【微分積分】8-3-2 3重積分の変数変換｜問題集
1.次の3重積分を求めなさい。
(1)\(\displaystyle \iiint_\Omega(x+y^2z)dxdydz\)
\(\Omega=\{(x,y,z)|0\leqq z\leqq \sqrt{x^2+y^2}\leqq 1,x\geqq0\}\)
\(\Omega=\{(x,y,z)|0\leqq z\leqq \sqrt{x^2+y^2}\leqq 1,x\geqq0\}\)
積分領域\(\Omega\)は
\(\Omega=\{(x,y,z)|0\leqq z\leqq 1,z\leqq \sqrt{x^2+y^2}\leqq 1,x\geqq0\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta)|z\leqq r\leqq 1,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \int_0^1\iint_\Omega(x+y^2z)dxdydz\)
\(\displaystyle =\int_0^1\iint_K(r\cos\theta+zr^2\sin^2\theta・rdrd\theta dz\)
\(\displaystyle =\int_0^1\int_z^1\int_{-\frac{\pi}{2}}^\frac{\pi}{2}(r^2\cos\theta+zr^3\sin^2\theta)d\theta drdz\)
\(\displaystyle =\int_0^1\int_z^1\left[2r^2\sin\theta+zr^3\left(\theta-\frac{\sin2\theta}{2}\right)\right]_0^\frac{\pi}{2}drdz\)
\(\displaystyle =\int_0^1\int_z^1\left(2r^2+\frac{\pi z}{2}r^3\right)drdz\)
\(\displaystyle =\int_0^1\left(\frac{2}{3}(1-z^3)+\frac{\pi}{8}(z-z^5)\right)dz\)
\(\displaystyle =\frac{1}{2}+\frac{\pi}{24}\)
\(\Omega=\{(x,y,z)|0\leqq z\leqq 1,z\leqq \sqrt{x^2+y^2}\leqq 1,x\geqq0\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta)|z\leqq r\leqq 1,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \int_0^1\iint_\Omega(x+y^2z)dxdydz\)
\(\displaystyle =\int_0^1\iint_K(r\cos\theta+zr^2\sin^2\theta・rdrd\theta dz\)
\(\displaystyle =\int_0^1\int_z^1\int_{-\frac{\pi}{2}}^\frac{\pi}{2}(r^2\cos\theta+zr^3\sin^2\theta)d\theta drdz\)
\(\displaystyle =\int_0^1\int_z^1\left[2r^2\sin\theta+zr^3\left(\theta-\frac{\sin2\theta}{2}\right)\right]_0^\frac{\pi}{2}drdz\)
\(\displaystyle =\int_0^1\int_z^1\left(2r^2+\frac{\pi z}{2}r^3\right)drdz\)
\(\displaystyle =\int_0^1\left(\frac{2}{3}(1-z^3)+\frac{\pi}{8}(z-z^5)\right)dz\)
\(\displaystyle =\frac{1}{2}+\frac{\pi}{24}\)
(2)\(\displaystyle \iiint_\Omega zdxdydz\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,x^2+y^2\leqq ax,z\geqq0\}\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,x^2+y^2\leqq ax,z\geqq0\}\)
積分領域\(\Omega\)は
\(\Omega=\{(x,y,z)|0\leqq z\leqq \sqrt{a^2-x^2-y^2},x^2+y^2\geqq ax\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta)|0\leqq r\leqq a\cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \iiint_\Omega zdxdydz\)
\(\displaystyle =\iint_K\frac{a^2-r^2}{2}・rdrd\theta\)
\(\displaystyle =\frac{1}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^{a\cos\theta}(a^2r-r^3)drd\theta\)
\(\displaystyle =\frac{1}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left[\frac{a^2r^2}{2}-\frac{r^4}{4}\right]_0^{a\cos\theta}d\theta\)
\(\displaystyle =\frac{a^4}{4}\int_0^\frac{\pi}{2}(2\cos^2\theta-\cos^4\theta)d\theta\)
\(\displaystyle =\frac{5}{64}\pi a^4\)
\(\Omega=\{(x,y,z)|0\leqq z\leqq \sqrt{a^2-x^2-y^2},x^2+y^2\geqq ax\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta)|0\leqq r\leqq a\cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \iiint_\Omega zdxdydz\)
\(\displaystyle =\iint_K\frac{a^2-r^2}{2}・rdrd\theta\)
\(\displaystyle =\frac{1}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^{a\cos\theta}(a^2r-r^3)drd\theta\)
\(\displaystyle =\frac{1}{2}\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left[\frac{a^2r^2}{2}-\frac{r^4}{4}\right]_0^{a\cos\theta}d\theta\)
\(\displaystyle =\frac{a^4}{4}\int_0^\frac{\pi}{2}(2\cos^2\theta-\cos^4\theta)d\theta\)
\(\displaystyle =\frac{5}{64}\pi a^4\)
(3)\(\displaystyle \iiint_\Omega zdxdydz\)
\(\Omega=\{(x,y,z)|x^2+y^2\leqq z^2,x^2+y^2+z^2\leqq 1,z\geqq0\}\)
\(\Omega=\{(x,y,z)|x^2+y^2\leqq z^2,x^2+y^2+z^2\leqq 1,z\geqq0\}\)
積分領域\(\Omega\)は
\(\Omega=\{(x,y,z)|0\leqq z\leqq 1,x^2+y^2\geqq z^2,x^2+y^2\geqq 1-z^2\}\)
よって
\(\displaystyle \iiint_\Omega zdxdydz\)
\(\displaystyle =\int_0^1(\iint_Dzdxdy)dz\)
\(\displaystyle =\int_0^\frac{1}{\sqrt{2}}\pi z^3dz+\int_{1}{\sqrt{2}}^1\pi z(1-z^2)dz\)
\(\displaystyle =\left[\frac{\pi z^4}{4}\right]_0^\frac{1}{\sqrt{2}}+\left[\pi\left(\frac{z^2}{2}-\frac{z^4}{4}\right)\right]_\frac{1}{\sqrt{2}}^1\)
\(\displaystyle =\frac{\pi}{16}+\frac{\pi}{16}\)
\(\displaystyle =\frac{\pi}{8}\)
\(\Omega=\{(x,y,z)|0\leqq z\leqq 1,x^2+y^2\geqq z^2,x^2+y^2\geqq 1-z^2\}\)
よって
\(\displaystyle \iiint_\Omega zdxdydz\)
\(\displaystyle =\int_0^1(\iint_Dzdxdy)dz\)
\(\displaystyle =\int_0^\frac{1}{\sqrt{2}}\pi z^3dz+\int_{1}{\sqrt{2}}^1\pi z(1-z^2)dz\)
\(\displaystyle =\left[\frac{\pi z^4}{4}\right]_0^\frac{1}{\sqrt{2}}+\left[\pi\left(\frac{z^2}{2}-\frac{z^4}{4}\right)\right]_\frac{1}{\sqrt{2}}^1\)
\(\displaystyle =\frac{\pi}{16}+\frac{\pi}{16}\)
\(\displaystyle =\frac{\pi}{8}\)
(4)\(\displaystyle \iiint_\Omega xyzdxdydz\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,x\geqq0,y\geqq0,z\geqq0\}\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,x\geqq0,y\geqq0,z\geqq0\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq a,0\leqq \theta\leqq \frac{\pi}{2},0\leqq \varphi\leqq \frac{\pi}{2}\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega xyzdxdydz\)
\(\displaystyle =\iiint_K(r\sin\theta\cos\varphi)(r\sin\theta\sin\varphi)r\cos\theta・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\iiint_Kr^5\sin^3\theta\cos\theta\sin\varphi\cos\varphi drd\theta d\varphi\)
\(\displaystyle =\int_0^ar^5dr\int_0^\frac{\pi}{2}\sin^3\theta\cos\theta d\theta\int_0^\frac{\pi}{2}\sin\varphi\cos\varphi d\varphi\)
\(\displaystyle =\left[\frac{r^6}{6}\right]_0^a\left[\frac{\sin^4\theta}{4}\right]_0^\frac{\pi}{2}\left[\frac{\sin^2\varphi}{2}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{a^6}{6}・\frac{1}{4}・\frac{1}{2}\)
\(\displaystyle =\frac{a^6}{48}\)
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq a,0\leqq \theta\leqq \frac{\pi}{2},0\leqq \varphi\leqq \frac{\pi}{2}\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega xyzdxdydz\)
\(\displaystyle =\iiint_K(r\sin\theta\cos\varphi)(r\sin\theta\sin\varphi)r\cos\theta・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\iiint_Kr^5\sin^3\theta\cos\theta\sin\varphi\cos\varphi drd\theta d\varphi\)
\(\displaystyle =\int_0^ar^5dr\int_0^\frac{\pi}{2}\sin^3\theta\cos\theta d\theta\int_0^\frac{\pi}{2}\sin\varphi\cos\varphi d\varphi\)
\(\displaystyle =\left[\frac{r^6}{6}\right]_0^a\left[\frac{\sin^4\theta}{4}\right]_0^\frac{\pi}{2}\left[\frac{\sin^2\varphi}{2}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{a^6}{6}・\frac{1}{4}・\frac{1}{2}\)
\(\displaystyle =\frac{a^6}{48}\)
(5)\(\displaystyle \iiint_\Omega\tan^{-1}\sqrt{x^2+y^2+z^2}dxdydz\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq 1\}\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq 1\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq 1,0\leqq \theta\leqq \pi,0\leqq \varphi\leqq 2\pi\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega\tan^{-1}\sqrt{x^2+y^2+z^2}dxdydz\)
\(\displaystyle =\iiint_K\tan^{-1}r・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\int_0^{2\pi}d\varphi\int_0^\pi\sin\theta d\theta\int_0^1r^2\tan^{-1}rdr\)
\(\displaystyle =2\pi・2\left(\left[\frac{r^3}{3}\tan^{-1}r\right]_0^1-\frac{1}{3}\int_0^1r^3・\frac{1}{1+r^2}dr\right)\)
\(\displaystyle =4\pi\left(\frac{\pi}{12}-\frac{1}{3}\left[\frac{r^2}{2}-\frac{1}{2}\log(1+r^2)\right]_0^1\right)\)
\(\displaystyle =\frac{\pi(\pi-2+2\log2)}{3}\)
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq 1,0\leqq \theta\leqq \pi,0\leqq \varphi\leqq 2\pi\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega\tan^{-1}\sqrt{x^2+y^2+z^2}dxdydz\)
\(\displaystyle =\iiint_K\tan^{-1}r・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\int_0^{2\pi}d\varphi\int_0^\pi\sin\theta d\theta\int_0^1r^2\tan^{-1}rdr\)
\(\displaystyle =2\pi・2\left(\left[\frac{r^3}{3}\tan^{-1}r\right]_0^1-\frac{1}{3}\int_0^1r^3・\frac{1}{1+r^2}dr\right)\)
\(\displaystyle =4\pi\left(\frac{\pi}{12}-\frac{1}{3}\left[\frac{r^2}{2}-\frac{1}{2}\log(1+r^2)\right]_0^1\right)\)
\(\displaystyle =\frac{\pi(\pi-2+2\log2)}{3}\)
(6)\(\displaystyle \iiint_\Omega\cos\frac{\pi}{4}(x^2+y^2+z^2)^\frac{3}{2}dxdydz\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq 1\}\)
\(\Omega=\{(x,y,z)|x^2+y^2+z^2\leqq 1\}\)
\(\Omega\)を極座標変換すると
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq 1,0\leqq \theta\leqq \pi,0\leqq \varphi\leqq 2\pi\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega\cos\frac{\pi}{4}(x^2+y^2+z^2)^\frac{3}{2}dxdydz\)
\(\displaystyle =\iiint_K\cos\frac{\pi r^3}{4}・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\int_0^{2\pi}d\varphi\int_0^\pi\sin\theta d\theta\int_0^1r^2\cos\frac{\pi r^3}{4}dr\)
\(\displaystyle =2\pi・2\left[\frac{4}{3\pi}\sin\frac{\pi r^3}{4}\right]_0^1\)
\(\displaystyle =\frac{8\sqrt{2}}{3}\)
\(\displaystyle K=\left\{(r,\theta,\varphi)|0\leqq r\leqq 1,0\leqq \theta\leqq \pi,0\leqq \varphi\leqq 2\pi\right\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y,z)}{\partial(r,\theta,\varphi)}=r^2\sin\theta\)
よって
\(\displaystyle \iiint_\Omega\cos\frac{\pi}{4}(x^2+y^2+z^2)^\frac{3}{2}dxdydz\)
\(\displaystyle =\iiint_K\cos\frac{\pi r^3}{4}・r^2\sin\theta drd\theta d\varphi\)
\(\displaystyle =\int_0^{2\pi}d\varphi\int_0^\pi\sin\theta d\theta\int_0^1r^2\cos\frac{\pi r^3}{4}dr\)
\(\displaystyle =2\pi・2\left[\frac{4}{3\pi}\sin\frac{\pi r^3}{4}\right]_0^1\)
\(\displaystyle =\frac{8\sqrt{2}}{3}\)
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