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7.6 3重積分

1. (a)
$\displaystyle \int_{0}^{a}\int_{0}^{b}\int_{0}^{c}dxdydz$ $\displaystyle =$ $\displaystyle \int_{0}^{a}\int_{0}^{b}x\mid_{0}^{c} dydz$  
  $\displaystyle =$ $\displaystyle c\int_{0}^{a}y\mid_{0}^{b}dz = cb\int_{0}^{a}dz = cbz\mid_{0}^{a} = cba$  

(b)

$\displaystyle \int_{0}^{1}\int_{0}^{x}\int_{0}^{y}ydzdydz$ $\displaystyle =$ $\displaystyle \int_{0}^{1}\int_{0}^{x}yz\mid_{0}^{y}dydx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}\int_{0}^{x}y^2 dydx = \int_{0}^{1}\frac{y^3}{3}\mid_{0}^{x}dx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{1}\frac{x^3}{3}dx = \frac{x^4}{12}\mid_{0}^{1} = \frac{1}{12}$  

2. (a) ボール $x^2 + y^2 + z^2 \leq r^2$の質量を表すには，微小体積に密度をかける必要がある．題意より密度は中心からの距離に比例することから，密度を$\rho$とすると， $\rho = k\sqrt{x^2 + y^2 + z^2}$，ただし，$k$は比例定数． まず，ボールを$xy$平面に正射影すると，$z = 0$より

$\displaystyle \Omega = \left\{(x,y): -r \leq x \leq r, -\sqrt{r^2 - x^2} \leq y \leq \sqrt{r^2 -x^2}\right\}$

ここで，球面座標変換を用いると,
$\displaystyle x$ $\displaystyle =$ $\displaystyle \rho\sin{\phi}\cos{\theta}$  
$\displaystyle y$ $\displaystyle =$ $\displaystyle \rho\sin{\phi}\sin{\theta}$  
$\displaystyle z$ $\displaystyle =$ $\displaystyle \rho\cos{\phi}$  

より，

$\displaystyle T = \left\{(\rho,\phi,\theta):0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi, 0 \leq \rho \leq r \right\}$

また， $x^2 + y^2 + z^2 = \rho^2$． したがって，
$\displaystyle M$ $\displaystyle =$ $\displaystyle \iiint_{V}k\sqrt{x^2 + y^2 + z^2}dxdydz$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{r}k\rho^3 \sin{\phi}d\rho d\phi d\theta$  

(b) 平面$z = 1$と曲面 $z = \sqrt{x^2 + y^2}$で囲まれた円錐の質量を求めるには，平面と円錐の交線を求める． $z = 1 = \sqrt{x^2 + y^2}$より，交線は $x^2 + y^2 = 1$．これより，

$\displaystyle \Omega_{xy} = \left\{(x,y):\sqrt{x^2 + y^2} \leq 1\right\}$

極座標変換を用いると，

$\displaystyle \Gamma = \left\{(r,\theta): 0 \leq \theta \leq 2\pi, r \leq 1\right\}$

密度 $\rho = k\sqrt{x^2 + y^2 + z^2}$より，
$\displaystyle M$ $\displaystyle =$ $\displaystyle \iint_{\Omega_{xy}}\int_{z = \sqrt{x^2 + y^2}}^{1}k\sqrt{x^2 + y^2 + z^2}dzdydx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{z=r}^{1}k\sqrt{r^2 + z^2}dzrdrd\theta$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{1}\int_{z=r}^{1}kr\sqrt{r^2 + z^2}dzdrd\theta$  

(c) 放物面 $z = 4-x^2 - y^2$と曲面 $z = 2 + y^2$で囲まれた部分の体積を求めるには，放物面と曲面の交線を求める． $z = 4-x^2-y^2 = 2 + y^2$より，交線は $x^2 + 2y^2 = 2$．これより，

$\displaystyle \Omega_{xy} = \left\{(x,y):x^2 + 2y^2 \leq 2\right\}$

ここで， $x = \sqrt{2}\cos{\theta}, y = \sin{\theta}$とおくと，

$\displaystyle \Gamma = \left\{(r,\theta):0 \leq \theta \leq 2\pi, 0 \leq r \leq \sqrt{\cos^{2}{\theta} + 1}\right\}$

これより，
$\displaystyle V$ $\displaystyle =$ $\displaystyle \iint_{\Omega_{xy}}\int_{z=2+y^2}^{4-x^2-y^2}dzdydx$  
  $\displaystyle =$ $\displaystyle \int_{0}^{2\pi}\int_{0}^{\sqrt{\cos^{2}{\theta}}}(2-r^2-\sin^{2}{\theta})rdrd\theta$  

3. (a) 縦線集合を用いて領域を求めると，

$\displaystyle \Omega = \left\{(x,y):0 \leq x \leq 1, x^2 \leq y \leq x\right \}$

次に，密度を一定として質量を求めると，

$\displaystyle \iint_{\Omega}\rho dx dy = \rho \int_{0}^{1}\int_{x^2}^{x}dy dx = \rho \int_{0}^{1}(x-x^2)dx = \rho(\frac{1}{2}-\frac{1}{3}) = \frac{\rho}{6}$

また，$y$軸に関する１次モーメントは
$\displaystyle \int_{\Omega}\rho xdydx$ $\displaystyle =$ $\displaystyle \rho \int_{0}^{1}\int_{x^2}^{x}xdy dx = \rho \int_{0}^{1}xy\mid_{x^2}^{x}dx$  
  $\displaystyle =$ $\displaystyle \rho\int_{0}^{1}(x^2 - x^3)dx = \rho(\frac{1}{3}-\frac{1}{4}) = \frac{\rho}{12}$  

これより， $\bar{x} = \frac{\rho/12}{\rho/6} = \frac{1}{2}$

$\bar{y}$を求めるには，$x$軸に関する１次モーメントを求める必要がある．

$\displaystyle \iint_{\Omega}\rho ydydx$ $\displaystyle =$ $\displaystyle \rho\int_{0}^{1}\int_{x^2}^{x}ydydx$  
  $\displaystyle =$ $\displaystyle \rho\int_{0}^{1}\frac{y^2}{2}\mid_{x^2}^{x}dx = \frac{\rho}{2} \int_{0}^{1}(x^2 - x^4)dx$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2}(\frac{1}{3}-\frac{1}{5}) = \frac{\rho}{15}$  

これより， $\bar{y} = \frac{\rho/15}{\rho/6} = \frac{2}{5}$. したがって，重心は $(\frac{1}{2},\frac{2}{5})$.

(b) 縦線集合を用いて領域を求めると，

$\displaystyle \Omega = \left\{(x,y):-2 \leq x \leq 4, \frac{x^2}{4} \leq y \leq \frac{x}{2}+2\right \}$

次に，密度を一定として質量を求めると，
$\displaystyle \iint_{\Omega}\rho dx dy$ $\displaystyle =$ $\displaystyle \rho \int_{-2}^{4}\int_{\frac{x^2}{4}}^{\frac{x}{2}+2}dy dx$  
  $\displaystyle =$ $\displaystyle \rho \int_{-2}^{4}(\frac{x}{2} + 2-\frac{x^2}{4})dx = \rho(\frac{x^2}{4} + 2x-\frac{x^3}{12}\mid_{-2}^{4}$  
  $\displaystyle =$ $\displaystyle \rho(\frac{4^2-(-2)^2}{4} + 2(4-(-2)) - \frac{4^3 - (-2)^3}{12} = 9\rho$  

また，$y$軸に関する１次モーメントは
$\displaystyle \int_{\Omega}\rho xdydx$ $\displaystyle =$ $\displaystyle \rho \int_{-2}^{4}\int_{\frac{x^2}{4}}^{\frac{x}{2}+2}xdy dx = \rho \int_{-2}^{4}xy\mid_{\frac{x^2}{4}}^{\frac{x}{2}+2}dx$  
  $\displaystyle =$ $\displaystyle \rho\int_{-2}^{4}(\frac{x^2}{2}+2x - \frac{x^3}{4})dx = \rho(\frac{x^3}{6} + x^2 -\frac{x^4}{16})\mid_{-2}^{4}$  
  $\displaystyle =$ $\displaystyle \rho(\frac{4^3 - (-2)^3}{6} + 4^2-(-2)^2 - \frac{4^4-(-2)^4}{16}) = 9\rho$  

これより， $\bar{x} = \frac{9\rho}{9\rho} = 1$.

$\bar{y}$を求めるには，$x$軸に関する１次モーメントを求める必要がある．

$\displaystyle \iint_{\Omega}\rho ydydx$ $\displaystyle =$ $\displaystyle \rho\int_{-2}^{4}\int_{\frac{x^2}{4}}^{\frac{x}{2}+2}ydydx$  
  $\displaystyle =$ $\displaystyle \rho\int_{-2}^{4}\frac{y^2}{2}\mid_{\frac{x^2}{4}}^{\frac{x}{2}+2}dx = \frac{\rho}{2} \int_{-2}^{4}(\frac{(x+4)^2}{4} - \frac{x^4}{16})dx$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2}(\frac{(x+4)^3}{12}-\frac{x^5}{80})\mid_{-2}^{4} = \frac{\rho}{2}(\frac{8^3 - 2^3}{12} - \frac{4^5 - (-2)^5}{80}$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2}(42-\frac{66}{5}) = \frac{72\rho}{5}$  

これより， $\bar{y} = \frac{72\rho/5}{9\rho} = \frac{8}{5}$. したがって，重心は $(1,\frac{8}{5})$.

(c) 縦線集合を用いて領域を求めると，

$\displaystyle \Omega = \left\{(x,y):0 \leq x \leq 4, x^2-2x \leq y \leq 6x-x^2\right \}$

次に，密度を一定として質量を求めると，
$\displaystyle \iint_{\Omega}\rho dx dy$ $\displaystyle =$ $\displaystyle \rho \int_{0}^{4}\int_{x^2-2x}^{6x-x^2}dy dx$  
  $\displaystyle =$ $\displaystyle \rho \int_{0}^{4}(8x-2x^2)dx = \rho(4x^2 - \frac{2x^3}{3})\mid_{0}^{4}$  
  $\displaystyle =$ $\displaystyle \rho(64 - \frac{128}{3}) = \frac{64\rho}{3}$  

また，$y$軸に関する１次モーメントは
$\displaystyle \int_{\Omega}\rho xdydx$ $\displaystyle =$ $\displaystyle \rho \int_{0}^{4}\int_{x^2 - 2x}^{6x-x^2}xdy dx = \rho \int_{0}^{4}xy\mid_{x^2-2x}^{6x-x^2}dx$  
  $\displaystyle =$ $\displaystyle \rho\int_{0}^{4}(8x^2-2x^3)dx = \rho(\frac{8x^3}{3} - \frac{x^4}{4})\mid_{0}^{4}$  
  $\displaystyle =$ $\displaystyle \rho(\frac{512-384}{3}) = \frac{128\rho}{3}$  

これより， $\bar{x} = \frac{\frac{128\rho}{3}}{\frac{64\rho}{3}} = 2$.

$\bar{y}$を求めるには，$x$軸に関する１次モーメントを求める必要がある．

$\displaystyle \iint_{\Omega}\rho ydydx$ $\displaystyle =$ $\displaystyle \rho\int_{0}^{4}\int_{x^2 -2x}^{6x-x^2}ydydx$  
  $\displaystyle =$ $\displaystyle \rho\int_{0}^{4}\frac{y^2}{2}\mid_{x^2 -2x}^{6x-x^2}dx = \frac{\rho}{2} \int_{0}^{4}((6x-x^2)^2 - (x^2 - 2x)^2)dx$  
  $\displaystyle =$ $\displaystyle \frac{\rho}{2}((4x)(8x-2x^2))\mid_{0}^{4} = \rho\int_{0}^{4}(16x^2 - 4x^3)dx$  
  $\displaystyle =$ $\displaystyle \rho(\frac{16\cdot 64}{3} - 256) = \frac{256\rho}{3}$  

これより， $\bar{y} = \frac{256\rho/3}{\frac{64\rho}{3}} = 4$. したがって，重心は$(2,4)$.

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Next: 演習問題詳解 Up: 確認問題詳解 Previous: 7.5 2重積分の応用   索引