=\left[-\cos\theta\right]_0^\pi=2

(2)

x2+y2+z2≤a2x2+y2≤axx^2+y^2+z^2\le a^2 \qquad x^2+y^2\le ax

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと 0≤r≤acos⁡θ0\le r\le a\cos\theta 0≤θ≤π0\le\theta\le\pi

z=a2−x2−y2z=\sqrt{a^2-x^2-y^2} から

2∫0π∫0acos⁡θa2−r2⋅r dr dθ=2∫0π[−13(a2−r2)3]0acos⁡θdθ=2∫0π−13a3sin⁡3θ+13a3 dθ2\int_0^\pi\int_0^{a\cos\theta}\sqrt{a^2-r^2}\cdot r\,dr\,d\theta=2\int_0^\pi\left[-\frac13\sqrt{(a^2-r^2)^3}\right]_0^{a\cos\theta}d\theta=2\int_0^\pi-\frac13a^3\sin^3\theta+\frac13a^3\,d\theta

=−23a3∫0π(sin⁡θ−sin⁡θ⋅cos⁡2θ)−1 dθ=-\frac23a^3\int_0^\pi(\sin\theta-\sin\theta\cdot\cos^2\theta)-1\,d\theta

第２項において cos⁡θ=t\cos\theta=t とおくと

=−23a3{[−cos⁡θ]0π−[−13t3]−11−[θ]0π}=−23a3(2−23−π)=43(π2−23)a3=-\frac23a^3\left\{\left[-\cos\theta\right]_0^\pi-\left[-\frac13t^3\right]_{-1}^1-\left[\theta\right]_0^\pi\right\}=-\frac23a^3\left(2-\frac23-\pi\right)=\frac43\left(\frac{\pi}{2}-\frac23\right)a^3

(3)

x2+y2≤a20≤z≤x+yx^2+y^2\le a^2 \qquad 0\le z\le x+y

x=rcos⁡θx=r\cos\theta y=rsin⁡θy=r\sin\theta とおくと 0≤r≤a0\le r\le a 0≤x+y0\le x+y から 0≤cos⁡θ+sin⁡θ0\le\cos\theta+\sin\theta −π4≤θ≤3π4-\dfrac{\pi}{4}\le\theta\le\dfrac{3\pi}{4}

∫−14π34π∫0ar2(cos⁡θ+sin⁡θ) dr dθ=∫−14π34π13a3(cos⁡θ+sin⁡θ) dθ=[13a3(sin⁡θ−cos⁡θ)]−14π34π\int_{-\frac14\pi}^{\frac34\pi}\int_0^a r^2(\cos\theta+\sin\theta)\,dr\,d\theta=\int_{-\frac14\pi}^{\frac34\pi}\frac13a^3(\cos\theta+\sin\theta)\,d\theta=\left[\frac13a^3(\sin\theta-\cos\theta)\right]_{-\frac14\pi}^{\frac34\pi}

=13a3(22+22)=223a3=\frac13a^3\left(\frac{2}{\sqrt2}+\frac{2}{\sqrt2}\right)=\frac{2\sqrt2}{3}a^3

(4)

x2+y2+z2≤2z≥x2+y2x^2 + y^2 + z^2 \le 2 \qquad z \ge x^2 + y^2

z2=2−x2−y2=2−zz2+z−2=0(z+2)(z−1)=0z=1,−2z^2 = 2 - x^2 - y^2 = 2 - z \qquad z^2 + z - 2 = 0 \qquad (z+2)(z-1) = 0 \qquad z = 1, -2

x2+y2≥0x^2 + y^2 \ge 0 から z=1z = 1として x2+y2≤1x^2 + y^2 \le 1

x=rcos⁡θx = r\cos\theta y=rsin⁡θy = r\sin\theta とおくと 0≤r≤10 \le r \le 1 0≤θ≤2π0 \le \theta \le 2\pi

∫02π∫01(2−r2−r2)r dr dθ=∫02π[−13(2−r2)3−14r4]01dθ=∫02π(−13−14+223)dθ\int_0^{2\pi}\int_0^1 (\sqrt{2-r^2} - r^2) r \, dr\, d\theta = \int_0^{2\pi} \left[-\frac{1}{3}\sqrt{(2-r^2)^3} - \frac{1}{4}r^4\right]_0^1 d\theta = \int_0^{2\pi} \left(-\frac{1}{3} - \frac{1}{4} + \frac{2\sqrt{2}}{3}\right) d\theta

=∫02π112(82−7) dθ=16(82−7)π= \int_0^{2\pi} \frac{1}{12}(8\sqrt{2} - 7)\, d\theta = \frac{1}{6}(8\sqrt{2} - 7)\pi

解説: r31bn1z