Converting bounds of double integral to polar coordinates

Question

I'm trying to convert the following to polar coordinates:

$$\int_0^\infty \int_{-\infty}^{-x}\frac{1}{2\pi}e^{-(x^2+y^2)/2}\,dx\,dy$$

After converting to polar coordinates, it should be:

$$\int_0^\infty \int_{(3/2)\pi}^{(7/4)\pi}\frac{1}{2\pi}e^{-r^2/2}\,dr\,d\theta$$

My question is how do we arrive at the bounds of $(3/2)\pi$ and $(7/4)\pi$.

Answer 1

The first integral should be

$$\large\int_{0}^{\infty}dx\int_{-\infty}^{-x}\frac{1}{2\pi}e^{-(x^2+y^2)/2}dy$$

wich represent the integral over the half of the 4 quadrant between $y$ axis and $y=-x$ that is for $\theta$ between $3\pi/2$ and $7\pi/4$.

Note also that here

$$\large\int_{0}^{\infty}dr\int_{(3/2)\pi}^{(7/4)\pi}\frac{1}{2\pi}\color{red}re^{-r^2/2}d\theta$$

we need an "$r$" extra term for polar coordinates jacobian wich makes this kind of integral easy to be solved by polar coordinates.

Answer 2

You wrote $$ \int_0^\infty \int_{-\infty}^{-x} \cdots \,dx\,dy. $$ That makes no sense: It says $x$ goes from something to $-x.$ From the polar form of your integral, it becomes clear that what you should have written instead is $$ \int_0^\infty \int_{-\infty}^{-x} \cdots \,dy\,dx. $$ Without attention to this sort of thing, you will not be able to understand how to get the proper bounds in polar form.

View the integral like this: $$ \int_0^\infty \left( \int_{-\infty}^{-x} \cdots \, dy \right) dx $$ The variable $x$ goes from $0$ to $\infty$ and that tells you that you're working in the right half of the plane. Then for each fixed value of $x$ you have $y$ going from $-\infty$ to $-x.$ Draw a picture of that. The line $y=-x$ goes through the origin and is at a $45^\circ$ angle to either the $x$-axis or the $y$-axis, and goes through the fourth quadrant. Since you're going from $h=-\infty$ to $y=-x,$ your going through all space below that line, up to that line. Again: draw the picture.

That tells you $\theta$ is going from $3\pi/2$ to $7\pi/4.$

Note that $dx\,dy$ needs to get replaced by $r\,dr\,d\theta.$ That is because in the direction at right angles to the radial direction, the element of distance is $r\,d\theta.$