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Question Number 192884 by 073 last updated on 30/May/23

Commented by witcher3 last updated on 03/Jun/23

let u=x+y,v=x−y g(u,v)=(x,y) J_g = ((((1/2) (1/2))),(((1/2) −(1/2))) ) ((x),(y) )=(1/2) (((u+v)),((u−v)) ) ∫∫_R^2 e^(−5(x^2 +y^2 )+8xy) dxdy =∫∫_R^2 e^(−(5/2)(u^2 +v^2 )+2(u^2 −v^2 )) .det determinant ((((1/2) (1/2))),(((1/2) −(1/2))))dudv =(1/2)∫∫_R^2 e^(−(u^2 /2)−((9v^2 )/2)) dudv A=(1/2)∫_R e^(−(u^2 /2)) du.∫_R e^(−(9/2)v^2 ) ∫_R e^(−ax^2 ) dx=(1/( (√a)))∫_R e^(−x^2 ) dx=((√π)/( (√a))) A=(1/2).(√2).(√(2/9)).π =(π/( (√9)))=(π/3)

$$\mathrm{let}\:\mathrm{u}=\mathrm{x}+\mathrm{y},\mathrm{v}=\mathrm{x}−\mathrm{y} \\ $$$$\mathrm{g}\left(\mathrm{u},\mathrm{v}\right)=\left(\mathrm{x},\mathrm{y}\right) \\ $$$$\mathrm{J}_{\mathrm{g}} =\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{y}}\end{pmatrix}=\frac{\mathrm{1}}{\mathrm{2}}\begin{pmatrix}{\mathrm{u}+\mathrm{v}}\\{\mathrm{u}−\mathrm{v}}\end{pmatrix} \\ $$$$ \\ $$$$\int\int_{\mathbb{R}^{\mathrm{2}} } \mathrm{e}^{−\mathrm{5}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} \right)+\mathrm{8xy}} \mathrm{dxdy} \\ $$$$=\int\int_{\mathbb{R}^{\mathrm{2}} } \mathrm{e}^{−\frac{\mathrm{5}}{\mathrm{2}}\left(\mathrm{u}^{\mathrm{2}} +\mathrm{v}^{\mathrm{2}} \right)+\mathrm{2}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{v}^{\mathrm{2}} \right)} .\mathrm{det}\begin{vmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:−\frac{\mathrm{1}}{\mathrm{2}}}\end{vmatrix}\mathrm{dudv} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\int_{\mathbb{R}^{\mathrm{2}} } \mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{9v}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{dudv} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathbb{R}} \mathrm{e}^{−\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{2}}} \mathrm{du}.\int_{\mathbb{R}} \mathrm{e}^{−\frac{\mathrm{9}}{\mathrm{2}}\mathrm{v}^{\mathrm{2}} } \\ $$$$\int_{\mathbb{R}} \mathrm{e}^{−\mathrm{ax}^{\mathrm{2}} } \mathrm{dx}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{a}}}\int_{\mathbb{R}} \mathrm{e}^{−\mathrm{x}^{\mathrm{2}} } \mathrm{dx}=\frac{\sqrt{\pi}}{\:\sqrt{\mathrm{a}}} \\ $$$$\mathrm{A}=\frac{\mathrm{1}}{\mathrm{2}}.\sqrt{\mathrm{2}}.\sqrt{\frac{\mathrm{2}}{\mathrm{9}}}.\pi \\ $$$$=\frac{\pi}{\:\sqrt{\mathrm{9}}}=\frac{\pi}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by 073 last updated on 03/Jun/23

answer is (π/3)??

$$\mathrm{answer}\:\mathrm{is}\:\frac{\pi}{\mathrm{3}}?? \\ $$

Commented by witcher3 last updated on 03/Jun/23

yes (5/2)v^2 +2v^2 =((9v^2 )/2) i have writen ((7v^2 )/2)sorry

$$\mathrm{yes}\:\frac{\mathrm{5}}{\mathrm{2}}\mathrm{v}^{\mathrm{2}} +\mathrm{2v}^{\mathrm{2}} =\frac{\mathrm{9v}^{\mathrm{2}} }{\mathrm{2}}\:\mathrm{i}\:\mathrm{have}\:\mathrm{writen}\:\frac{\mathrm{7v}^{\mathrm{2}} }{\mathrm{2}}\mathrm{sorry} \\ $$

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