Area under a normal distribution: Why is my answer wrong?

I was presented to the following integral:

$$(1)$$ $$\int_{-\infty}^{\infty}e^{-x^{2}}dx$$

Let's call the value of this integral $$A$$ for "Answer".

I was told that it would help to think of a tridimensional analog to the curve in $$(1)$$ by thinking about this other integral:

$$(2)$$ $$\iint_R e^{-(x^2+y^2)}dA$$, where $$R$$ is the whole $$xy$$ plane

So I made the following substitution to polar coordinates:

$$(3)$$ $$\int_0^{2\pi} \int_0^\infty e^{-r^2} rdrd\theta$$

Which indeed got me the correct result for $$(2)$$, which is $$\pi$$.

Then this is how my thinking went:

Well, the area that $$(1)$$ is computing (i.e, the area under the curve $$y = e^{-x^{2}}$$) is just an infinitesimally thin slice of $$(3)$$ (more precisely, it's the area under the curve that lies on the intersection of the plane $$y=0$$ with the surface $$z=e^{-(x^2+y^2)}$$).

Because of rotational symmetry, it must be the case that if I were to generate a solid by rotating the area described by $$(1)$$ around the $$y$$-axis, I will get a solid whose volume is $$\pi$$, because that's what $$(3)$$ was computing.

That is, rotating the whole area of $$(1)$$ around the $$y$$-axis is the same as computing $$\int_0^{2\pi}\int_0^{\infty}e^{-r^2} rdrd\theta$$

But the inner integral $$\int_0^{\infty}e^{-r^2} rdr$$ is computing the area under the surface from the origin to infinity (if I understand polar coordinates well), which is half the area of $$(1)$$, since in $$(1)$$ the area goes from negative infinity to positive infinity.

So it must be the case that $$\int_0^\infty e^{-r^2}rdr$$ is half the value of $$A$$.

So it must be true that:

$$(4)$$ $$\int_0^\infty e^{-r^2}rdr$$ = $$\frac{A}{2}$$

Substituing this into $$(3)$$ we get:

$$(5)$$ $$\int_0^{2\pi} \frac{A}{2} d\theta= \pi$$

Solving for $$A$$, we get $$A = 1$$, which is wrong. The correct answer is $$A = \sqrt{\pi}$$

I was then presented to the correct way of solving the problem, but I still couldn't see where my flaw was. Could someone clarify?

• You have $\int \mathrm{e}^{-r^2}dr$ and $\int \mathrm{e}^{-r^2}rdr$ being equivalent. Apr 25 '21 at 21:03
• Fixed that typo. It was not part of the assumptions I made. Apr 25 '21 at 21:13
• It doesn’t matter if $r$ is radius and $x$ is Cartesian - the fact you are still integrating along $r$ is the same way so $\int_0^\infty f(x) dx = \int_0^\infty f(r)dr$ if you have the same function. Further more $\int r\mathrm{e}^{-r^2}dr =-\frac{1}{2} \int \frac{d}{dr} \mathrm{e}^{-r^2}dr$ which you don’t need me to tell you what that is. Apr 25 '21 at 21:35

$$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)/2} \, dx \, dy = \left(\int_{-\infty}^\infty e^{-x^2/2} \, dx\right) \left(\int_{-\infty}^\infty e^{-y^2/2} \, dy\right)$$
• @DiegoOliveira As Chinny84 points out, the inner integral $\int_0^\infty e^{-r^2/2} r \, dr$ is not the area under the surface of $e^{-(x^2+y^2)}$ along some ray to infinity; it is the area under $e^{-(x^2+y^2)/2} \sqrt{x^2+y^2}$ along the ray, which is altogether different. The extra $r$ only enters as a scaling factor to compute the area of a little wedge at radius $r$ and angle $d\theta$ when making the rotation argument, and has no interpretation when thinking about the original integrand $e^{-(x^2+y^2)/2}$ along a ray. Apr 25 '21 at 22:05