【微分積分】8-5-1 体積｜問題集

\(z=x^2+y^2\)が\(z=2y\)より下にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|-1\leqq x\leqq 1,1-\sqrt{1-x^2}\leqq y\leqq 1+\sqrt{1-x^2}\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2\sin\theta,0\leqq \theta\leqq \pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_D\{2y-(x^2+y^2)\}dxdy\)
\(\displaystyle =\iint_E(2r\sin\theta-r^2)・rdrd\theta\)
\(\displaystyle =2\int_0^\frac{\pi}{2}\int_0^{2\sin\theta}(2r^2\sin\theta-r^3)drd\theta\)
\(\displaystyle =2\int_0^\frac{\pi}{2}\left(\frac{16\sin^4\theta}{3}-\frac{16\sin^4\theta}{4}\right)d\theta\)
\(\displaystyle =\frac{8}{3}\int_0^\frac{\pi}{2}\sin^4\theta d\theta\)
\(\displaystyle =\frac{\pi}{2}\)

\(z=\sqrt{1-x^2-y^2}\)が\(z=-\sqrt{1-x^2-y^2}\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|-1\leqq x\leqq 1,-\sqrt{1-x^2}\leqq y\leqq \sqrt{1-x^2}\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(x^2+y^2)dxdy\)
\(\displaystyle =\iint_Er^2・rdrd\theta\)
\(\displaystyle =\int_0^1r^3dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\pi\left[\frac{1}{4}\theta\right]_0^{2\pi}\)
\(\displaystyle =\frac{\pi}{2}\)

\(x^2+y^2\leqq a^2\)が\(z=x,z=0\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq a^2,x\geqq0\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq a,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(2-y)dxdy\)
\(\displaystyle =\iint_Er\cos\theta・rdrd\theta\)
\(\displaystyle =2\int_0^ar^2dr\int_0^\frac{\pi}{2}\cos\theta d\theta\)
\(\displaystyle =2\left[\frac{r^3}{3}\right]_0^a[\sin\theta]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{2a^3}{3}\)

積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1-y^2,0\leqq y\leqq 1\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(1-x^2)dxdy\)
\(\displaystyle =\int_0^1\int_0^{1-y^2}(1-x^2)dxdy\)
\(\displaystyle =\int_0^1\left(\frac{2}{3}-y^4+\frac{1}{3}y^6\right)dy\)
\(\displaystyle =\frac{18}{35}\)

\(z=\sqrt{a^2-(x^2+y^2)}\)が\(z=-\sqrt{a^2-(x^2+y^2)}\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq ax\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|r\leqq a\cos\theta,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\}\)
よって、求める体積は
\(\displaystyle V=2\iint_D\sqrt{a^2-(x^2+y^2)}dxdy\)
\(\displaystyle =2\iint_E\sqrt{a^2-r^2}・rdrd\theta\)
\(\displaystyle =4\int_0^\frac{\pi}{2}\int_0^{a\cos\theta}\sqrt{a^2-r^2}rdrd\theta\)
\(\displaystyle =-\frac{4}{3}\int_0^\frac{\pi}{2}(a^3\sin^3\theta-a^3)d\theta\)
\(\displaystyle =-\frac{4a^3}{3}\left(\frac{2}{3}-\frac{\pi}{2}\right)\)

\(z=1-\sqrt{x^2+y^2}\)が\(z=x\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1-\sqrt{x^2+y^2}\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq \frac{1}{1+\cos\theta},-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(1-\sqrt{x^2+y^2}-x)dxdy\)
\(\displaystyle =\iint_E(1-r-r\cos\theta)・rdrd\theta\)
\(\displaystyle =2\int_0^\frac{\pi}{2}\int_0^\frac{1}{1+\cos\theta}\{r-r^2(1\cos\theta)\}drd\theta\)
\(\displaystyle =\frac{1}{3}\int_0^\frac{\pi}{2}\frac{1}{4\cos^4\theta}d\theta\)
\(\displaystyle =\frac{1}{12}\int_0^\frac{\pi}{2}\left(1+\tan^2\frac{\theta}{2}\right)\sec^2\frac{\theta}{2}d\theta\)
\(\displaystyle =\frac{1}{6}\left[t+\frac{t^3}{3}\right]_0^1\)
\(\displaystyle =\frac{2}{9}\)