integrate $x^{2}-xy + y^{2}$ over the region $x^{2}-xy + y^{2}leq2$. use the transformation…

Explanation:

Step1: Calculate the Jacobian

The Jacobian matrix (J) of the transformation (x = \sqrt{2}u-\sqrt{\frac{2}{3}}v), (y=\sqrt{2}u+\sqrt{\frac{2}{3}}v) is: [ 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}=\begin{vmatrix} \sqrt{2}&-\sqrt{\frac{2}{3}}\ \sqrt{2}&\sqrt{\frac{2}{3}} \end{vmatrix}=\sqrt{2}\times\sqrt{\frac{2}{3}}-\left(-\sqrt{\frac{2}{3}}\times\sqrt{2}\right)=\frac{4}{\sqrt{3}} ]

Step2: Transform the integrand and the region

Substitute (x) and (y) into (x^{2}-xy + y^{2}): [ \begin{align*} x^{2}-xy+y^{2}&=(\sqrt{2}u-\sqrt{\frac{2}{3}}v)^{2}-(\sqrt{2}u-\sqrt{\frac{2}{3}}v)(\sqrt{2}u+\sqrt{\frac{2}{3}}v)+(\sqrt{2}u+\sqrt{\frac{2}{3}}v)^{2}\ &=(2u^{2}-\frac{4}{\sqrt{3}}uv+\frac{2}{3}v^{2})-(2u^{2}-\frac{2}{3}v^{2})+(2u^{2}+\frac{4}{\sqrt{3}}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) or (u^{2}+v^{2}\leq1)

Step3: Set up and evaluate the double - integral

Using polar coordinates (u = r\cos\theta), (v = r\sin\theta), (dA = |J|dudv=\frac{4}{\sqrt{3}}dudv) and (u^{2}+v^{2}=r^{2}) The double - integral (\iint_{R}(x^{2}-xy + y^{2})dxdy=\iint_{S}(2u^{2}+2v^{2})\times\frac{4}{\sqrt{3}}dudv) [ \begin{align*} &=\frac{8}{\sqrt{3}}\int_{0}^{2\pi}\int_{0}^{1}r^{2}\times rdr d\theta\ &=\frac{8}{\sqrt{3}}\int_{0}^{2\pi}d\theta\int_{0}^{1}r^{3}dr\ &=\frac{8}{\sqrt{3}}\times2\pi\times\left[\frac{r^{4}}{4}\right]_{0}^{1}\ &=\frac{4\pi}{\sqrt{3}} \end{align*} ]

Answer:

(\frac{4\pi}{\sqrt{3}})