【微分積分】8-2-2 円の中心が原点の極座標変換｜問題集
1.次の2重積分を求めなさい。
(1)\(\displaystyle \iint_D(x^2+y^2)dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq 4\}\)
\(D=\{(x,y)|x^2+y^2\leqq 4\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D(x^2+y^2)dxdy\)
\(\displaystyle =\iint_Er^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_0^2r^3dr\)
\(\displaystyle =2\pi\left[\frac{r^4}{4}\right]_0^2\)
\(=8\pi\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D(x^2+y^2)dxdy\)
\(\displaystyle =\iint_Er^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_0^2r^3dr\)
\(\displaystyle =2\pi\left[\frac{r^4}{4}\right]_0^2\)
\(=8\pi\)
(2)\(\displaystyle \iint_D\frac{1}{x^2+y^2}dxdy\)
\(D=\{(x,y)|1\leqq x^2+y^2\leqq 4,y\geqq 0\}\)
\(D=\{(x,y)|1\leqq x^2+y^2\leqq 4,y\geqq 0\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|1\leqq r\leqq 2,0\leqq \theta\leqq \pi\}\)
よって
\(\displaystyle \iint_D\frac{1}{x^2+y^2}dxdy\)
\(\displaystyle =\iint_E\frac{1}{r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{\pi}d\theta\int_1^2\frac{1}{r}dr\)
\(\displaystyle =\pi[\log r]_1^2\)
\(=\pi\log2\)
\(\displaystyle E=\{(r,\theta)|1\leqq r\leqq 2,0\leqq \theta\leqq \pi\}\)
よって
\(\displaystyle \iint_D\frac{1}{x^2+y^2}dxdy\)
\(\displaystyle =\iint_E\frac{1}{r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{\pi}d\theta\int_1^2\frac{1}{r}dr\)
\(\displaystyle =\pi[\log r]_1^2\)
\(=\pi\log2\)
(3)\(\displaystyle \iint_Dy^2dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq 1\}\)
\(D=\{(x,y)|x^2+y^2\leqq 1\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_Dy^2dxdy\)
\(\displaystyle =\iint_Er^2\sin^2\theta・rdrd\theta\)
\(\displaystyle =\int_0^1r^3dr\int_0^{2\pi}d\theta\)
\(\displaystyle =\frac{1}{4}\left[\frac{\theta}{2}-\frac{\sin2\theta}{4}\right]_0^{2\pi}\)
\(\displaystyle =\frac{\pi}{4}\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_Dy^2dxdy\)
\(\displaystyle =\iint_Er^2\sin^2\theta・rdrd\theta\)
\(\displaystyle =\int_0^1r^3dr\int_0^{2\pi}d\theta\)
\(\displaystyle =\frac{1}{4}\left[\frac{\theta}{2}-\frac{\sin2\theta}{4}\right]_0^{2\pi}\)
\(\displaystyle =\frac{\pi}{4}\)
(4)\(\displaystyle \iint_Dx^2dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq 4\}\)
\(D=\{(x,y)|x^2+y^2\leqq 4\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_Dx^2dxdy\)
\(\displaystyle =\iint_Er^2\cos^2\theta・rdrd\theta\)
\(\displaystyle =\int_0^2r^3dr\int_0^{2\pi}\cos^2\theta d\theta\)
\(\displaystyle =4・\frac{1}{2}\left[\theta+\frac{\sin2\theta}{2}\right]_0^{2\pi}\)
\(=4\pi\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_Dx^2dxdy\)
\(\displaystyle =\iint_Er^2\cos^2\theta・rdrd\theta\)
\(\displaystyle =\int_0^2r^3dr\int_0^{2\pi}\cos^2\theta d\theta\)
\(\displaystyle =4・\frac{1}{2}\left[\theta+\frac{\sin2\theta}{2}\right]_0^{2\pi}\)
\(=4\pi\)
(5)\(\displaystyle \iint_D\log(x^2+y^2)dxdy\)
\(D=\{(x,y)|1\leqq x^2+y^2\leqq 4\}\)
\(D=\{(x,y)|1\leqq x^2+y^2\leqq 4\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|1\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D\log(x^2+y^2)dxdy\)
\(\displaystyle =\iint_E\log r^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_1^2r\log r^2dr\)
\(\displaystyle =2\pi\left[r^2\log r-\frac{r^2}{2}\right]_1^2\)
\(=8\pi\log2-3\pi\)
\(\displaystyle E=\{(r,\theta)|1\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D\log(x^2+y^2)dxdy\)
\(\displaystyle =\iint_E\log r^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_1^2r\log r^2dr\)
\(\displaystyle =2\pi\left[r^2\log r-\frac{r^2}{2}\right]_1^2\)
\(=8\pi\log2-3\pi\)
(6)\(\displaystyle \iint_De^{x^2+y^2}dxdy\)
\(D=\{(x,y)|1< x^2+y^2< 4\}\)
\(D=\{(x,y)|1< x^2+y^2< 4\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|1< r< 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_De^{x^2+y^2}dxdy\)
\(\displaystyle =\iint_Ee^{r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_1^2e^{r^2}rdr\)
\(\displaystyle =2\pi\left[\frac{1}{2}e^{r^2}\right]_1^2\)
\(=\pi(e^4-e)\)
\(\displaystyle E=\{(r,\theta)|1< r< 2,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_De^{x^2+y^2}dxdy\)
\(\displaystyle =\iint_Ee^{r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_1^2e^{r^2}rdr\)
\(\displaystyle =2\pi\left[\frac{1}{2}e^{r^2}\right]_1^2\)
\(=\pi(e^4-e)\)
(7)\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq 1\}\)
\(D=\{(x,y)|x^2+y^2\leqq 1\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_0^1\sqrt{1-r^2}rdr\)
\(\displaystyle =2\pi\left[-\frac{1}{3}(1-r^2)^\frac{3}{2}\right]_0^1\)
\(\displaystyle =\frac{2\pi}{3}\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって
\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}d\theta\int_0^1\sqrt{1-r^2}rdr\)
\(\displaystyle =2\pi\left[-\frac{1}{3}(1-r^2)^\frac{3}{2}\right]_0^1\)
\(\displaystyle =\frac{2\pi}{3}\)
(8)\(\displaystyle \iint_D(1-x-2y)dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq 1,x\geqq 0,y\geqq 0\}\)
\(D=\{(x,y)|x^2+y^2\leqq 1,x\geqq 0,y\geqq 0\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq \frac{\pi}{2}\}\)
よって
\(\displaystyle \iint_D(1-x-2y)dxdy\)
\(\displaystyle =\iint_E(1-r\cos\theta-2r\sin\theta)・rdrd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^1(r-r^2\cos\theta-2r^2\sin\theta)drd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\left[\frac{r^2}{2}-\frac{r^3\cos\theta}{3}-\frac{2r^3\sin\theta}{3}\right]_0^1d\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\left(\frac{1}{2}-\frac{\cos\theta}{3}-\frac{2\sin\theta}{3}\right)d\theta\)
\(\displaystyle =\left[\frac{\theta}{2}-\frac{\sin\theta}{3}+\frac{2\cos\theta}{3}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{4}-1\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq \frac{\pi}{2}\}\)
よって
\(\displaystyle \iint_D(1-x-2y)dxdy\)
\(\displaystyle =\iint_E(1-r\cos\theta-2r\sin\theta)・rdrd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^1(r-r^2\cos\theta-2r^2\sin\theta)drd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\left[\frac{r^2}{2}-\frac{r^3\cos\theta}{3}-\frac{2r^3\sin\theta}{3}\right]_0^1d\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\left(\frac{1}{2}-\frac{\cos\theta}{3}-\frac{2\sin\theta}{3}\right)d\theta\)
\(\displaystyle =\left[\frac{\theta}{2}-\frac{\sin\theta}{3}+\frac{2\cos\theta}{3}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{4}-1\)
次の学習に進もう！