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

Commented by witcher3 last updated on 03/Jun/23

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

Commented by 073 last updated on 03/Jun/23

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

Commented by witcher3 last updated on 03/Jun/23

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

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