Surface: intersection of 2 polar curves

Question

I have these two polar curves:

$$ C_1: r = 2 - \cos(\theta)\\ C_2: r = 3 \cos(\theta) $$

Plots: C1 and C2.

I need to find the surface of $D = C_1 \cap C_2$.

I started by finding the solution to $C_1 = C_2$. I found $\pm \pi\over{3}$.

I tried this integral:

$$ \int_{-\pi\over{3}}^{\pi\over{3}} \int_{3\cos(\theta)}^{2-\cos(\theta)} r \;\operatorname d\!r \operatorname d\!\theta $$

However, I'm not sure about that $r$, and I get a negative answer.

I'd appreciate if someone could point me in the right direction!

Answer 1

For $-\frac{\pi}{3} \le \theta \le \frac{\pi}{3}$, we have $3\cos \theta \ge 2-\cos\theta$, and so, your integral yields a negative value.

Draw both graphs on the same region. You'll see that $\theta$ can range from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$.

If $0 \le |\theta| \le \frac{\pi}{3}$, and the point $(r,\theta)$ is in $D$, then we must have $0 \le r \le 2-\cos \theta$.

If $\frac{\pi}{3} \le |\theta| \le \frac{\pi}{2}$, and the point $(r,\theta)$ is in $D$, then we must have $0 \le r \le 3\cos \theta$.

So, the area should be $\displaystyle 2\int_{0}^{\pi/3}\int_{0}^{2-\cos\theta}r\,dr\,d\theta + 2\int_{\pi/3}^{\pi/2}\int_{0}^{3\cos\theta}r\,dr\,d\theta$.

Note: I have exploited symmetry about the $x$-axis, which is why I only integrated over $0 \le \theta \le \frac{\pi}{2}$ and multiplied the result by $2$.

Answer 2

The problem arises because the angular interval over which you want to integrate the area between the two curves does not correspond for each curve. This is because the curve $r = 3 \cos \theta$ is traversed twice when $r = 2 - \cos \theta$ is traversed only once for $\theta \in [0,2\pi)$:

To rectify this, we would use the method shown in JimmK4542's answer: over $\theta \in [0,\pi/3]$, we integrate the sector traced by the blue curve, and then from $\theta \in [\pi/3, \pi/2]$, we integrate the slice traced by the red curve. However, you can "rescue" your calculation by computing the negative area as you did, and then simply add the area of the red circle: $$A = \pi \bigl(\tfrac{3}{2}\bigr)^2 + \int_{\theta = -\pi/3}^{\pi/3} \int_{r = 3 \cos \theta}^{2 - \cos \theta} r \, dr \, d\theta.$$