Translating to polar I got a double integral with r=0 to 3 and theta = 0 to arccos(sqrt(0.9)/3) of
5rcos(theta)*rsin(theta)-rsin(theta)
And when I throw that in Wolfram alpha to evaluate I get ~14 which it says is wrong when I submit it
At this point I'm more concerned with understanding where I went wrong than with the right answer....any ideas?
Using polar coordinates $x = r\cos \theta, y = r \sin \theta$ we have, that circle $x^2+y^2=9$ goes to $r=3$. $y=0$ gives $\theta = 0$ and $y=3x$ gives $\theta =\arctan 3$, so for integral in first quadrant we have $$\int\limits_{0}^{\arctan 3}\int\limits_{0}^{3}r\cdot(5r^2\sin \theta \cos \theta-r \sin \theta)drd\theta$$
Addition