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?
[Math] Find the area under the curve: $r = 2 \sin (3\theta)$
calculusintegrationmultivariable-calculuspolar coordinates
Related Solutions
There are two distinct regions where your curves overlap. They do overlap for the intervals of $\theta$ that you give, but that gives only the large overlap at the upper right of the origin. You left out the smaller overlap at the lower left of the origin.
That overlap is covered by $\frac{5\pi}4\le \theta\le \frac{3\pi}2$ for the $r=1+\cos\theta$ curve and by $\pi\le \theta\le \frac{5\pi}4$ for the $r=1+\sin\theta$ curve. To get the correct answer you will need to include that little piece.
Area of a curve given in polar coordinate is given by $$\rm{A} = \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
We can obtain this formula by dividing the curve into infinitely many thin sectors. (each subtending angle $\rm d \theta$ at the origin)
Then use the formula of area of sector to obtain $$dA=\frac{1}{2} r^2 d \theta$$
Now integrate both the sides,
$$\int \rm dA=A= \frac 12\int_{\theta_1}^{\theta_2} r^2(\theta) \:\rm{d} \theta$$
Best Answer
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.