Originally Posted by

**HallsofIvy** The "2- 6y" is the correct integrand. As for the limits of integration, "$\displaystyle \pi$" would apply **if** you were integrating in polar coordinates!

Or, if you really wanted to (I don't recommend it) you could integrate in polar coordinates: the angle $\displaystyle \theta$ goes from 0 to $\displaystyle \pi$ and r goes from 0 to a, independently. The "hard part" is that you need to convert the integrand into polar coordinates. $\displaystyle x- 6y= rcos(\theta)- 6rsin(\theta)= r(cos(\theta- 6sin(\theta)$. And, of course, the "differential of area" in polar coordinates is $\displaystyle rdrd\theta$

$\displaystyle \int_0^\pi\int_0^a r^2(cos(\theta)- 6sin(\theta))drd\theta= \int_0^\pi (cos(\theta)- 6sin(\theta))d\theta\int_0^a r^2 dr$