Polar coordinates and extremes of integration

Question

I have trouble understanding how finding the extremes of integration of $\theta$ when I pass in polar coordinates.

1° example - Let $(X,Y)$ a random vector with density $f(x,y)=\frac{1}{2\pi}e^{-\frac{(x^2+y^2)}{2}}$.

Using the transformation $g=\left\{\begin{matrix} x=rcos\theta\\ y=rsin\theta \end{matrix}\right.$ and after calculating the determinant of Jacobian matrix, I have $dxdy=rdrd\theta$ from which

$\mathbb{E}[g(X^2+Y^2)]=\int_{\mathbb{R}^2}g(x^2+y^2)f(x,y)dxdy=\frac{1}{2\pi}\int_{0}^{+\infty}g(r^2)e^{-\frac{r^2}{2}}\int_{0}^{2\pi}d\theta$ $\Rightarrow X^2+Y^2\sim Exp(\frac{1}{2})$

2° example - Why for $\int\int_{B}\frac{y}{x^2+y^2}dxdy$ with $B$ annulus of centre $(0,0)$ and radius $1$ and $2$ the extremes of integration of $\theta$ are $(0,\pi)$?

3° example - Why for $\int\int_{B}\sqrt{{x^2+y^2}}dxdy$ with $B$ segment of circle $(0,0)$ and radius $1$ and $2$ the extremes of integration of $\theta$ are $(0,\frac{\pi}{2})$?

4° example - Why for $\int\int_{S}(x-y)dxdy$ with $S={((x,y)\in \mathbb{R}:x^2+y^2=r^2; y\geq 0)}$ the extremes of integration of $\theta$ are $(0,\pi)$?

I hope I have made clear my difficulties. Thanks in advance for any answer!

Answer 1

We have that

• for example $2$: we are dealing with one half of annulus ($1^{st}$ and $2^{nd}$quadrants)
• for example $3$: we are dealing with a quarter of annulus ($1^{st}$ quadrant)
• for example $4$: we are dealing with one half of circle ($2^{nd}$ and $3^{rd}$quadrants)

Answer 2

In your first example, you don't say what region you are integrating over. Since r is going from 0 to infinity and $\theta$ is going from 0 to $2\pi$ the integration is over the entire plane. Is that correct?

In your second example, your region of integration is "the annulus with center at the origin and radii 1 and 2" and ask "why are the extremes of integration of $\theta$ 0 and $\pi$?" They aren't! Since a full annulus goes "all the way around", $\theta$ will go from 0 to $2\pi$. (I see no mention of the "half annulus" that gimusi refers to.)

In example three, you are integrating only in the first quadrant. You should know that, in polar coordinates, the positive x-axis has $\theta= 0$. Since the y-axis is at right angles to the x-axis, the positive y-axis has $\theta= \frac{\pi}{2}$. We add $\frac{\pi}{2}$ as we go around each quadrant. The negative x-axis is $\theta= \frac{\pi}{2}+ \frac{\pi}{2}= \pi$ and the negative y-axis is $\theta= \pi+ \frac{\pi}{2}= \frac{3\pi}{2}$.