【微分積分】8-2-4 一般の変数変換｜問題集
1.次の2重積分を求めなさい。
(1)\(\displaystyle \iint_D(x-y)^2\sin(x+y)dxdy\)
\(D=\{(x,y)|0\leqq x-y\leqq \pi,0\leqq x+y\leqq \pi\}\)
\(D=\{(x,y)|0\leqq x-y\leqq \pi,0\leqq x+y\leqq \pi\}\)
\(D\)を\(s=x-y,t=x+y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq \pi,0\leqq t\leqq \pi\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\displaystyle \frac{1}{2} & \displaystyle \frac{1}{2} \\ \displaystyle -\frac{1}{2} & \displaystyle \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x-y)^2\sin(x+y)dxdy\)
\(\displaystyle =\iint_Es^2\sin t・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^\pi s^2ds\int_0^\pi\sin tdt\)
\(\displaystyle =\frac{1}{2}・\frac{\pi^2}{3}・2\)
\(\displaystyle =\frac{\pi^3}{3}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq \pi,0\leqq t\leqq \pi\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\displaystyle \frac{1}{2} & \displaystyle \frac{1}{2} \\ \displaystyle -\frac{1}{2} & \displaystyle \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x-y)^2\sin(x+y)dxdy\)
\(\displaystyle =\iint_Es^2\sin t・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^\pi s^2ds\int_0^\pi\sin tdt\)
\(\displaystyle =\frac{1}{2}・\frac{\pi^2}{3}・2\)
\(\displaystyle =\frac{\pi^3}{3}\)
(2)\(\displaystyle \iint_De^{x-y}\sin(x+y)dxdy\)
\(D=\{(x,y)|0\leqq x-y\leqq 2,0\leqq x+y\leqq \pi\}\)
\(D=\{(x,y)|0\leqq x-y\leqq 2,0\leqq x+y\leqq \pi\}\)
\(D\)を\(s=x-y,t=x+y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 2,0\leqq t\leqq \pi\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\displaystyle \frac{1}{2} & \displaystyle \frac{1}{2} \\ \displaystyle -\frac{1}{2} & \displaystyle \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_De^{x-y}\sin(x+y)dxdy\)
\(\displaystyle =\iint_Ee^s\sin t・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^2 e^sds\int_0^\pi\sin tdt\)
\(\displaystyle =\frac{1}{2}・(e^2-1)・2\)
\(=e^2-1\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 2,0\leqq t\leqq \pi\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\displaystyle \frac{1}{2} & \displaystyle \frac{1}{2} \\ \displaystyle -\frac{1}{2} & \displaystyle \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_De^{x-y}\sin(x+y)dxdy\)
\(\displaystyle =\iint_Ee^s\sin t・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^2 e^sds\int_0^\pi\sin tdt\)
\(\displaystyle =\frac{1}{2}・(e^2-1)・2\)
\(=e^2-1\)
(3)\(\displaystyle \iint_Dydxdy\)
\(D=\{(x,y)|y\leqq x^2\leqq 8y,x\leqq y^2\leqq 8x,x>0\}\)
\(D=\{(x,y)|y\leqq x^2\leqq 8y,x\leqq y^2\leqq 8x,x>0\}\)
\(D\)を\(\displaystyle s=\frac{x^2}{y},t=\frac{y^2}{x}\)で変数変換すると
\(\displaystyle E=\{(s,t)|1\leqq s\leqq 8,1\leqq t\leqq 8\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{2}{3}s^{-\frac{1}{3}}t^{\frac{1}{3}} & \frac{1}{3}s^{\frac{2}{3}}t^{-\frac{2}{3}} \\ \frac{1}{3}s^{-\frac{2}{3}}t^{\frac{2}{3}} & \frac{2}{3}s^{\frac{1}{3}}t^{-\frac{1}{3}}\end{vmatrix}=\frac{1}{3}\)
よって
\(\displaystyle \iint_Dydxdy\)
\(\displaystyle =\iint_Es^\frac{1}{3}t^\frac{2}{3}・\frac{1}{3}dsdt\)
\(\displaystyle =\frac{1}{3}\int_1^8 s^\frac{1}{3}ds\int_1^8 t^\frac{2}{3}dt\)
\(\displaystyle =\frac{1}{3}\left[\frac{3}{4}s^\frac{4}{3}\right]_1^8・\left[\frac{3}{5}t^\frac{5}{3}\right]_1^8\)
\(\displaystyle =\frac{279}{4}\)
\(\displaystyle E=\{(s,t)|1\leqq s\leqq 8,1\leqq t\leqq 8\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{2}{3}s^{-\frac{1}{3}}t^{\frac{1}{3}} & \frac{1}{3}s^{\frac{2}{3}}t^{-\frac{2}{3}} \\ \frac{1}{3}s^{-\frac{2}{3}}t^{\frac{2}{3}} & \frac{2}{3}s^{\frac{1}{3}}t^{-\frac{1}{3}}\end{vmatrix}=\frac{1}{3}\)
よって
\(\displaystyle \iint_Dydxdy\)
\(\displaystyle =\iint_Es^\frac{1}{3}t^\frac{2}{3}・\frac{1}{3}dsdt\)
\(\displaystyle =\frac{1}{3}\int_1^8 s^\frac{1}{3}ds\int_1^8 t^\frac{2}{3}dt\)
\(\displaystyle =\frac{1}{3}\left[\frac{3}{4}s^\frac{4}{3}\right]_1^8・\left[\frac{3}{5}t^\frac{5}{3}\right]_1^8\)
\(\displaystyle =\frac{279}{4}\)
(4)\(\displaystyle \iint_D(y^2-x^2)e^{x^2+y^2}dxdy\)
\(D=\{(x,y)|y\geqq x\geqq 0,x+y\leqq 1\}\)
\(D=\{(x,y)|y\geqq x\geqq 0,x+y\leqq 1\}\)
\(D\)を\(s=x+y,t=-x+y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq s\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(y^2-x^2)e^{x^2+y^2}dxdy\)
\(\displaystyle =\iint_Este^\frac{s^2+t^2}{2}・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_0^s ste^\frac{s^2+t^2}{2}dtds\)
\(\displaystyle =\frac{1}{2}\int_0^1s(e^{s^2}-e^\frac{s^2}{2})ds\)
\(\displaystyle =\frac{(\sqrt{e}-1)^2}{4}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq s\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(y^2-x^2)e^{x^2+y^2}dxdy\)
\(\displaystyle =\iint_Este^\frac{s^2+t^2}{2}・\frac{1}{2}dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_0^s ste^\frac{s^2+t^2}{2}dtds\)
\(\displaystyle =\frac{1}{2}\int_0^1s(e^{s^2}-e^\frac{s^2}{2})ds\)
\(\displaystyle =\frac{(\sqrt{e}-1)^2}{4}\)
(5)\(\displaystyle \iint_D(x+y)dxdy\)
\(D=\{(x,y)|0\leqq x+y\leqq 1,|x-y|\leqq 1\}\)
\(D=\{(x,y)|0\leqq x+y\leqq 1,|x-y|\leqq 1\}\)
\(D\)を\(s=x+y,t=x-y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,-1\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x+y)dxdy\)
\(\displaystyle =\iint_Es・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_{-1}^1 sdtds\)
\(\displaystyle =\frac{1}{2}\int_0^1 2sds\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,-1\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x+y)dxdy\)
\(\displaystyle =\iint_Es・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_{-1}^1 sdtds\)
\(\displaystyle =\frac{1}{2}\int_0^1 2sds\)
\(\displaystyle =\frac{1}{2}\)
(6)\(\displaystyle \iint_D(2x+3y)dxdy\)
\(D=\{(x,y)|0\leqq x+y\leqq 1,0\leqq x-y\leqq 1\}\)
\(D=\{(x,y)|0\leqq x+y\leqq 1,0\leqq x-y\leqq 1\}\)
\(D\)を\(s=x-y,t=x+y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(2x+3y)dxdy\)
\(\displaystyle =\iint_E\frac{-s+5t}{2}・\frac{1}{2}dsdt\)
\(\displaystyle =\int_0^1\int_0^1\frac{-s+5t}{4}dsdt\)
\(\displaystyle =\frac{1}{4}\int_0^1\left(-\frac{1}{2}+5t\right)dt\)
\(\displaystyle =\frac{1}{2}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2}\end{vmatrix}=\frac{1}{2}\)
よって
\(\displaystyle \iint_D(2x+3y)dxdy\)
\(\displaystyle =\iint_E\frac{-s+5t}{2}・\frac{1}{2}dsdt\)
\(\displaystyle =\int_0^1\int_0^1\frac{-s+5t}{4}dsdt\)
\(\displaystyle =\frac{1}{4}\int_0^1\left(-\frac{1}{2}+5t\right)dt\)
\(\displaystyle =\frac{1}{2}\)
(7)\(\displaystyle \iint_D2xdxdy\)
\(D=\{(x,y)|0\leqq x-y\leqq 1,0\leqq x+2y\leqq 1\}\)
\(D=\{(x,y)|0\leqq x-y\leqq 1,0\leqq x+2y\leqq 1\}\)
\(D\)を\(s=x-y,t=x+2y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3}\end{vmatrix}=\frac{1}{3}\)
よって
\(\displaystyle \iint_D2xdxdy\)
\(\displaystyle =\iint_E\frac{4s+2t}{3}・\frac{1}{3}dtds\)
\(\displaystyle =\int_0^1\int_0^1\frac{4s+2t}{9}dtds\)
\(\displaystyle =\frac{2}{9}\int_0^1\left(\frac{1}{2}+2s\right)ds\)
\(\displaystyle =\frac{1}{3}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,0\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3}\end{vmatrix}=\frac{1}{3}\)
よって
\(\displaystyle \iint_D2xdxdy\)
\(\displaystyle =\iint_E\frac{4s+2t}{3}・\frac{1}{3}dtds\)
\(\displaystyle =\int_0^1\int_0^1\frac{4s+2t}{9}dtds\)
\(\displaystyle =\frac{2}{9}\int_0^1\left(\frac{1}{2}+2s\right)ds\)
\(\displaystyle =\frac{1}{3}\)
(8)\(\displaystyle \iint_De^{\frac{y-x}{y+x}}dxdy\)
\(D=\{(x,y)|x+y\leqq 1,x\geqq0,y\geqq0\}\)
\(D=\{(x,y)|x+y\leqq 1,x\geqq0,y\geqq0\}\)
\(D\)を\(s=x+y,t=x-y\)で変数変換すると
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,-s\leqq t\leqq s\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_De^{\frac{y-x}{y+x}}dxdy\)
\(\displaystyle =\iint_Ee^{-\frac{t}{s}}・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_{-s}^s e^{-\frac{t}{s}}dtds\)
\(\displaystyle =\frac{1}{2}\int_0^1(-se^{-1}+se)ds\)
\(\displaystyle =-\frac{1}{4e}+\frac{e}{4}\)
\(\displaystyle E=\{(s,t)|0\leqq s\leqq 1,-s\leqq t\leqq s\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_De^{\frac{y-x}{y+x}}dxdy\)
\(\displaystyle =\iint_Ee^{-\frac{t}{s}}・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{2}\int_0^1\int_{-s}^s e^{-\frac{t}{s}}dtds\)
\(\displaystyle =\frac{1}{2}\int_0^1(-se^{-1}+se)ds\)
\(\displaystyle =-\frac{1}{4e}+\frac{e}{4}\)
(9)\(\displaystyle \iint_D(x^2+y^2)e^{-x+y}dxdy\)
\(D=\{(x,y)|-1\leqq x+y\leqq 1,-1\leqq x-y\leqq 1\}\)
\(D=\{(x,y)|-1\leqq x+y\leqq 1,-1\leqq x-y\leqq 1\}\)
\(D\)を\(s=x+y,t=x-y\)で変数変換すると
\(\displaystyle E=\{(s,t)|-1\leqq s\leqq 1,-1\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x^2+y^2)e^{-x+y}dxdy\)
\(\displaystyle =\iint_E\frac{s^2+t^2}{2}e^{-t}・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{4}\int_{-1}^1\int_{-1}^1(s^2+t^2)e^{-t}dtds\)
\(\displaystyle =\frac{1}{4}\int_{-1}^1\{s^2\left(e-\frac{1}{e}\right)+\left(e-\frac{5}{e}\right)\right\}ds\)
\(\displaystyle =\frac{2e}{3}-\frac{4}{3e}\)
\(\displaystyle E=\{(s,t)|-1\leqq s\leqq 1,-1\leqq t\leqq 1\}\)
ヤコビアンは
\(\displaystyle \frac{\partial(x,y)}{\partial(s,t)}=\begin{vmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{vmatrix}=-\frac{1}{2}\)
よって
\(\displaystyle \iint_D(x^2+y^2)e^{-x+y}dxdy\)
\(\displaystyle =\iint_E\frac{s^2+t^2}{2}e^{-t}・\left|-\frac{1}{2}\right|dsdt\)
\(\displaystyle =\frac{1}{4}\int_{-1}^1\int_{-1}^1(s^2+t^2)e^{-t}dtds\)
\(\displaystyle =\frac{1}{4}\int_{-1}^1\{s^2\left(e-\frac{1}{e}\right)+\left(e-\frac{5}{e}\right)\right\}ds\)
\(\displaystyle =\frac{2e}{3}-\frac{4}{3e}\)
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