Relation between the volume under $e^{-(x^2+y^2)}$ and the area under $e^{-x^2}$.

Question

I've been having some trouble identifying where my reasoning is wrong in this problem.

I have 2 premises:

$\int_{-\infty}^\infty e^{-x^2} dx=\sqrt\pi=A $

$V=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{-x^2-y^2} dx dy=\pi $

Being A the area under the function $y=e^{-x^2}$

and V the volume under the function $z= e^{-x^2-y^2}$

$\downarrow$ $\mathbf {Here's\;the\;BIG\;error}$

So, if $e^{-x^2-y^2}$ is the solid of revolution of $e^{-x^2}$ and the volume of a revolution solid is equal to pi times the area of the section why isn't $V=\pi A$?

$\uparrow$ $\mathbf {Here's\;the\;BIG\;error}$

Where is the error?

Answer 1

Revolving around the $y$ axis gives $\pi\int_0^1 x^2dy=-\pi\int_0^1 \ln ydy=-\pi\left[y\ln y-y\right]_0^1=\pi.$

Answer 2

HINT:

Amplitude $a$ of Bell curve and standard deviation ( inflection radius $\sigma$) in

$$ r= \sqrt{x^2+y^2} = a e^{-z^2/ {2\sigma^2}}$$

should be taken to obtain area and voume.

$$ A= \int_{-\infty}^\infty a e^{-z^2/ {2\sigma^2} }dz $$

$$ V=\int_{-\infty}^\infty \pi r^2 dz$$

Please check integrated area and volume to have physical dimensions $(2,3)$ respectively.