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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

enter image description here

At this point I'm more concerned with understanding where I went wrong than with the right answer....any ideas?

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  • $\begingroup$ Second angle you should obtain from $\tan \theta =3$. $\endgroup$ Commented Oct 31, 2021 at 1:10
  • $\begingroup$ I had a typo in the OP (which I've fixed) I originally had arccos(sqrt(9)/3) when it should have read arccos(sqrt(0.9)/3) $\endgroup$ Commented Oct 31, 2021 at 1:21
  • $\begingroup$ I wrote in the answer which curve turns into which one and which integral is obtained $\endgroup$ Commented Oct 31, 2021 at 1:23
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    $\begingroup$ Also you missed Jacobian for polar coordinates in integrand. $\endgroup$ Commented Oct 31, 2021 at 1:28
  • $\begingroup$ arccos(sqrt(0.9)/3) and arctan(3) are the same, so at least I got the bounds correct. I added the screenshot from Wolfram Alpha...I've been staring at this so long I'm afraid I'm making a very silly mistake $\endgroup$ Commented Oct 31, 2021 at 1:33

1 Answer 1

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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 enter image description here

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