Integration - finding the area between two polar curves I have attempted to solve a polar curves question and I keep getting the wrong answer. 

The question is:
Find the area enclosed by between the curves 

$$
r = 3 - 3\cos\theta 
$$
and 
$$
r =4\cos\theta
$$

So far I have forked out the angle that they interest at so:

$$
4cos\theta = 3 - 3cos\theta 
$$
$$
\therefore \theta = 1.127885 radians
$$
Then I integrating with the upper bound as 1.127885 and the lower bound as zero.
$$
\int_{0}^{1.127885}{\frac{1}{2}*(4\cos\theta)^2}d\theta = 6.06037
$$

$$
\int_{0}^{1.127885}{\frac{1}{2}*(3-3\cos\theta)^2} d\theta  = 0.3527
$$

Then the difference in area is 6.06037 - 0.3527 = 5.70767.

Then by symmetry the total area is 5.70567*2 = 11.4

The answer I get is 11.4 but apparently the answer 1.15. Could you help me with this?
Well, that first integral is certainly suspect. Observe that when $t=0$, $r=4$, so you’ve computing the area of most of the upper black semicircle in the diagram below: 



Your first integral computes the area of that semicircle less the segment bounded by the chord from the origin to point $A$ (the dotted line). Your value is plausible for this area—it’s a bit under $2\pi$, the semicircle’s area. The second integral computes the area between the blue curve and the dotted chord, so when you subtract its value, you take another little bite out of the semicircle. All in all, you’ve computed an area outside of the blue cardioid instead of inside it. What you need to do instead is compute the area of that segment of the circle bounded by the chord $OA$ and add that to the area between the cardioid and $OA$, which you’re already computing correctly.
grapg

I think it's will help you
You need to calc the diff between this areas