How to find polar coordinates

Question

I am having some trouble figuring out how to get the polar coordinates for the following integral:

$\int e^{-x^2} dx$ on $\Bbb R=[0,\infty) \times [0,\infty) $

First I have found g such that g'= $e^{-x^2}$

So now I have

$\int e^{-(x^2+y^2)} dx\,dy$

I know that $r{^2}$ = $x^2+y^2$

so

$\int e^{-r^2} r drdθ$

I am not sure how I can find the polar coordinates for this.

The answer says that the integral is from $0< r<\infty$ and $0<\theta<\pi/2$ but I don't know how to get that answer.

Answer 1

Your question is why

$$\int \int_ {[0,\infty)\times [0,\infty)} e^{-(x^2+y^2)} dxdy =\int _{0}^{\infty} \int _{0}^{\infty}e^{-(x^2+y^2)} dxdy$$

$$\int_{0}^{\pi/2} \int _{0}^{\infty}e^{-r^2} r drdθ$$

Well, the first equality is by Fubini's Theorem which describes double integrals into iterated integras.

The second equality is describing the first quadrant, in polar coordinates.

If you visulize the first quadrant as the interior of a quarter circle with radius infinity and the center at the origin, you can see the range of $0<\theta<\pi /2$ and $0<r<\infty$.

Answer 2

You have to compute the double integral $$\iint_{\mathbf R_+\times\mathbf R_+}\mkern-27mu\mathrm e^{-x^2-y^2}\mathrm d x\,\mathrm d y=$$ in polar coordinates, and apply Fubini's theorem.

Answer 3

$$I=\int_0^\infty e^{-x^2} dx$$ Since all the values are positive, we can calculate $I^2$ instead, then take the square root $$I^2=\left(\int_0^\infty e^{-x^2} dx\right)\left(\int_0^\infty e^{-y^2} dy\right)$$ I choose different variable name for the second term. Now we can write this as a double integral: $$I^2=\int_0^\infty\int_0^\infty e^{-(x^2+y^2)} dx\ dy$$ Now we transform in polar coordinated. $x^2+y^2=r^2$ and $dxdy=r dr d\theta$. Also the limits are now $r\in[0,\infty)$ and $\theta\in[0,\pi/2]$ So $$I^2=\int_0^\infty rdr\int_0^{\pi/2} d\theta e^{-r^2} $$ Notice that the exponential does not depend on $\theta$, so we can take that outside, and using $$\int_0^{\pi/2}d\theta=\frac{\pi}{2}$$ you get $$I^2=\frac{\pi}{2}\int_0^\infty e^{-r^2}rdr$$ We can now change the variable $r^2=z$ and $2rdr=dz$ so $$I^2=\frac{\pi}{4}\int_0^\infty dz e^{-z}=\frac{\pi}{4}$$ Therefore $I=\frac{\sqrt\pi}{2}$