I am working on a Calculus IV problem and just want to check if my argument is accurate ? Sorry if the question seems basic but I am self-studying Multiple Integral at the moment and would really appreciate any feedback.
$$\iint_S e^{-x}dxdy\quad \text{ where }\; S = \left\{(x,y) \in \mathbb{R}^2 : 0 < |y| < x^2 \right\}$$
Attempted Solution:
Notice $|y| < x^2 \iff -x^2 < y < x^2$, the area of integration is "sandwiched" between $-x^2$ and $x^2$. Because of symmetry, we have, $$\iint_S e^{-x}dxdy = 2\int_{0}^{\infty}\int_0^{x^2} e^{-x} dy dx = 4$$ (After two steps of integration by parts).
$$\int\int_S e^{-x}dxdy=\int_{-\infty}^{\infty}e^{-x}\Big(\int_{-x^2}^{x^2}dy\Big)dx=\int_{-\infty}^{\infty}2x^2e^{-x}dx$$
This integral is $+\infty$ since $\Big(\int_{-\infty}^{-1}+\int_{1}^{\infty}\Big)e^{-x}dx$ is $+\infty$
If you give extra condition $x>0$, then this integral $\int_{0}^{\infty}2x^2e^{-x}dx=2\Gamma(3)=2\times2!=4$.
Then $\int\int_S e^{-x}dxdy=4$
Note: $e^{-x}$ is not symmetric about $0$ function. $e^{-|x|}$ is.
$$\int\int_S e^{-|x|}dxdy=\int_{-\infty}^{\infty}e^{-|x|}\Big(\int_{-x^2}^{x^2}dy\Big)dx=\int_{-\infty}^{\infty}2x^2e^{-|x|}dx\\=2\cdot 2 \int_{0}^{\infty}x^2e^{-|x|}dx=4\cdot\Gamma(3)=8$$