integrate ( x^{2}-xy + y^{2} ) over the region ( x^{2}-xy + y^{2}leq2 ).\nuse the transformation (…

Explanation:

Step1: Jacobian determinant calculation

The transformation is (x = \sqrt{2}u-\sqrt{\frac{2}{3}}v), (y=\sqrt{2}u + \sqrt{\frac{2}{3}}v). The Jacobian matrix (J=\begin{vmatrix}\frac{\partial x}{\partial u}&\frac{\partial x}{\partial v}\\frac{\partial y}{\partial u}&\frac{\partial y}{\partial v}\end{vmatrix}), where (\frac{\partial x}{\partial u}=\sqrt{2}), (\frac{\partial x}{\partial v}=-\sqrt{\frac{2}{3}}), (\frac{\partial y}{\partial u}=\sqrt{2}), (\frac{\partial y}{\partial v}=\sqrt{\frac{2}{3}}). [J=\sqrt{2}\times\sqrt{\frac{2}{3}}-\left(-\sqrt{\frac{2}{3}}\right)\times\sqrt{2}=2\sqrt{\frac{2}{3}}]

Step2: Transform the integrand and region

The integrand (x^{2}-xy + y^{2}): [ \begin{align*} x^{2}&=(\sqrt{2}u-\sqrt{\frac{2}{3}}v)^{2}=2u^{2}-\frac{4}{\sqrt{6}}uv+\frac{2}{3}v^{2}\ y^{2}&=(\sqrt{2}u+\sqrt{\frac{2}{3}}v)^{2}=2u^{2}+\frac{4}{\sqrt{6}}uv+\frac{2}{3}v^{2}\ xy&=(\sqrt{2}u-\sqrt{\frac{2}{3}}v)(\sqrt{2}u+\sqrt{\frac{2}{3}}v)=2u^{2}-\frac{2}{3}v^{2}\ x^{2}-xy + y^{2}&=(2u^{2}-\frac{4}{\sqrt{6}}uv+\frac{2}{3}v^{2})-(2u^{2}-\frac{2}{3}v^{2})+(2u^{2}+\frac{4}{\sqrt{6}}uv+\frac{2}{3}v^{2})\ &=2u^{2}+2v^{2} \end{align*} ] The region (x^{2}-xy + y^{2}\leq2) becomes (2u^{2}+2v^{2}\leq2), i.e., (u^{2}+v^{2}\leq1)

Step3: Polar - coordinate transformation

Use polar coordinates (u = r\cos\theta), (v=r\sin\theta), (dudv = rdr d\theta), and the region (u^{2}+v^{2}\leq1) corresponds to (0\leq r\leq1), (0\leq\theta\leq2\pi) The double - integral (\iint_{u^{2}+v^{2}\leq1}(2u^{2}+2v^{2})\vert J\vert dudv) Substitute (u = r\cos\theta), (v = r\sin\theta) and (\vert J\vert=2\sqrt{\frac{2}{3}}) [ \begin{align*} \iint_{u^{2}+v^{2}\leq1}(2u^{2}+2v^{2})\vert J\vert dudv&=2\sqrt{\frac{2}{3}}\int_{0}^{2\pi}\int_{0}^{1}2r^{2}\cdot rdr d\theta\ &=4\sqrt{\frac{2}{3}}\int_{0}^{2\pi}d\theta\int_{0}^{1}r^{3}dr \end{align*} ]

Step4: Evaluate the integrals

First, (\int_{0}^{2\pi}d\theta=2\pi), and (\int_{0}^{1}r^{3}dr=\left[\frac{r^{4}}{4}\right]{0}^{1}=\frac{1}{4}) [ \begin{align*} 4\sqrt{\frac{2}{3}}\int{0}^{2\pi}d\theta\int_{0}^{1}r^{3}dr&=4\sqrt{\frac{2}{3}}\times2\pi\times\frac{1}{4}\ &=2\sqrt{\frac{2}{3}}\pi \end{align*} ]

Answer:

(2\sqrt{\frac{2}{3}}\pi)