【微分積分】8-3-1 3重積分の定義｜問題集
1.次の3重積分を求めなさい。
(1)\(\displaystyle \iiint_Kdxdydz\)
\(K=[0,a]\times[0,b]\times[0,c]\)
\(K=[0,a]\times[0,b]\times[0,c]\)
\(\displaystyle =\int_0^c\int_0^b\int_0^a dxdydz\)
\(\displaystyle =\int_0^c\int_0^b[x]_0^adydz\)
\(\displaystyle =a\int_0^c[y]_0^bdz\)
\(\displaystyle =ab[z]_0^c\)
\(\displaystyle =abc\)
\(\displaystyle =\int_0^c\int_0^b[x]_0^adydz\)
\(\displaystyle =a\int_0^c[y]_0^bdz\)
\(\displaystyle =ab[z]_0^c\)
\(\displaystyle =abc\)
(2)\(\displaystyle \iiint_Kydzdydx\)
\(K=[0,1]\times[0,x]\times[0,y]\)
\(K=[0,1]\times[0,x]\times[0,y]\)
\(\displaystyle =\int_0^1\int_0^x\int_0^yydzdydx\)
\(\displaystyle =\int_0^1\int_0^x[yz]_0^ydydx\)
\(\displaystyle =\int_0^1\int_0^xy^2dydx\)
\(\displaystyle =\int_0^1\left[\frac{y^3}{3}\right]_0^xdx\)
\(\displaystyle =\int_0^1\frac{x^3}{3}dx\)
\(\displaystyle =\left[\frac{x^4}{12}\right]_0^1\)
\(\displaystyle =\frac{1}{12}\)
\(\displaystyle =\int_0^1\int_0^x[yz]_0^ydydx\)
\(\displaystyle =\int_0^1\int_0^xy^2dydx\)
\(\displaystyle =\int_0^1\left[\frac{y^3}{3}\right]_0^xdx\)
\(\displaystyle =\int_0^1\frac{x^3}{3}dx\)
\(\displaystyle =\left[\frac{x^4}{12}\right]_0^1\)
\(\displaystyle =\frac{1}{12}\)
(3)\(\displaystyle \iiint_\Omega dxdydz\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq y\leqq z\leqq 1\}\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq y\leqq z\leqq 1\}\)
積分領域\(\Omega\)は
\(\Omega=\{(x,y,z)|0\leqq x\leqq 1,x\leqq y\leqq 1,y\leqq z\leqq 1\}\)
と表されるので
\(\displaystyle \iiint_\Omega dxdydz\) \(\displaystyle =\int_0^1\int_x^1\int_y^1dzdydx\)
\(\displaystyle =\int_0^1\int_x^1[z]_y^1dydx\)
\(\displaystyle =\int_0^1\int_x^1(1-y)dydx\)
\(\displaystyle =\int_0^1\left[y-\frac{y^2}{2}\right]_x^1dx\)
\(\displaystyle =\int_0^1\left(\frac{1}{2}-x+\frac{1}{2}x^2\right)dx\)
\(\displaystyle =\left[\frac{x}{2}-\frac{x^2}{2}+\frac{x^3}{6}\right]_0^1\)
\(\displaystyle =\frac{1}{6}\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq 1,x\leqq y\leqq 1,y\leqq z\leqq 1\}\)
と表されるので
\(\displaystyle \iiint_\Omega dxdydz\) \(\displaystyle =\int_0^1\int_x^1\int_y^1dzdydx\)
\(\displaystyle =\int_0^1\int_x^1[z]_y^1dydx\)
\(\displaystyle =\int_0^1\int_x^1(1-y)dydx\)
\(\displaystyle =\int_0^1\left[y-\frac{y^2}{2}\right]_x^1dx\)
\(\displaystyle =\int_0^1\left(\frac{1}{2}-x+\frac{1}{2}x^2\right)dx\)
\(\displaystyle =\left[\frac{x}{2}-\frac{x^2}{2}+\frac{x^3}{6}\right]_0^1\)
\(\displaystyle =\frac{1}{6}\)
(4)\(\displaystyle \iiint_\Omega e^{x+y+z}dxdydz\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq y\leqq z\leqq 1\}\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq y\leqq z\leqq 1\}\)
積分領域\(\Omega\)は
\(\Omega=\{(x,y,z)|0\leqq x\leqq 1,x\leqq y\leqq 1,y\leqq z\leqq 1\}\)
と表されるので
\(\displaystyle \iiint_\Omega e^{x+y+z}dxdydz\) \(\displaystyle =\int_0^1\int_x^1\int_y^1e^{x+y+z}dzdydx\)
\(\displaystyle =\int_0^1\int_x^1[e^{x+y+z}]_y^1dydx\)
\(\displaystyle =\int_0^1\int_x^1(e^{x+y+1}-e^{x+2y})dydx\)
\(\displaystyle =\int_0^1\left[e^{x+y+1}-\frac{1}{2}e^{x+2y}\right]_x^1dx\)
\(\displaystyle =\int_0^1\left(e^{x+2}-\frac{1}{2}e^{x+2}-e^{2x+1}+\frac{1}{2}e^{3x}\right)dx\)
\(\displaystyle =\left[\frac{1}{2}e^{x+2}-\frac{1}{2}e^{2x+1}+\frac{1}{6}e^{3x}\right]_0^1\)
\(\displaystyle =\frac{1}{6}(e-1)^3\)
\(\Omega=\{(x,y,z)|0\leqq x\leqq 1,x\leqq y\leqq 1,y\leqq z\leqq 1\}\)
と表されるので
\(\displaystyle \iiint_\Omega e^{x+y+z}dxdydz\) \(\displaystyle =\int_0^1\int_x^1\int_y^1e^{x+y+z}dzdydx\)
\(\displaystyle =\int_0^1\int_x^1[e^{x+y+z}]_y^1dydx\)
\(\displaystyle =\int_0^1\int_x^1(e^{x+y+1}-e^{x+2y})dydx\)
\(\displaystyle =\int_0^1\left[e^{x+y+1}-\frac{1}{2}e^{x+2y}\right]_x^1dx\)
\(\displaystyle =\int_0^1\left(e^{x+2}-\frac{1}{2}e^{x+2}-e^{2x+1}+\frac{1}{2}e^{3x}\right)dx\)
\(\displaystyle =\left[\frac{1}{2}e^{x+2}-\frac{1}{2}e^{2x+1}+\frac{1}{6}e^{3x}\right]_0^1\)
\(\displaystyle =\frac{1}{6}(e-1)^3\)
次の学習に進もう！