【微分積分】8-2-3 円の中心が原点以外の極座標変換|要点まとめ
このページでは、大学数学の2重積分で頻出となる「円の中心が原点以外の極座標変換」について解説します。中心が原点からずれた円領域における変数変換(極座標表示)の考え方から、具体的な2重積分の計算例題・解答までを分かりやすくまとめています。
極座標変換
【例題】次の2重積分を求めなさい。
(1)\(\displaystyle \iint_Dxydxdy\)
\(D=\{(x,y)|x^2+y^2\leqq x,y\geqq 0\}\)
\(D=\{(x,y)|x^2+y^2\leqq x,y\geqq 0\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq \cos\theta,0\leqq \theta\leqq \frac{\pi}{2}\}\)
よって
\(\displaystyle \iint_Dxydxdy\)
\(\displaystyle =\iint_Er^2\cos\theta\sin\theta・rdrd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^{\cos\theta}r^3\cos\theta\sin\theta drd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\frac{1}{4}\cos^5\theta\sin\theta d\theta\)
\(\displaystyle =\left[-\frac{\cos^6\theta}{24}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{1}{24}\)
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq \cos\theta,0\leqq \theta\leqq \frac{\pi}{2}\}\)
よって
\(\displaystyle \iint_Dxydxdy\)
\(\displaystyle =\iint_Er^2\cos\theta\sin\theta・rdrd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^{\cos\theta}r^3\cos\theta\sin\theta drd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\frac{1}{4}\cos^5\theta\sin\theta d\theta\)
\(\displaystyle =\left[-\frac{\cos^6\theta}{24}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{1}{24}\)
(2)\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(D=\{(x,y)|x^2+y^2\leqq x\}\)
\(D=\{(x,y)|x^2+y^2\leqq x\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq \cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^{\cos\theta}\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left[-\frac{1}{3}(1-r^2)^\frac{3}{2}\right]_0^{\cos\theta}d\theta\)
\(\displaystyle =\frac{2}{3}\int_0^\frac{\pi}{2}(1-\sin^3\theta)d\theta\)
\(\displaystyle =\frac{2}{3}\left[\theta+\cos\theta-\frac{\cos^3\theta}{3}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{3}-\frac{4}{9}\)
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq \cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって
\(\displaystyle \iint_D\sqrt{1-x^2-y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^{\cos\theta}\sqrt{1-r^2}・rdrd\theta\)
\(\displaystyle =\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\left[-\frac{1}{3}(1-r^2)^\frac{3}{2}\right]_0^{\cos\theta}d\theta\)
\(\displaystyle =\frac{2}{3}\int_0^\frac{\pi}{2}(1-\sin^3\theta)d\theta\)
\(\displaystyle =\frac{2}{3}\left[\theta+\cos\theta-\frac{\cos^3\theta}{3}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{\pi}{3}-\frac{4}{9}\)
次の学習に進もう!