# Find the area lying inside the cardioid $r=1+\cos\theta$ and outside the parabola $r(1+ \cos\theta)=1$

I need to find the area lying inside the cardioid $$r=1+ \cos\theta$$ and outside the parabola $$r(1+ \cos\theta)=1$$.

ATTEMPT

First I found the intersection point of two curves which comes out to be $$-\frac{\pi}{2}$$ and $$\frac{\pi}{2}$$.

The integral setup will be $$\int_{\theta =\frac{-\pi}{2}}^{\frac{\pi}{2}}\int_{r=\frac{1}{1+\cos\theta}}^{1+\cos \theta}\ dr\ d\theta.$$On integrating this, I got the answer as $$\pi$$ but answer was given to be $$\frac{3\pi}{4}-\frac 43$$.

Can anybody check my integral setup?

• I think neither answer is right. I believe it is $3\pi/4+4/3$. It is a good idea to graph and at least make a reasonable guess and see if it "matches" your work. That's what I did before hitting algebra Commented Oct 1, 2017 at 16:07
• Is the integral setup is fine?
– User
Commented Oct 1, 2017 at 16:20

Actually, it's neither $$\pi$$ nor $$\frac{3\pi}4-\frac43$$.
You are right when you claim that the possible values for $$\theta$$ go from $$-\frac\pi2$$ to $$\frac\pi2$$; see the picture below. But you seem to have forgotten that, when you compute an area in polar coordinates, the function that you should be integrating is $$r$$, not $$1$$. So, you should compute$$\int_{-\pi/2}^{\pi/2}\int_{1/(1+\cos(\theta))}^{1+\cos(\theta)}r\,\mathrm dr\,\mathrm d\theta,\tag1$$which is equal to $$\frac{3\pi}4+\frac43$$. However, if, in $$(1)$$, you forget the $$r$$ inside the integral, then the integral will be equal to $$\pi$$.
Besides, you can see from the picture below that the area that you are trying to compute is more than half of the area of the region bounded by the cardioid. But the area of the region bounded by the cardioid is $$\frac{3\pi}2$$, and therefore that area that you're after must be greater than $$\frac{3\pi}4$$.