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Question Number 34714 by abdo mathsup 649 cc last updated on 10/May/18

$$calculate\int {\int }_{x^2+2y^2⩽1}(x^2-y^2)dxdy$$

Commented by math khazana by abdo last updated on 10/May/18

$$letconsiderthediffeomorphism$$$$(r,\theta )\to (x,y)=(rcos\theta ,\frac{r}{\sqrt{2}}sin\theta )=({\phi }_1(r,\theta ),{\phi }_2(r,\theta ))$$$$M_j=\left(\begin{array}{c}\frac{\partial {\phi }_1}{\partial r}\frac{\partial {\phi }_1}{\partial \theta }\\ \frac{\partial {\phi }_2}{\partial r}\frac{\partial {\phi }_2}{\partial \theta }\end{array}\right)$$$$=\left(\begin{array}{c}cos\theta -rsin\theta \\ \frac{sin\theta }{\sqrt{2}}\frac{r}{\sqrt{2}}cos\theta \end{array}\right)anddet\left(M_j\right)=$$$$\frac{r}{\sqrt{2}}{cos}^2\theta +\frac{r}{\sqrt{2}}{sin}^2\theta =\frac{1}{\sqrt{2}}r$$$$I=\int {\int }_{0⩽r⩽1,-\pi ⩽\theta ⩽\pi }(r^2{cos}^2\theta -\frac{r^2}{2}{sin}^2\theta )\frac{r}{\sqrt{2}}drd\theta$$$$I=\frac{1}{\sqrt{2}}{\int }_0^1{r}^{3}dr{\int }_{-\pi }^{\pi }{cos}^2\theta d\theta -\frac{1}{2\sqrt{2}}{\int }_0^1{r}^{3}dr{\int }_{-\pi }^{\pi }{sin}^2\theta d\theta$$$$=\frac{1}{4\sqrt{2}}{\int }_{-\pi }^{\pi }{cos}^2\theta d\theta -\frac{1}{8\sqrt{2}}{\int }_{-\pi }^{\pi }{sin}^2\theta d\theta but$$$${\int }_{-\pi }^{\pi }{cos}^2\theta d\theta =2{\int }_0^{\pi }\frac{1+cos\left(2\theta \right)}{2}d\theta$$$$={\int }_0^{\pi }(1+cos\left(2\theta \right))d\theta =\pi$$$${\int }_{-\pi }^{\pi }{sin}^2\theta d\theta =2{\int }_0^{\pi }\frac{1-cos\left(2\theta \right)}{2}d\theta$$$$={\int }_0^{\pi }(1-cos\left(2\theta \right))d\theta =\pi \Rightarrow$$$$I=\frac{\pi }{4\sqrt{2}}-\frac{\pi }{8\sqrt{2}}=\frac{2\pi -\pi }{8\sqrt{2}}=\frac{\pi }{8\sqrt{2}}$$$$I=\frac{\pi }{8\sqrt{2}}.$$

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