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.
$$3x^2 +2xy +y^2 = 3(x + y/3)^2 + \frac{2 y^2}{3}$$ So try the substitution $u = \sqrt{3} x + y/\sqrt{3}, v = \sqrt{2/3} y$
This should transform it into an integral over the unit circle. Then you can use polar coordinates.
Referring to another answer and transforming the ellipse into:
$$(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2=1$$
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 $$\iint_{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2<1} e^{(2+\sqrt{2})X^2+(2-\sqrt{2})Y^2} dX \, dY =\int_{0}^{2\pi} \int_{0}^{1} \frac{e^{u^2}}{\sqrt{2}} u\, du \, dv$$
Let $g(x,y) = 3x^2+2xy+y^2$ and let $h(u) = e^u$. Notice that $h(g(x,y)) = e^{3x^2+2xy+y^2}$.
In this case, it can be a good idea to integrate along level surfaces. Let $A(u)$ denote the area of the region contained within $g(x,y) \leq u$.
This is basically like calculating the area of the ellipse. By using the transformation given by $r = \sqrt{3}x+y/\sqrt{3}$ and letting $p = \sqrt{2/3}y$, we get that our region in the $r-p$ plane is a circle with $r^2+p^2 \leq u$, whose area can easily be calculated as $\pi u$. Notice that this area becomes $\pi u / \sqrt{2}$ in the $x-y$ plane to account for Jacobian of our transformation.
Now $A(u) := \pi / \sqrt{2} u$ from above. We can then calculate the value of the double integral as a single integral of the form:
$$\int_{0}^{1} A'(u)h(u) du = \pi / \sqrt{2} \int_{0}^{1} e^u du = \pi (e-1) / \sqrt{2}$$
Which was the result that was sought.
