Find the area under the curve: $r = 2 \sin (3\theta)$ using the polar coordinate system.
Since we are working in the polar coordinate system, the radius may not be negative, and so we will need to take the collective area from three regions of integration: $[0, \pi/3] \cup[2\pi/3, \pi] \cup [4\pi/3, 5\pi/3]$.
Since this region is not normal with respect to $\theta(r)$, it can only be iterated with respect to $r(\theta)$.
And so, the integral will be: $$3\int_{\theta=0}^{\theta= \pi/3}\int_{r=0}^{r=2 \sin(3\theta)}drd\theta$$ Have I set up the correct integral for this problem?


Since you are working with polar coordinates you need to use the correct formula for the area traced by your function.

The formula is $$ A= 1/2 \int r^2(\theta ) d\theta $$

Your function is $$r(\theta ) = 2 \sin (3\theta)$$

Be careful with the limits of integration for the area of each loop.

  • $\begingroup$ I don't know this formula. What's wrong with my solution? $\endgroup$
    – Aemilius
    Jun 10 '18 at 12:59
  • $\begingroup$ Your formula works for rectangular coordinates. You need to learn the formula which I gave you for the area of polar curves. It is not too complicated, just integrate your $r ^2(\theta) $ from $0$ to $\pi/3$ for one loop. $\endgroup$ Jun 10 '18 at 13:05
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    $\begingroup$ @Aemilius You may fix your method by replacing your $drd\theta$ with $ rdrd\theta $ $\endgroup$ Jun 10 '18 at 13:09
  • $\begingroup$ But of course, I need to use the Jacobian! now i get it! $\endgroup$
    – Aemilius
    Jun 10 '18 at 13:24

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