3.6 演習問題 〜 多重積分,累次積分,積分の順序の交換

3.34 (累次積分)   下図の領域 $ D$ は (i) $ x$ に関する単純な領域, (ii) $ y$ に関する単純な領域とみることができる. (i), (ii) の場合それぞれにおいて次の多重積分を計算せよ.
    (1)   $ \displaystyle{\iint_Ddxdy}$     (2)   $ \displaystyle{\iint_Dx^2y\,dxdy}$
\includegraphics[width=.25\textwidth]{sekibun-D1.eps}

3.35 (累次積分)   次の多重積分を計算せよ.
    (1)   $ \displaystyle{\iint_{D_{1}}(2-x+y)\,dxdy}$     (2)   $ \displaystyle{\iint_{D_{2}}ye^{xy}\,dxdy}$     (3)   $ \displaystyle{\iint_{D_3}x^2y\,dxdy}$     (4)   $ \displaystyle{\iint_{D_4}(x^3-3xy)\,dxdy}$
\includegraphics[width=.25\textwidth]{enshu/D_1.eps} \includegraphics[width=.25\textwidth]{enshu/D_2.eps} \includegraphics[width=.25\textwidth]{enshu/hcir.eps} \includegraphics[width=.25\textwidth]{enshu/D_4.eps}

3.36 (累次積分)   積分領域を図示し,多重積分を計算せよ.
    (1)   $ \displaystyle{\iint_D(x^2+y^2+1)\,dxdy,
D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq 1,\,x\leq y\leq 2x}\,\right\}}$
    (2)   $ \displaystyle{\iint_D\frac{x}{y}dxdy,
D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{1\leq y\leq 2,\,0\leq x\leq y^2}\,\right\}}$
    (3)   $ \displaystyle{\iint_Dx^2y\,dxdy,
D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x^2+y^2\leq 1,\,y\geq 0}\,\right\}}$
    (4)   $ \displaystyle{\iiint_Dxyz\,dxdydz,
D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq 1,\,0\leq y\leq 1-x,\,0\leq z\leq 2-y}\,\right\}}$
    (5)   $ \displaystyle{\iiint_Dz\,dxdydz,
D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq 1,\,0\leq y\leq x+1,\,0\leq z\leq x+y}\,\right\}}$
    (6)   $ \displaystyle{\int_{0}^{2}\!\!dx\int_{x^2}^{2x}\!\!\!\!xe^{y}\,dy}$     (7)   $ \displaystyle{\int_{0}^{1}\!\!dy\int_{0}^{\pi/2}\!\!\!\! y\sin xy\,dx}$
    (8)   $ \displaystyle{\iint_{D}(2x-y)\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq 1,\,\, 1\leq y\leq 2}\,\right\}$
    (9)   $ \displaystyle{\iint_{D}(3x-2y)\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq1, 1\leq y\leq 2}\,\right\}$
    (10)   $ \displaystyle{\iint_{D}(x^2+y^2)\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq1,\, 0\leq y\leq 1}\,\right\}$
    (11)   $ \displaystyle{\iint_{D}(x^2y+y^2)\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{1\leq x\leq2,\, 2\leq y\leq 3}\,\right\}$
    (12)   $ \displaystyle{\iint_{D}xe^{x+y}\,dxdy}$, $ \displaystyle{D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq 3,\: 0\leq y\leq 2}\,\right\}}$
    (13)   $ \displaystyle{\iint_{D}\sin(2x+y)\,dxdy}$, $ \displaystyle{D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq \pi/2,\: 0\leq y\leq \pi/2}\,\right\}}$
    (14)   $ \displaystyle{\iint_{D}x^2\sin(xy)\,dxdy}$, $ \displaystyle{D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq \pi/2,\: 0\leq y\leq 1}\,\right\}}$
    (15)   $ \displaystyle{\iint_{D}x\cos(x+y)\,dxdy}$, $ \displaystyle{D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq \pi/2,\: 0\leq y\leq 1}\,\right\}}$
    (16)   $ \displaystyle{\iint_{D}xe^y\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\leq x\leq 2,\,\, x^2\leq y\leq 2x}\,\right\}$
    (17)   $ \displaystyle{\iint_{D}xy^2\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{0\le y\le x\le 1}\,\right\}$
    (18)   $ \displaystyle{\iint_{D}(2x-y)\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x\le y\le 2x,\, x+y\le 3}\,\right\}$
    (19)   $ \displaystyle{\iint_{D}r\,drd\theta}$, $ \displaystyle{D=\left\{\left.\,{(r,\theta)}\,\,\right\vert\,\,{0\leq r\leq a\sin\theta,\:0\leq \theta\leq\pi/2}\,\right\}}$
    (20)   $ \displaystyle{\iint_{D}\,dxdy=\pi}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x^2+y^2\leq1}\,\right\}$ となることを示せ.
    (21)   $ \displaystyle{\iint_{D}x\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x^2+y^2\leq 1,\, x\ge 0}\,\right\}$
    (22)   $ \displaystyle{\iint_{D}\sqrt{a^2-y^2}\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x^2+y^2\le a^2}\,\right\}\, (a>0)$
    (23)   $ \displaystyle{\iint_{D}x^2y\,dxdy}$, $ D=\left\{\left.\,{(x,y)}\,\,\right\vert\,\,{x^2+y^2\le a^2,\, y\ge 0}\,\right\}\, (a>0)$
    (24)   $ \displaystyle{\iiint_{D}z\,dxdydz}$, $ D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq 1,\,0\leq y\leq 1-x,\,0\leq z\leq 1-x-y}\,\right\}$
    (25)   $ \displaystyle{\iiint_{D}y\,dxdydz}$, $ D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{x\ge0,\,y\ge0,\,z\ge0,\,x+2y+3z\le6}\,\right\}$
    (26)   $ \displaystyle{\iiint_{D}y\sin(x+z)\,dxdydz}$, $ \displaystyle{D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq \pi/2,\,
0\leq y\leq\sqrt{x},\,
0\leq z\leq \pi/2-x}\,\right\}}$
    (27)   $ \displaystyle{\iiint_{D}(xy+yz+zx)\,dxdydz}$, $ D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq 1,\,0\leq y\leq x,\,0\leq z\leq y}\,\right\}$
    (28)   $ \displaystyle{\iiint_{D}2xz\,dxdydz}$, $ D=\left\{\left.\,{(x,y,z)}\,\,\right\vert\,\,{0\leq x\leq 1,\,
\,0\leq y\leq x,\,
0\leq z\leq \sqrt{2-x^2-y^2}}\,\right\}$

3.37 (積分の順序の交換)   $ \Box$ を埋めて次の式を完成させよ. ただし $ D_{4}$ は上記の図の領域とする.

$\displaystyle \iint_{D_{4}}f(x,y)\,dxdy =\int^{\Box}_{\Box}\!dx\int^{\Box}_{\Box}\!dy\, f(x,y) =\int^{\Box}_{\Box}\!dy\int^{\Box}_{\Box}\!dx\, f(x,y)$    

3.38 (積分の順序の交換)   次の積分の領域を図示し,積分の順序を変更せよ.
    (1)   $ \displaystyle{\int_{0}^{1}dx\int_{0}^{x}dy\,f(x,y)}$     (2)   $ \displaystyle{\int_{0}^{1}dx\int_{x^2}^{x}dy\,f(x,y)}$     (3)   $ \displaystyle{\int_{0}^{4}dy\int_{y-2}^{\sqrt{y}}f(x,y)\,dx}$
    (4)   $ \displaystyle{\int_{0}^{1}dy\int_{y-1}^{-y+1} f(x,y)\,dx}$     (5)   $ \displaystyle{\int^{1}_{-1}dx\int^{2\sqrt{1-x^2}}_{0} f(x,y)\,dy}$     (6)   $ \displaystyle{\int_{-2}^{1}dx\int_{x^2}^{-x+2} f(x,y)\,dy}$
    (7)   $ \displaystyle{\int_{0}^{4}dy\int_{y}^{2\sqrt{y}}f(x,y)\,dx}$


平成21年1月14日