Integral of Heaviside($R^2-x^2-y^2$)

Question

I have this integral (where $R$ is a positive constant) $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} H(R^2-x^2-y^2)dxdy$$ and I'm pretty lost when trying to calculate it.

I don't know if it's smart to do so, but I tried to convert it to polar coordinates and got $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty}H(R^2-r^2)rdrd\theta$$ By definition the Heavside function should be one everywhere, where $r<R$ so i thought about writing the following $$\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{R}rdrd\theta$$ But now I don't know what to do next or if this has been correct so far. Cause the result of this would be just infinity I guess... I would be really grateful if someone could give me a little help or a hint on how to approach this. Thanks!

Answer 1

The function you are integrating is zero outside the circle centered in $(0,0)$ with radius $R$ and is one inside that circle. Therefore, the integral corresponds to the area of the circle: $\pi R^2$. No need for polar coordinates. But if you really want to use them,

$$ \iint_{\mathbb{R}^2} H(R^2-x^2-y^2) dxdy = \int_{-R}^R \int_{-\sqrt{R^2-x^2}}^{\sqrt{R^2-x^2}} dxdy = \int_{0}^{2\pi} \int_0^R \rho d\rho d\theta = 2 \pi [\frac 12 \rho^2]_0^R = \pi R^2. $$

Answer 2

When you pass to the polar coordinates, you have to change the limits appropriately: when $x$ and $y$ run from $-\infty$ to $\infty$, the radius $r$ runs from $0$ to $\infty$ and the polar angle $\theta$ runs from $0$ to $2\pi$ (or from $-\pi$ to $\pi$, say). Adjusting your calculations accordingly will allow you to get the correct answer, which will probably look somewhat familiar.

Answer 3

$$I=\int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} H(R^2-x^2-y^2)dx dy =\int_{0}^{2\pi} \int_{0}^{R}H(R^2-r^2) r dr d\theta=\int_{0}^{2\pi} \int_{0}^{R} r dr d\theta= \pi R^2$$