4. Evaluate \( \iint_{R}\left(x^{2}-x y+y^{2}\right) d A \), where \( R \) is the region bounded by the ellipse \( x^{2}-x y+y^{2}=2 \) by using the transformation \( x=\sqrt{2} u-\sqrt{2 / 3} v, \quad y=\sqrt{2} u+\sqrt{2 / 3} v \)

We start with the integral \[ I=\iint_{R}\left(x^{2}-xy+y^{2}\right)\,dA, \] where \( R \) is the region bounded by the ellipse \[ x^{2}-xy+y^{2}= 2. \] A transformation is given by \[ x=\sqrt{2}\,u-\sqrt{\frac{2}{3}}\,v,\quad y=\sqrt{2}\,u+\sqrt{\frac{2}{3}}\,v. \] We'll follow these steps: --- **1. Expressing the Quadratic Form in \((u,v)\):** Define \[ A=\sqrt{2}\,u \quad \text{and} \quad B=\sqrt{\frac{2}{3}}\,v. \] Then \[ x=A-B,\quad y=A+B. \] Compute \[ \begin{aligned} x^2 &= (A-B)^2 = A^2-2AB+B^2,\\[1mm] y^2 &= (A+B)^2 = A^2+2AB+B^2,\\[1mm] xy &= (A-B)(A+B)= A^2-B^2. \end{aligned} \] Substitute these into the quadratic form: \[ \begin{aligned} x^2-xy+y^2 &= \Bigl[A^2-2AB+B^2\Bigr] - \Bigl[A^2-B^2\Bigr] + \Bigl[A^2+2AB+B^2\Bigr]\\[1mm] &= \left(A^2-2AB+B^2 -A^2+B^2+A^2+2AB+B^2\right)\\[1mm] &= A^2 + 3B^2. \end{aligned} \] Recall that \[ A^2 = 2u^2 \quad \text{and} \quad B^2=\frac{2}{3}v^2. \] Thus, \[ x^2-xy+y^2 = 2u^2+ 3\left(\frac{2}{3}v^2\right)= 2u^2+2v^2= 2\left(u^2+v^2\right). \] --- **2. Transforming the Region:** The original region is defined by \[ x^2-xy+y^2=2. \] Using the new variables, this becomes \[ 2\left(u^2+v^2\right)=2 \quad \Longrightarrow \quad u^2+v^2=1. \] Thus the region transforms into the unit disk in the \((u,v)\)-plane. --- **3. Computing the Jacobian:** The transformation is \[ x=\sqrt{2}\,u-\sqrt{\frac{2}{3}}\,v,\quad y=\sqrt{2}\,u+\sqrt{\frac{2}{3}}\,v. \] The Jacobian matrix is \[ J=\begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\[1mm] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix}= \begin{pmatrix} \sqrt{2} & -\sqrt{\frac{2}{3}}\\[1mm] \sqrt{2} & \sqrt{\frac{2}{3}} \end{pmatrix}. \] Its determinant is \[ \begin{aligned} \det(J)&=\sqrt{2}\cdot\sqrt{\frac{2}{3}} - \left(-\sqrt{\frac{2}{3}}\cdot\sqrt{2}\right)\\[1mm] &=\sqrt{2}\sqrt{\frac{2}{3}}+\sqrt{2}\sqrt{\frac{2}{3}} =2\sqrt{2}\sqrt{\frac{2}{3}}. \end{aligned} \] Since \[ \sqrt{2}\sqrt{\frac{2}{3}}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}, \] the determinant becomes \[ \det(J)=2\cdot\frac{2}{\sqrt{3}}=\frac{4}{\sqrt{3}}. \] Thus, the area element transforms as \[ dA = \left|\det(J)\right|\,du\,dv=\frac{4}{\sqrt{3}}\,du\,dv. \] --- **4. Rewriting the Integral:** The integrand becomes \[ x^2-xy+y^2=2\left(u^2+v^2\right), \] so the integral is \[ \begin{aligned} I &=\iint_{R} \left(x^2-xy+y^2\right)\,dA\\[1mm] &=\iint_{u^2+v^2\le 1} 2\left(u^2+v^2\right)\cdot\frac{4}{\sqrt{3}}\,du\,dv\\[1mm] &=\frac{8}{\sqrt{3}} \iint_{u^2+v^2\le 1}\left(u^2+v^2\right)\,du\,dv. \end{aligned} \] --- **5. Converting to Polar Coordinates:** Let \[ u=r\cos\theta,\quad v=r\sin\theta, \] with \(r\in [0,1]\) and \(\theta\in [0,2\pi]\). The area element becomes \(du\,dv = r\,dr\,d\theta\) and \[ u^2+v^2=r^2. \] Thus the integral becomes \[ \begin{aligned} I &=\frac{8}{\sqrt{3}} \int_{0}^{2\pi}\int_{0}^{1} r^2\cdot r\,dr\,d\theta\\[1mm] &=\frac{8}{\sqrt{3}}\int_{0}^{2\pi}\int_{0}^{1} r^3\,dr\,d\theta. \end{aligned} \] Evaluate the \(r\)-integral: \[ \int_{0}^{1} r^3\,dr=\left.\frac{r^4}{4}\right|_{0}^{1}=\frac{1}{4}. \] Then the \(\theta\)-integral: \[ \int_{0}^{2\pi} d\theta=2\pi. \] So we have \[ I=\frac{8}{\sqrt{3}}\cdot\frac{1}{4}\cdot2\pi = \frac{8\cdot2\pi}{4\sqrt{3}} = \frac{16\pi}{4\sqrt{3}} = \frac{4\pi}{\sqrt{3}}. \] --- **Final Answer:** \[ \boxed{\frac{4\pi}{\sqrt{3}}} \]