# calculus , double integral at ellipse region

Let S be the region satisfying $3x^2 + 2xy + y^2 \leq\ 1$ in the real plane $R^{2}$. Then compute the value of the double integral

$$\int\int_S\ e^{(3x^2+2xy+y^2)}dxdy$$

i learned double integral at polar coordinates. but i didn't solve this problem. Please give me hint or answer.

• You can use something similar to polar coordinates but transformed in such a way that the circle becomes the ellipse – Paul Castle Jul 11 '16 at 6:42

$$3x^2 +2xy +y^2 = 3(x + y/3)^2 + \frac{2 y^2}{3}$$ So try the substitution $u = \sqrt{3} x + y/\sqrt{3}, v = \sqrt{2/3} y$
$$(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2=1$$
Let $(X,Y)=\displaystyle \left( \frac{u\cos v}{\sqrt{2+\sqrt{2}}}, \frac{u\sin v}{\sqrt{2-\sqrt{2}}} \right)$, then $dX \, dY=\displaystyle \frac{u\, du \, dv}{\sqrt{2}}$
Now $$\iint_{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2<1} e^{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2} dX \, dY =\int_{0}^{2\pi} \int_{0}^{1} \frac{e^{u^2}}{\sqrt{2}} u\, du \, dv$$