【微分積分】8-5-2 曲面積｜要点まとめ

積分領域は
\(D=\{(x,y)|x^2+y^2\leqq 2\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq \sqrt{2},0\leqq \theta\leqq 2\pi\}\)
よって、求める曲面積は
\(\displaystyle S=\iint_D\sqrt{1+4x^2+4y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1+4r^2}・rdrd\theta\)
\(\displaystyle =\int_0^\sqrt{2}r\sqrt{1+4r^2}dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\pi\left[\frac{1}{12}(1+4r^2)^\frac{3}{2}\right]_0^\sqrt{2}\)
\(\displaystyle =\frac{26}{3}\pi\)

積分領域は
\(D=\{(x,y)|x^2+y^2\leqq 8\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2\sqrt{2},0\leqq \theta\leqq 2\pi\}\)
よって、求める曲面積は
\(\displaystyle S=\iint_D\frac{3}{\sqrt{9-x^2-y^2}}dxdy\)
\(\displaystyle =\iint_E\frac{3}{\sqrt{9-r^2}}・rdrd\theta\)
\(\displaystyle =3\int_0^{2\sqrt{2}}\frac{r}{\sqrt{9-r^2}}dr\int_0^{2\pi}d\theta\)
\(\displaystyle =6\pi[-\sqrt{9-r^2}]_0^{2\sqrt{2}}\)
\(\displaystyle =12\pi\)

積分領域は
\(D=\{(x,y)|x^2+y^2\leqq 2x\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq 2\cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって、求める曲面積は
\(\displaystyle S=\iint_D\frac{2}{\sqrt{4-x^2-y^2}}dxdy\)
\(\displaystyle =\iint_E\frac{2}{\sqrt{4-r^2}}・rdrd\theta\)
\(\displaystyle =2\int_{-\frac{\pi}{2}}^\frac{\pi}{2}\int_0^{2\cos\theta}\frac{r}{\sqrt{4-r^2}}drd\theta\)
\(\displaystyle =8\int_0^\frac{\pi}{2}(1-\sin\theta)d\theta\)
\(\displaystyle =8[\theta+\cos\theta]_0^\frac{\pi}{2}\)
\(\displaystyle =4\pi-8\)

\(D\)上の回転放物面\(z=x^2+y^2\)の曲面積を\(S_1\)、半球面\(z=\sqrt{2-x^2-y^2}\)の曲面積を\(S_2\)とおく。
\(\displaystyle S_1=\iint_D\sqrt{1+4x^2+4y^2}dxdy\)
\(\displaystyle =\iint_E\sqrt{1+4r^2}・rdrd\theta\)
\(\displaystyle =\int_0^1r\sqrt{1+4r^2}dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\pi\left[\frac{1}{12}(1+4r^2)^\frac{3}{2}\right]_0^1\)
\(\displaystyle =\frac{5\sqrt{5}-1}{6}\pi\)

\(\displaystyle S_2=\iint_D\frac{\sqrt{2}}{\sqrt{2-x^2-y^2}}dxdy\)
\(\displaystyle =\iint_E\frac{\sqrt{2}}{\sqrt{2-r^2}}・rdrd\theta\)
\(\displaystyle =\sqrt{2}\int_0^1\frac{r}{\sqrt{2-r^2}}dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\sqrt{2}\pi[-\sqrt{2-r^2}]_0^1\)
\(\displaystyle =(4-2\sqrt{2})\pi\)

よって、求める表面積は
\(\displaystyle S=S_1+S_2\)
\(\displaystyle =\frac{23+5\sqrt{5}-12\sqrt{2}}{6}\pi\)

積分領域は
\(\displaystyle D=\{(x,y)|(x^2+y^2)^2\leqq a^2(x^2-y^2)\}\)
\(x\)軸\(y\)軸に関して対称なので、4倍すればよい。
\(\displaystyle D'=\{(x,y)|(x^2+y^2)^2\leqq a^2(x^2-y^2),x\geqq0,y\geqq0\}\)
\(D'\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq a\sqrt{\cos2\theta},0\leqq \theta\leqq \frac{\pi}{4}\right\}\)
よって、求める表面積は
\(\displaystyle S=\iint_{D'}\frac{\sqrt{a^2+x^2+y^2}}{a}dxdy\)
\(\displaystyle =4\iint_E\frac{\sqrt{a^2+r^2}}{a}・rdrd\theta\)
\(\displaystyle =\frac{4}{a}\int_0^\frac{\pi}{4}\int_0^{a\sqrt{\cos2\theta}}r\sqrt{a^2+r^2}drd\theta\)
\(\displaystyle =\frac{4a^2}{3}\int_0^\frac{\pi}{4}(2\sqrt{2}\cos^3\theta-1)d\theta\)
\(\displaystyle =\frac{8\sqrt{2}a^2}{3}・\frac{5}{6\sqrt{2}}-\frac{a^2}{3}\pi\)
\(\displaystyle =\frac{20-3\pi}{9}a^2\)