【微分積分】8-5-2 曲面積｜問題集

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

積分領域は
\(\displaystyle D=\{(x,y)|x+y\leqq 2,x\geqq0,y\geqq0\}\)
よって、求める曲面積は
\(\displaystyle S=\iint_D\sqrt{3}dxdy\)
\(\displaystyle =\sqrt{3}\int_0^2\int_0^{2-x}dydx\)
\(\displaystyle =\sqrt{3}\int_0^2(2-x)dx\)
\(\displaystyle =2\sqrt{3}\)

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

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

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

積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq a^2\}\)
よって、求める曲面積は
\(\displaystyle S=2\iint_D\frac{a}{\sqrt{a^2-(x^2+y^2)}}dxdy\)
\(\displaystyle =2\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\frac{a}{\sqrt{a^2-x^2}}dydx\)
\(\displaystyle =4\int_{-a}^aadx\)
\(\displaystyle =8a^2\)