重積分の基礎

■問題

次の2重積分を求めよ．その際領域${\displaystyle D}$ の図を描くこと．

${\displaystyle {\iint }_D{y}^{2}dxdy}$　　${\displaystyle \left(D:x^2+y^2\leqq 9,x\geqq 0\right)}$

■答

${\displaystyle \frac{81}{8}\pi }$  

■ヒント

極座標表示に変数変換する．

■解説

${\displaystyle \{\begin{array}{l}x=r\cos \theta \\ y=r\sin \theta \end{array}}$ とおき変数変換する．このときヤコビアン J は

${\displaystyle J=\frac{\partial \left(x,y\right)}{\partial \left(r,\theta \right)}=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|}$${\displaystyle =\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|}$${\displaystyle =r{\cos }^2\theta +r{\sin }^2\theta =r}$  

よって，

${\displaystyle dxdy\to }$ ${\displaystyle \left|J\right|drd\theta =\left|r\right|drd\theta =rdrd\theta }$

となる．

積分領域は

${\displaystyle D:x^2+y^2\leqq 9,x\geqq 0}$ 　 →　${\displaystyle D^{\prime }:}$ ${\displaystyle 0\leqq r\leqq 3}$，${\displaystyle -\frac{\pi }{2}\leqq \theta \leqq \frac{\pi }{2}}$

となる．

${\displaystyle {\iint }_D{y}^{2}dxdy}$

${\displaystyle ={{\displaystyle \int }}_0^3\left({{\displaystyle \int }}_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{\left(r\sin \theta \right)}^{2}rd\theta \right)dr}$

${\displaystyle ={\displaystyle {\int }_0^3\left({\displaystyle {\int }_{-\frac{\pi }{2}}^{\frac{\pi }{2}}{r}^3{\sin }^2\theta d\theta }\right)}dr}$

${\displaystyle ={\displaystyle {\int }_0^3\left(2r^3{\displaystyle {\int }_0^{\frac{\pi }{2}}{\sin }^2\theta d\theta }\right)}dr}$  

${\displaystyle ={\displaystyle {\int }_0^{3}2r^3\cdot \frac{1}{2}\cdot \frac{\pi }{2}}dr}$  

${\displaystyle =\frac{\pi }{2}{\displaystyle {\int }_0^3{r}^{3}dr}}$

${\displaystyle =\frac{\pi }{2}{\displaystyle {\int }_0^3{r}^{3}dr}}$  

${\displaystyle =\frac{\pi }{2}{\left[\frac{1}{4}{r}^4\right]}_0^3}$  

${\displaystyle =\frac{81}{8}\pi }$  

　

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最終更新日： 2026年10月6日