【微分積分】8-5-1 体積｜要点まとめ

\(z=x^2+y^2\)が\(z=2x+y+1\)より下にあるので、
\(x^2+y^2\leqq 2x+y+1\)
\(\displaystyle (x-1)^2+\left(y-\frac{1}{2}\right)^2\leqq \frac{9}{4}\)
積分領域は
\(\displaystyle D=\left\{(x,y)|(x-1)^2+\left(y-\frac{1}{2}\right)^2\leqq \frac{9}{4}\right\}\)
\(D\)を\(\displaystyle x=1+r\cos\theta,y=\frac{1}{2}+r\sin\theta\)で極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq \frac{3}{2},0\leqq \theta\leqq 2\pi\right\}\)
よって、求める体積は
\(\displaystyle V=\iint_D\{(2x+y+1)-(x^2+y^2)\}dxdy\)
\(\displaystyle =\iint_E\left(\frac{9}{4}-r^2\right)・rdrd\theta\)
\(\displaystyle =\int_0^\frac{3}{2}\left(\frac{9}{4}r-r^3\right)dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\pi\left[\frac{9}{8}r^2-\frac{r^4}{4}\right]_0^\frac{3}{2}\)
\(\displaystyle =\frac{81}{32}\pi\)

\(x^2+y^2\leqq4\)が\(z=2-y,z=0\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq4\}\)
\(D\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(2-y)dxdy\)
\(\displaystyle =\iint_E(2-r\sin\theta)・rdrd\theta\)
\(\displaystyle =\int_0^2\int_0^{2\pi}(2r-r^2\sin\theta)d\theta dr\)
\(\displaystyle =2\pi\int_0^22rdr\)
\(\displaystyle =8\pi\)

\(z=\sqrt{3-x^2-y^2}\)が\(z=-\sqrt{3-x^2-y^2}\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq1\}\)
\(D\)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
\(\displaystyle V=\iint_D\{\sqrt{3-x^2-y^2}-(-\sqrt{3-x^2-y^2})\}dxdy\)
\(\displaystyle =\iint_E2\sqrt{3-r^2}・rdrd\theta\)
\(\displaystyle =2\int_0^1r\sqrt{3-r^2}dr\int_0^{2\pi}d\theta\)
\(\displaystyle =4\pi\left[-\frac{1}{3}(3-r^2)^\frac{3}{2}\right]_0^1\)
\(\displaystyle =\frac{4\pi}{3}(3\sqrt{3}-2\sqrt{2})\)

\((x-3)^2+(y-2)^2\leqq4\)が\(z=xy,z=0\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|(x-3)^2+(y-2)^2\leqq4\}\)
\(D\)を\(\displaystyle x=3+r\cos\theta,y=2+r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_Dxydxdy\)
\(\displaystyle =\iint_E(3+r\cos\theta)(2+r\sin\theta)・rdrd\theta\)
\(\displaystyle =\int_0^2\int_0^{2\pi}\left(6r+2r^2\cos\theta+3r^2\sin\theta+\frac{r^3}{2}\sin2\theta\right)d\theta dr\)
\(\displaystyle =2\pi\int_0^26rdr\)
\(\displaystyle =24\pi\)

\(z=x^2-y^2\)が\(z=18-2(x-3)^2-4y^2\)より下にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|(x-2)^2+y^2\leqq4\}\)
\(D\)を\(\displaystyle x=2+r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\{(r,\theta)|0\leqq r\leqq 2,0\leqq \theta\leqq 2\pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_D\{18-2(x-3)^2-4y^2-(x^2-y^2)\}dxdy\)
\(\displaystyle =\iint_E3(4-r^2)・rdrd\theta\)
\(\displaystyle =3\int_0^2(4r-r^3)dr\int_0^{2\pi}d\theta\)
\(\displaystyle =6\pi\left[2r^2-\frac{r^4}{4}\right]_0^2\)
\(\displaystyle =24\pi\)

\(z=\sqrt{a^2-x^2}\)が\(z=-\sqrt{a^2-x^2}\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq a^2\}\)
よって、求める体積は
\(\displaystyle V=\iint_D\{\sqrt{a^2-x^2}-(-\sqrt{a^2-x^2})\}dxdy\)
\(\displaystyle =\int_{-a}^a\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}2\sqrt{a^2-x^2}dydx\)
\(\displaystyle =\int_{-a}^a4(a^2-x^2)dx\)
\(\displaystyle =\frac{16}{3}a^3\)

積分領域は
\(\displaystyle D=\{(x,y)|(x^2+y^2)^2\leqq2a^2xy\}\)
\(D\)上で\(z=xy,z=0\)より上にあるので、
積分領域\(D'\)は
\(\displaystyle D'=\{(x,y)|(x^2+y^2)^2\leqq2a^2xy,x\geqq0,y\geqq0\}\)
\(D'\)を\(\displaystyle x=r\cos\theta,y=r\sin\theta\)で極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|0\leqq r\leqq a\sqrt{\sin2\theta},0\leqq \theta\leqq \frac{\pi}{2}\right\}\)
よって、求める体積は
\(\displaystyle V=2\iint_{D'}xydxdy\)
\(\displaystyle =2\iint_Er^2\cos\theta\sin\theta・rdrd\theta\)
\(\displaystyle =\int_0^\frac{\pi}{2}\int_0^{a\sqrt{\sin2\theta}}r^3\sin2\theta drd\theta\)
\(\displaystyle =\frac{a^4}{16}\int_0^\frac{\pi}{2}(3\sin2\theta-\sin6\theta)d\theta\)
\(\displaystyle =\frac{a^4}{16}\left[-\frac{3}{2}\cos2\theta+\frac{1}{6}\cos6\theta\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{a^4}{6}\)

\(z=\sqrt{2-x^2-y^2}\)が\(z=x^2+y^2\)より上にあるので、
積分領域は
\(\displaystyle D=\{(x,y)|x^2+y^2\leqq1\}\)
\(D\)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
\(\displaystyle V=\iint_D\{\sqrt{2-x^2-y^2}-(x^2+y^2)\}dxdy\)
\(\displaystyle =\iint_E(\sqrt{2-r^2}-r^2)・rdrd\theta\)
\(\displaystyle =\int_0^1(r\sqrt{2-r^2}-r^3)dr\int_0^{2\pi}d\theta\)
\(\displaystyle =2\pi\left[-\frac{1}{3}(2-r^2)^\frac{3}{2}-\frac{r^4}{4}\right]_0^1\)
\(\displaystyle =\frac{8\sqrt{2}-7}{6}\pi\)