<s>[INST] Calculate $\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$ Calculate $\displaystyle\iint_D 3dxdy$ where $D = \{(x, y) : (x+y)^2 + (2x - y)^2 \le 4 \}$.
I tried to solve this can I failed.
Can you please give me some hints? 
 [/INST] Assume that $D:5x^2+2y^2-2xy=4$. Since $4-10<0$, $D$ is an ellipse and it is sufficient that we calculate area of it. Now, consider $A=\left[ \begin{array}{cc} 5 & -1  \\ -1 & 2 \end{array} \right]$. Let us call $\lambda_1$ and $\lambda_2$ the eigenvalues of $A$. One can see, $D$ in new coordinate is $\lambda_1X^2+\lambda_2Y^2=4$. So, the area of $D$ is $\pi\sqrt{\frac4{\lambda_1}}\sqrt{\frac{4}{\lambda_2}}=\frac{4\pi}{\sqrt{\lambda_1\lambda_2}}$. The product of the eigenvalues is the determinant of the matrix $A$. Therefore the solution is $$3\times\frac{4\pi}{\sqrt{9}}=4\pi.$$ 
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