I have this graph here, but how do I find out the area that's inside the circle but outside the 3sin2θ rose shaped curve? I'm lost, please help me solve this
How to determine the area that is inside the circle r=3cos(θ) and outside the r=3sin(2θ) curve (For the first quadrant)
calculusintegrationpolar coordinatestrigonometry
Related Solutions
If $$r_1(\theta) = 3a \cos \theta$$ is the equation of the circle, and $$r_2\theta) = a(1+\cos \theta)$$ is the equation for the cardioid, then these curves intersect when $r_1(\theta) = r_2(\theta)$, or $3a \cos \theta = a(1+\cos \theta)$, or $$2 \cos \theta = 1.$$ This gives us the intersection points $$\theta = \pm \frac{\pi}{3}.$$ There is another point of intersection at the origin, because the circle is traversed twice for a single traversal of the cardioid. But for our purposes, we can integrate on $\theta \in [-\pi/3, \pi/3]$. The area enclosed between $r_1$ and $r_2$ is then given by $$A = \int_{\theta = -\pi/3}^{\pi/3} \frac{1}{2}\left(r_1^2(\theta) - r_2^2(\theta)\right) \, d\theta.$$ Note this is not the same as $$\int_{\theta = -\pi/3}^{\pi/2} \frac{1}{2}\left(r_1(\theta) - r_2(\theta)\right)^2 \, d\theta.$$ The second integral is incorrect for the area enclosed.
Note that the integral limits for the circle is $[-\frac\pi2, \frac\pi2]$. Thus, the integral should be set up as
$$\frac12\int^{\pi}_{-\pi}\left(1+\cos\theta\right)^2d\theta -\frac12\int^{\pi/2}_{-\pi/2}\cos^2\theta\ d\theta = \frac{5\pi}4$$
Best Answer
The area of a region in polar coordinates defined by the equation $r=f(θ)$ with $\alpha\leq\theta\leq\beta$ is given by:$$\int_{\alpha}^{\beta} \frac{1}{2}{r}^2 \; d\theta$$
The area of a region between two functions in polar coordinates is just the difference in area between the two functions in that region . First, find $\alpha$ and $\beta$, which is where $r$ is equal in the two functions: $$3\cos{\theta}=3\sin{2\theta} \implies \theta=\pm\frac{\pi}{6}$$ Note that you're looking for the area only in the first quadrant. Try finishing the rest given what I have provided here.