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!
Best Answer
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 $$\int_{\theta=0}^{\theta=2\pi}\int_{r=0}^{r=1} \left(18+r^2\right)\;r\;dr\;d\theta.$$