【微分積分】8-5-4 図形の重心と回転体の体積｜問題集
1.次の体積を求めなさい。
(1)曲面\(z=x+y\)の上に有界で3点\((0,0),(0,1),(1,0)\)を頂点とする三角面の下に有界な立体
積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq 1-x\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(x+y)dxdy\)
\(\displaystyle =\int_0^1\int_0^{1-x}(x+y)dydx\)
\(\displaystyle =\int_0^1\left(-\frac{x^2}{2}+\frac{1}{2}\right)dx\)
\(\displaystyle =\frac{1}{3}\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq 1-x\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(x+y)dxdy\)
\(\displaystyle =\int_0^1\int_0^{1-x}(x+y)dydx\)
\(\displaystyle =\int_0^1\left(-\frac{x^2}{2}+\frac{1}{2}\right)dx\)
\(\displaystyle =\frac{1}{3}\)
(2)曲面\(z=2x+3y\)の上に有界で4点\((0,0),(0,1),(1,0),(1,1)\)を頂点とする正方面の下に有界な立体
積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq 1\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(2x+3y)dxdy\)
\(\displaystyle =\int_0^1\int_0^1(2x+3y)dydx\)
\(\displaystyle =\int_0^1\left(2x+\frac{3}{2}\right)dx\)
\(\displaystyle =\frac{5}{2}\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 1,0\leqq y\leqq 1\}\)
よって、求める体積は
\(\displaystyle V=\iint_D(2x+3y)dxdy\)
\(\displaystyle =\int_0^1\int_0^1(2x+3y)dydx\)
\(\displaystyle =\int_0^1\left(2x+\frac{3}{2}\right)dx\)
\(\displaystyle =\frac{5}{2}\)
(3)曲面\(z=x^2+y^2\)の上に有界で単位円\(x^2+y^2\leqq1\)の下に有界な立体
\(\displaystyle D=\{(r,\theta)|0\leqq r\leqq 1,0\leqq \theta\leqq 2\pi\}\)
よって、求める体積は
\(\displaystyle V=\iint_Dr^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}\int_0^1r^3drd\theta\)
\(\displaystyle =\int_0^{2\pi}\frac{1}{4}d\theta\)
\(\displaystyle =\frac{\pi}{2}\)
よって、求める体積は
\(\displaystyle V=\iint_Dr^2・rdrd\theta\)
\(\displaystyle =\int_0^{2\pi}\int_0^1r^3drd\theta\)
\(\displaystyle =\int_0^{2\pi}\frac{1}{4}d\theta\)
\(\displaystyle =\frac{\pi}{2}\)
2.密度が一定であるとき、次の平面図形\(D\)の重心を求めなさい。
(1)\(y=x\)と\(y=x^2\)で囲まれた領域
\(D\)の面積は
\(\displaystyle |D|=\int_0^1\int_{x^2}^xdydx=\frac{1}{6}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =6\int_0^1\int_{x^2}^xxdydx\)
\(\displaystyle =6\int_0^1x(x-x^2)dx\)
\(\displaystyle =6・\frac{1}{12}\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =6\int_0^1\int_{x^2}^xydydx\)
\(\displaystyle =6\int_0^1\frac{1}{2}(x^2-x^4)dx\)
\(\displaystyle =6・\frac{1}{15}\)
\(\displaystyle =\frac{2}{5}\)
よって、重心の座標は\(\displaystyle \left(\frac{1}{2},\frac{2}{5}\right)\)
\(\displaystyle |D|=\int_0^1\int_{x^2}^xdydx=\frac{1}{6}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =6\int_0^1\int_{x^2}^xxdydx\)
\(\displaystyle =6\int_0^1x(x-x^2)dx\)
\(\displaystyle =6・\frac{1}{12}\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =6\int_0^1\int_{x^2}^xydydx\)
\(\displaystyle =6\int_0^1\frac{1}{2}(x^2-x^4)dx\)
\(\displaystyle =6・\frac{1}{15}\)
\(\displaystyle =\frac{2}{5}\)
よって、重心の座標は\(\displaystyle \left(\frac{1}{2},\frac{2}{5}\right)\)
(2)\(x^2=4y\)と\(x-2y+4=0\)で囲まれた領域
積分領域は
\(\displaystyle D=\left\{(x,y)|-2\leqq x\leqq 4,\frac{x^2}{4}\leqq y\leqq \frac{x}{2}+2\right\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}dydx=9\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}xdydx\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\left(\frac{x^2}{2}+2x-\frac{x^3}{4}\right)dx\)
\(\displaystyle =\frac{1}{9}・9\)
\(\displaystyle =1\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}ydydx\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\left(\frac{(x+4)^2}{4}-\frac{x^4}{16}\right)dx\)
\(\displaystyle =\frac{1}{9}・\frac{72}{5}\)
\(\displaystyle =\frac{8}{5}\)
よって、重心の座標は\(\displaystyle \left(1,\frac{8}{5}\right)\)
\(\displaystyle D=\left\{(x,y)|-2\leqq x\leqq 4,\frac{x^2}{4}\leqq y\leqq \frac{x}{2}+2\right\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}dydx=9\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}xdydx\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\left(\frac{x^2}{2}+2x-\frac{x^3}{4}\right)dx\)
\(\displaystyle =\frac{1}{9}・9\)
\(\displaystyle =1\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\int_\frac{x^2}{4}^{\frac{x}{2}+2}ydydx\)
\(\displaystyle =\frac{1}{9}\int_{-2}^4\left(\frac{(x+4)^2}{4}-\frac{x^4}{16}\right)dx\)
\(\displaystyle =\frac{1}{9}・\frac{72}{5}\)
\(\displaystyle =\frac{8}{5}\)
よって、重心の座標は\(\displaystyle \left(1,\frac{8}{5}\right)\)
(3)\(y=x^2-2x\)と\(y=6x-x^2\)で囲まれた領域
積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 4,x^2-2x\leqq y\leqq 6x-x^2\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_0^4\int_{x^2-2x}^{6x-x^2}dydx=\frac{64}{3}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{3}{64}\int_0^4\int_{x^2-2x}^{6x-x^2}xdydx\)
\(\displaystyle =\frac{3}{64}\int_0^4(8x^2-2x^3)dx\)
\(\displaystyle =\frac{3}{64}・\frac{128}{3}\)
\(\displaystyle =2\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{3}{64}\int_0^4\int_{x^2-2x}^{6x-x^2}ydydx\)
\(\displaystyle =\frac{3}{64}\int_0^4(16x^2-4x^3)dx\)
\(\displaystyle =\frac{3}{64}・\frac{256}{3}\)
\(\displaystyle =4\)
よって、重心の座標は\((2,4)\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 4,x^2-2x\leqq y\leqq 6x-x^2\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_0^4\int_{x^2-2x}^{6x-x^2}dydx=\frac{64}{3}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{3}{64}\int_0^4\int_{x^2-2x}^{6x-x^2}xdydx\)
\(\displaystyle =\frac{3}{64}\int_0^4(8x^2-2x^3)dx\)
\(\displaystyle =\frac{3}{64}・\frac{128}{3}\)
\(\displaystyle =2\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{3}{64}\int_0^4\int_{x^2-2x}^{6x-x^2}ydydx\)
\(\displaystyle =\frac{3}{64}\int_0^4(16x^2-4x^3)dx\)
\(\displaystyle =\frac{3}{64}・\frac{256}{3}\)
\(\displaystyle =4\)
よって、重心の座標は\((2,4)\)
(4)\(y=x\)と\(y=6x-x^2\)で囲まれた領域
積分領域は
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 5,x\leqq y\leqq 6x-x^2\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_0^5\int_x^{6x-x^2}dydx=\frac{125}{6}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{6}{125}\int_0^5\int_x^{6x-x^2}xdydx\)
\(\displaystyle =\frac{6}{125}\int_0^5x(5x-x^2)dx\)
\(\displaystyle =\frac{6}{125}・\frac{625}{12}\)
\(\displaystyle =\frac{5}{2}\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{6}{125}\int_0^5\int_x^{6x-x^2}ydydx\)
\(\displaystyle =\frac{6}{125}\int_0^5\left(\frac{1}{2}x^4-6x^3+\frac{35}{2}x^2\right)dx\)
\(\displaystyle =\frac{6}{125}・\frac{625}{6}\)
\(\displaystyle =5\)
よって、重心の座標は\(\displaystyle \left(\frac{5}{2},5\right)\)
\(\displaystyle D=\{(x,y)|0\leqq x\leqq 5,x\leqq y\leqq 6x-x^2\}\)
\(D\)の面積は
\(\displaystyle |D|=\int_0^5\int_x^{6x-x^2}dydx=\frac{125}{6}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{6}{125}\int_0^5\int_x^{6x-x^2}xdydx\)
\(\displaystyle =\frac{6}{125}\int_0^5x(5x-x^2)dx\)
\(\displaystyle =\frac{6}{125}・\frac{625}{12}\)
\(\displaystyle =\frac{5}{2}\)
\(\displaystyle y_G=\frac{1}{|D|}\iint_Dydxdy\)
\(\displaystyle =\frac{6}{125}\int_0^5\int_x^{6x-x^2}ydydx\)
\(\displaystyle =\frac{6}{125}\int_0^5\left(\frac{1}{2}x^4-6x^3+\frac{35}{2}x^2\right)dx\)
\(\displaystyle =\frac{6}{125}・\frac{625}{6}\)
\(\displaystyle =5\)
よって、重心の座標は\(\displaystyle \left(\frac{5}{2},5\right)\)
(5)\(ax\leqq x^2+y^2\leqq a^2\)の領域
\(D\)の面積は
\(\displaystyle |D|=\pi a^2-\pi\left(\frac{a}{2}\right)^2=\frac{3\pi a^2}{4}\)
図形の対称性より\(y_G=0\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|a\cos\theta\leqq r\leqq a,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
\(\displaystyle \ \ \ \ \ \cup\left\{(r,\theta)|0\leqq r\leqq a,\frac{\pi}{2}\leqq \theta\leqq \frac{3\pi}{2}\right\}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{8}{3\pi a^2}\left(\int_0^\frac{\pi}{2}\int_{a\cos\theta}^ar^2\cos\theta drd\theta+\int_\frac{\pi}{2}^\pi\int_0^ar^2\cos\theta drd\theta\right)\)
\(\displaystyle =\frac{8}{3\pi a^2}\left(\int_0^\frac{\pi}{2}\frac{a^3}{3}(\cos\theta-\cos^4\theta)d\theta+\int_\frac{\pi}{2}^\pi\frac{a^3\cos\theta}{3}d\theta\right)\)
\(\displaystyle =\frac{8}{3\pi a^2}・\left(-\frac{\pi a^3}{16}\right)\)
\(\displaystyle =-\frac{a}{6}\)
よって、重心の座標は\(\displaystyle \left(-\frac{a}{6},0\right)\)
\(\displaystyle |D|=\pi a^2-\pi\left(\frac{a}{2}\right)^2=\frac{3\pi a^2}{4}\)
図形の対称性より\(y_G=0\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|a\cos\theta\leqq r\leqq a,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
\(\displaystyle \ \ \ \ \ \cup\left\{(r,\theta)|0\leqq r\leqq a,\frac{\pi}{2}\leqq \theta\leqq \frac{3\pi}{2}\right\}\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dxdxdy\)
\(\displaystyle =\frac{8}{3\pi a^2}\left(\int_0^\frac{\pi}{2}\int_{a\cos\theta}^ar^2\cos\theta drd\theta+\int_\frac{\pi}{2}^\pi\int_0^ar^2\cos\theta drd\theta\right)\)
\(\displaystyle =\frac{8}{3\pi a^2}\left(\int_0^\frac{\pi}{2}\frac{a^3}{3}(\cos\theta-\cos^4\theta)d\theta+\int_\frac{\pi}{2}^\pi\frac{a^3\cos\theta}{3}d\theta\right)\)
\(\displaystyle =\frac{8}{3\pi a^2}・\left(-\frac{\pi a^3}{16}\right)\)
\(\displaystyle =-\frac{a}{6}\)
よって、重心の座標は\(\displaystyle \left(-\frac{a}{6},0\right)\)
3.密度が原点からの距離に比例するとき、次の重心を求めなさい。
(1)半球\(x^2+y^2+z^2\leqq a^2,z\geqq0\)
密度を\(\rho(x,y,z)=k\sqrt{x^2+y^2+z^2}\)とおく。
\(\displaystyle D=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,z\geqq0\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(\rho,\theta,\varphi)|\rho\leqq a,0\leqq \theta\leqq 2\pi,0\leqq \varphi\leqq \frac{\pi}{2}\right\}\)
\(D\)の質量は
\(\displaystyle |D|=\iiint_Ek\rho・\rho^2\sin\varphi d\rho d\theta d\varphi=\frac{k\pi a^4}{2}\)
図形の対称性より\(x_G=y_G=0\)
\(\displaystyle z_G=\frac{1}{|D|}\iiint_Ez・k\rho・\rho^2\sin\varphi d\rho d\theta d\varphi\)
\(\displaystyle =\frac{2}{k\pi a^4}・k\int_0^a\rho^4d\rho\int_0^{2\pi}d\theta\int_0^\frac{\pi}{2}\sin\varphi\cos\varphi d\varphi\)
\(\displaystyle =\frac{2}{k\pi a^4}・2k\pi\left[\frac{\rho^5}{5}\right]_0^a\left[\frac{\sin^2\varphi}{2}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{2}{k\pi a^4}・\frac{k\pi a^5}{5}\)
\(\displaystyle =\frac{2a}{5}\)
よって、重心の座標は\(\displaystyle \left(0,0,\frac{2a}{5}\right)\)
\(\displaystyle D=\{(x,y,z)|x^2+y^2+z^2\leqq a^2,z\geqq0\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(\rho,\theta,\varphi)|\rho\leqq a,0\leqq \theta\leqq 2\pi,0\leqq \varphi\leqq \frac{\pi}{2}\right\}\)
\(D\)の質量は
\(\displaystyle |D|=\iiint_Ek\rho・\rho^2\sin\varphi d\rho d\theta d\varphi=\frac{k\pi a^4}{2}\)
図形の対称性より\(x_G=y_G=0\)
\(\displaystyle z_G=\frac{1}{|D|}\iiint_Ez・k\rho・\rho^2\sin\varphi d\rho d\theta d\varphi\)
\(\displaystyle =\frac{2}{k\pi a^4}・k\int_0^a\rho^4d\rho\int_0^{2\pi}d\theta\int_0^\frac{\pi}{2}\sin\varphi\cos\varphi d\varphi\)
\(\displaystyle =\frac{2}{k\pi a^4}・2k\pi\left[\frac{\rho^5}{5}\right]_0^a\left[\frac{\sin^2\varphi}{2}\right]_0^\frac{\pi}{2}\)
\(\displaystyle =\frac{2}{k\pi a^4}・\frac{k\pi a^5}{5}\)
\(\displaystyle =\frac{2a}{5}\)
よって、重心の座標は\(\displaystyle \left(0,0,\frac{2a}{5}\right)\)
(2)\(ax\leqq x^2+y^2\leqq a^2\)
密度を\(\rho(x,y)=k\sqrt{x^2+y^2}\)とおく。
\(\displaystyle D=\{(x,y)|ax\leqq x^2+y^2\leqq a^2\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|a\cos\theta\leqq r\leqq a,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
\(\displaystyle \ \ \ \ \ \cup\left\{(r,\theta)|0\leqq r\leqq a,\frac{\pi}{2}\leqq \theta\leqq \frac{3\pi}{2}\right\}\)
\(D\)の質量は
\(\displaystyle |D|=\iint_Ekr^2drd\theta=\frac{2ka^3(3\pi-2)}{9}\)
図形の対称性より\(y_G=0\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dx\rho(x,y)dxdy\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・2k\left(\int_0^\frac{\pi}{2}\int_{a\cos\theta}^a r^3\cos\theta drd\theta+\int_\frac{\pi}{2}^\pi\int_0^ar^3\cos\theta drd\theta\right)\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・2k\left(\int_0^\frac{\pi}{2}\frac{a^4}{4}(\cos\theta-\cos^5\theta)d\theta+\int_\frac{\pi}{2}^\pi\frac{a^4}{4}\cos\theta d\theta\right)\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・\frac{k\pi a^5}{5}\)
\(\displaystyle =-\frac{6a}{5(3\pi-2)}\)
よって、重心の座標は\(\displaystyle \left(-\frac{6a}{5(3\pi-2)},0\right)\)
\(\displaystyle D=\{(x,y)|ax\leqq x^2+y^2\leqq a^2\}\)
\(D\)を極座標変換すると
\(\displaystyle E=\left\{(r,\theta)|a\cos\theta\leqq r\leqq a,-\frac{\pi}{2}\leqq \theta\leqq \frac{\pi}{2}\right\}\)
\(\displaystyle \ \ \ \ \ \cup\left\{(r,\theta)|0\leqq r\leqq a,\frac{\pi}{2}\leqq \theta\leqq \frac{3\pi}{2}\right\}\)
\(D\)の質量は
\(\displaystyle |D|=\iint_Ekr^2drd\theta=\frac{2ka^3(3\pi-2)}{9}\)
図形の対称性より\(y_G=0\)
\(\displaystyle x_G=\frac{1}{|D|}\iint_Dx\rho(x,y)dxdy\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・2k\left(\int_0^\frac{\pi}{2}\int_{a\cos\theta}^a r^3\cos\theta drd\theta+\int_\frac{\pi}{2}^\pi\int_0^ar^3\cos\theta drd\theta\right)\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・2k\left(\int_0^\frac{\pi}{2}\frac{a^4}{4}(\cos\theta-\cos^5\theta)d\theta+\int_\frac{\pi}{2}^\pi\frac{a^4}{4}\cos\theta d\theta\right)\)
\(\displaystyle =\frac{9}{2ka^3(3\pi-2)}・\frac{k\pi a^5}{5}\)
\(\displaystyle =-\frac{6a}{5(3\pi-2)}\)
よって、重心の座標は\(\displaystyle \left(-\frac{6a}{5(3\pi-2)},0\right)\)
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