重積分の計算問題

1. 次の重積分を計算せよ．

   • •	${\displaystyle {\int }_0^2{\int }_{\frac{y}{2}}^1\left(x+y\right)dxdy}$ ⇒ 解答

   • •	${\displaystyle {\int }_0^1{\int }_0^{1-x}\left(x^2-2y\right)dydx}$ ⇒ 解答

   • •	${\displaystyle {\int }_0^{\pi }{\int }_0^{\pi }sin\left(x+y\right)dxdy}$ ⇒ 解答
2. 次の重積分を計算せよ．
   • ${\displaystyle {\displaystyle {\iint }_{D}ydxdy}}$ ${\displaystyle (D:0\leqq x\leqq 1,0\leqq y\leqq x^2)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(x^3+xy\right)dxdy}}$ ${\displaystyle (D:0\leqq y\leqq 1,0\leqq x\leqq \sqrt{y})}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(x^2-2y\right)dxdy}}$ ${\displaystyle (D:x+y\leqq 1,x\geqq 0,y\geqq 0)}$ ⇒ 解答　
   • ${\displaystyle {\iint }_D\log xy}dxdy$ ${\displaystyle \left(D:1\leqq x\leqq 2,1\leqq y\leqq 2\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(2x-y\right)dxdy}$ ${\displaystyle \left(D:-3\leqq x\leqq 3,0\leqq y\leqq 3\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}x{y}^{2}dxdy}$ ${\displaystyle \left(D:0\leqq x\leqq 2,-2\leqq y\leqq 1\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(x-2xy+3y\right)dxdy}$ ${\displaystyle \left(D:-1\leqq x\leqq 1,-2\leqq y\leqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(x^2+y^2\right)dxdy}$ ${\displaystyle \left(D:-2\leqq x\leqq 2,-2\leqq y\leqq 2\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}sin\left(x+y\right)dxdy}$ ${\displaystyle \left(D:0\leqq x\leqq \pi ,0\leqq y\leqq \pi \right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}2xcosydxdy}$ ${\displaystyle \left(D:0\leqq x\leqq \frac{\pi }{2},\frac{\pi }{6}\leqq y\leqq \frac{\pi }{2}\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}rsin\theta drd\theta }$ ${\displaystyle \left(D:\frac{\pi }{3}\leqq \theta \leqq \pi ,0\leqq r\leqq 4\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}xdxdy}$ ${\displaystyle \left(D:\frac{x}{2}+\frac{y}{4}\leqq 1,x\geqq 0,y\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(x+y\right)dxdy}$ ${\displaystyle \left(D:\frac{x}{3}+y\leqq 2,x\geqq 0,y\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}xydxdy}$ ${\displaystyle \left(D:2x+\frac{y}{2}\leqq 1,x\geqq 0,y\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(x^2+y^2\right)dxdy}$ ${\displaystyle \left(D:x+y\leqq 2,x\geqq 0,y\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}sin\left(x+y\right)dxdy}$ ${\displaystyle \left(D:x+y\leqq \pi ,x\geqq 0,y\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(4-x-y\right)dxdy}$ ${\displaystyle \left(D:0\leqq x\leqq 2,-x\leqq y\leqq x\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D{\left(x-y\right)}^{2}dxdy}$ ${\displaystyle \left(D:\left|x+2\right|\leqq 1,\left|x-2y\right|\leqq 1\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}xdxdy}$ ${\displaystyle \left(D:0\leqq y\leqq 3x-x^2\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(x+2\right)dxdy}$ ${\displaystyle \left(D:-1\leqq x\leqq 1,0\leqq y\leqq x^2+1\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(2xy+3y^2\right)dxdy}$ ${\displaystyle \left(D:x^2\leqq y\leqq x\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D\left(2x+y\right)dxdy}$ ${\displaystyle \left(D:x^2\leqq y\leqq x+2\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_{D}xydxdy}$ ${\displaystyle \left(D:x^2\leqq y\leqq x\right)}$ ⇒ 解答
   • ${\displaystyle {\iint }_D{y}^{2}dxdy}$ ${\displaystyle \left(D:x^2+y^2\leqq 9,x\geqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(x^2-y^2\right)}\left(x+y\right)dxdy}$ ${\displaystyle \left(D:0\leqq x+y\leqq 2,-1\leqq x-y\leqq 2\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_{D}4\left(x^2+2xy+y^2\right)}\left(x^2-y^2\right)dxdy}$ ${\displaystyle \left(D:0\leqq x+y\leqq 2,-1\leqq x-y\leqq 2\right)}$ ⇒ 解答
   • ${\displaystyle {{\displaystyle \iint }}_D\left(x^3+y^3\right)dxdy}$ ${\displaystyle \left(D:2\leqq x+y\leqq 5,-6\leqq x-2y\leqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_{D}x\sin xydxdy}}$     ${\displaystyle \left(D:0\leqq x\leqq 1,0\leqq y\leqq \frac{\pi }{2}\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D{e}^x\cos ydxdy}}$     ${\displaystyle \left(D:0\leqq x\leqq 1,0\leqq y\leqq \frac{\pi }{4}\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D{e}^x\sin ydxdy}}$     ${\displaystyle \left(D:0\leqq x\leqq 1,0\leqq y\leqq \pi x\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\frac{x-y}{1+x+y}}dxdy}$ ${\displaystyle \left(D:0\leqq x+y\leqq 1,0\leqq x-y\leqq 1\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_{D}64xydxdy}}$ ${\displaystyle \left(D:1\leqq x+y\leqq 2,-2\leqq x-3y\leqq 0\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(2x-y\right)dx}dy}$ ${\displaystyle \left(D:0\leqq x\leqq 1,x\leqq y\leqq \sqrt{x}\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\sqrt{a^2-x^2-y^2}dxdy}}$ ${\displaystyle (D:x^2+y^2\leqq ax)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(x^2+y^2\right)e^{-x-y}}dxdy}$ ${\displaystyle (D:-2<x+y<2,-2<x-y<2)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_{D}4\left(x^2+2xy+y^2\right)}\left(x^2-y^2\right)dxdy}$ ${\displaystyle \left(D:0\leqq x+y\leqq 2,-1\leqq x-y\leqq 2\right)}$ ⇒ 解答
   • ${\displaystyle {\displaystyle {\iint }_D\left(x^3+y^3\right)}dxdy}$ ${\displaystyle \left(D:0\leqq x+y\leqq 1,-1\leqq x-2y\leqq 2\right)}$ ⇒ 解答
3. 次の問題に答えよ．
   • 次の変数変換の場合のヤコビアン ${\displaystyle J\left(u,v\right)}$ を計算せよ．

     ${\displaystyle x=6u-7v}$ ， ${\displaystyle y=3u+5y}$ ⇒ 解答

   • 積分の順序を変更して次の重積分を求めよ．

     ${\displaystyle {{\displaystyle \int }}_0^{2a}\left({{\displaystyle \int }}_{\frac{x^2}{a}}^{6a-x}ydy\right)dx}$ ⇒ 解答

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最終更新日： 2025年3月28日

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