Electric charge on a disk

Question

Electric charge is distributed over the disk $x^2+y^2\leq1$ so that the charge density at $(x,y)$ is $\sigma(x,y)=18+x^2+y^2$ coulombs per square meter.

How can I find the total charge on the disk?

I know that I have to use the integral but setting the limits is my issue!

Answer 1

The total charge would be the double integral of the density function over the region D which is the unit disk. By using $x = r\cos(t)$, $y = r\sin(t)$, we easily get

$$\int_0^{2\pi}\int_0^1(18 + r^2)\space r\space dr\space dt = \frac{37\pi}2$$

Answer 2

You have several choices about how to handle the limits of integration.

The most straightforward thing to do is to observe that for any given $x$, the $y$-coordinate of a point in the disk must lie between $-\sqrt{1-x^2}$ and $+\sqrt{1-x^2}$, since $y$ outside this range makes $x^2+y^2 > 1$. So the integral is:

$$\int_{x=-1}^{x=1}\int_{y={-\sqrt{1-x^2}}}^{y=\sqrt{1-x^2}} \sigma(x,y)\;dy\;dx$$

(Note the minus sign on the lower limit of integration.)

You can simplify this a bit by calculating the charge just for the upper half of the disk, and multiplying that by 2. This works because the charge density function is symmetric across the $x$-axis. (That is, $\sigma(x, -y) = \sigma(x, y)$.) Then the integral is $$\color{blue}{2}\int_{x=-1}^{x=1}\int_{\color{blue}{y=0}}^{y=\sqrt{1-x^2}} \sigma(x,y)\;dy\;dx$$

Or similarly you could cut the region into fourths:

$$\color{blue}{4}\int_{\color{blue}{x=0}}^{x=1}\int_{{y=0}}^{y=\sqrt{1-x^2}} \sigma(x,y)\;dy\;dx$$

But probably the best thing do to is to transform the problem into polar coordinates, because it is circularly symmetric, and then the integral becomes $$\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1} \left(18+r^2\right)\;r\;dr\;d\theta.$$