# Find the area under the curve: $r = 2 \sin (3\theta)$

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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.

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### Aemilius

Updated on June 11, 2020

• Aemilius over 3 years

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?

• Aemilius over 5 years
I don't know this formula. What's wrong with my solution?
• Mohammad Riazi-Kermani over 5 years
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.
• Mohammad Riazi-Kermani over 5 years
@Aemilius You may fix your method by replacing your $drd\theta$ with $rdrd\theta$
• Aemilius over 5 years
But of course, I need to use the Jacobian! now i get it!