Is the set $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$ bounded? 
Is the set $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$ bounded?

My thoughts:
Let's consider the function $g:\Bbb R\to\Bbb R, g(x)=e^x-x-2.$ Checking its derivative, we see $g$ is strictly increasing on $[0,+\infty)$. Furthermore, $g(0)=-1$ and $\lim\limits_{x\to+\infty}g(x)=+\infty,$ which means $g$ changes signs and has a unique root, call it $r_0$ in $[0,+\infty).$
My conclusion is that $S$ is a circle with a radius $r_0$ centered at the origin, hence bounded.
I'm sceptical about my deduction and answer. Is there any room for improvement?
 A: What you did are all correct. But if all you want is the boundedness, then $\lim_{x\rightarrow\infty} g(x)=+\infty$ is all you need. Indeed, this implies that there exists $r>0$, such that for all $x>r$, $g(x)>0$. Hence any point $(x,y)$ on $S$ must satisfy $x^2+y^2\le r$, thus $S$ is bounded.
This is helpful for more complicated problems. For example, let $S=\{(x,y)\in\mathbb R^2 : e^{x^2+y^2} = 2 + x^2 + 10000y\}$ which is not any familiar curve. We still consider $g$, and use the fact that when both $y$ and $x^2+y^2$ are sufficiently large, $e^{x^2+y^2}>2+x^2+y^2>2+x^2+10000y$. Hence for $(x,y)$ on $S$, $x^2+y^2$ (and $y$) has to be small.
It's often beneficial to think roughly than precisely in analysis.
A: Looks good other than the crime of not converting $x^2 + y^2$ to polar coordinates. Substituting $x^2 + y^2$ for x isn't something I'd do.
Let $x=r\cos(\theta), y=r\sin(\theta)$
$\implies x^2 + y^2 = r^2\cos^2(\theta) + r^2\sin^2(\theta) = r^2(\cos^2(\theta) + \sin^2(\theta))=r^2$.
Then $S=\{(x,y)\in\Bbb R^2\mid e^{x^2+y^2}=2+x^2+y^2\}$
$=\{(r, \theta) \in\Bbb R^2 \mid e^{r^2}=2+r^2\}$.
Let $g(r)=e^{r^2}-2-r^2$
$\implies g'( r) = e^{r^2}\cdot 2r - 2r = 2r(e^{r^2} -1)$.
Since $\forall r\ge 0,  g'(r)\ge0$
$\implies S$ is bounded.
But I'm open to others' opinions.
