From $$\int_{-a}^{a} \,\int_{0}^{\sqrt{a^2-x^2}} \,(x^2+y^2)\,\text{e}^{x^4+2x^2y^2+y^4} \, \text{d}y \,\text{d}x=\int_{-a}^{a}\, \int_{0}^{\sqrt{a^2-x^2}} \,(x^2+y^2)\,\text{e}^{(x^2+y^2)^2} \, \text{d}y \, \text{d}x\,,$$ so $-a\leq x\leq a$ or $-a\sec{\theta}\leq r \leq a\sec{\theta}$, and since the region is between $y=0$ and $y=\sqrt{a^2-x^2} \ $ , so $0\leq \theta\leq \pi$.
I integrate $$\int_{0}^{\pi} \,\int_{-a\sec{\theta}}^{a\sec{\theta}} r^3\,\text{e}^{r^4} \, \text{d}r \, \text{d}\theta\,.$$ Solving it I get the area to be zero. If my calculations are correct, is it safe to assume that whenever I am integrating over a region that is symmetric against $y$-axis the area should be zero?
The limits of the integral in $r$ are wrong. The area of integration is half a disk of radius $a$ centered at the origin. The graph of $y=\sqrt{a^2-x^2}$ is the semicircle $x^2+y^2=a^2$, $y\ge0$. The region of integration is $0\le\theta\le\pi$, $0\le r\le a$.