This is done by setting up a right triangle.  The length of the leg along the x axis is 1 and the length of the leg along the y axis is \(\displaystyle \sqrt{3}\).  

Thus the length of the hypotenuse can be found using Pythagorean Theorem.

\(\displaystyle r^2=\sqrt{3}^2+1^2=4\rightarrow r=2\)

We can then find theta by solving

\(\displaystyle tan(\theta )=\frac{\sqrt{3}}{1}\rightarrow \theta=\frac{\pi }{3}\)

Whenever we are finding the area between two curves, we need to sketch or graph the two curves to get an idea of what they look like. The graph is as follows.

We need to find the area inside the circle (blue) and outside the rose (red).

Looking at the graph, we can see that the two graphs are symmetric to both x and y axes. So we can find the desired area in the first quadrant and then multiply by 4 to get the total area. To get area of the first quadrant, we simply use \(\displaystyle \theta_1 = 0\) and \(\displaystyle \theta_2 = \pi/2\) for the bounds of integration.

Writing out the formula for the area between two polar curves, and multiplying by 4 for symmetry, we get

\(\displaystyle 4[\frac{1}{2} \int_{0}^{\pi/2} (1)^2 - (\sin2\theta)^2 d\theta]\)

Lets simplify my multiplying the 4 and 1/2 outside the integral and then using the Pythagorean Trig Identity \(\displaystyle 1 - \sin^2 x = \cos^2 x\). Remember that in this case, \(\displaystyle x = 2\theta\) for the sine AND the cosine.

\(\displaystyle 2 \int_{0}^{\pi/2} \cos^2 2\theta d\theta\)

As it is now, we cannot integrate. We first must use the double-angle identity, \(\displaystyle \cos^2 x = \frac{1}{2}(1 + \cos 2x)\). Again, remember that in this problem, \(\displaystyle x = 2\theta\), so \(\displaystyle 2x = 4\theta\).

Using the double-angle identity, we get

\(\displaystyle 2 \int_{0}^{\pi/2} \frac{1}{2} (1 + \cos 4\theta) d\theta\)

The 2 and the 1/2 cancel leaving,

\(\displaystyle \int_{0}^{\pi/2} 1 + \cos 4\theta d\theta\)

Now we can integrate both terms. Remember that we have to account for the 4 inside the cosine.

\(\displaystyle [\theta +\frac{1}{4}\sin4\theta]\vert_{0}^{\pi/2}\)

Now we plug in the bounds and simplify.

\(\displaystyle [(\frac{\pi}{2} + \frac{1}{4}\sin 4(\frac{\pi}{2})) - (0 + \frac{1}{4}\sin 4(0))]\)

\(\displaystyle \frac{\pi}{2} + \frac{1}{4} \sin 2\pi - \frac{1}{4} \sin 0\)

\(\displaystyle \sin 2\pi\) and \(\displaystyle \sin 0\) both equal zero, so we can drop those terms leaving the final answer.

\(\displaystyle \frac{\pi}{2}\)

The first thing we should do is graph the two curves to see what they look like.

The red curve is the limacon, \(\displaystyle R_2 = 2 + \cos \theta\). The blue curve is the circle, \(\displaystyle R_1 = 2\)

From the graph we can see the area inside the circle (blue) and outside the limacon (red) appears to fall between \(\displaystyle \frac{\pi}{2}\) and \(\displaystyle \frac{3 \pi}{2}\). To be certain, we must set the two equations equal two each other.

\(\displaystyle R_1 = R_2\)

\(\displaystyle 2 = 2+ \cos\theta\)

Subtract \(\displaystyle 2\) from both sides to isolate \(\displaystyle \cos \theta\)

\(\displaystyle 0 = \cos \theta\)

\(\displaystyle \cos \theta\) equals zero at \(\displaystyle \theta = \frac{\pi}{2}, \frac{3\pi}{2}\).

Now we know the bounds for the following integration.

To find the area between two polar curves, we use the following formula.

\(\displaystyle \frac{1}{2} \int_{\theta_1}^{\theta_2} (R_{outer})^2 - (R_{inner})^2 d\theta\), where \(\displaystyle \theta_1\) and \(\displaystyle \theta_2\) are the bounds, \(\displaystyle R_{outer}\) is the curve with the larger radius, and \(\displaystyle R_{inner}\) is the curve with the smaller radius.

Now we plug in the bounds and the radius curves to get,

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} (2)^2 - (2+\cos\theta)^2 d\theta\)

Now we simplify the inside of the integral. Exponents are the first thing we deal with. Write the \(\displaystyle (2 + \cos \theta)^2\) as \(\displaystyle (2+ \cos\theta)(2 + \cos\theta)\) and multiply it out. Then distribute the negative sign through the result and combine like terms.

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} 4 - (2+\cos\theta)(2 + \cos \theta) d\theta\)

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} 4 -(4 + 4\cos\theta + \cos^2\theta)d\theta\)

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} 4 - 4 - 4\cos\theta - \cos^2\theta d\theta\)

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} - 4\cos\theta - \cos^2\theta d\theta\)

Now we have two terms inside the integral. \(\displaystyle -4 \cos \theta\) follows a basic integral form of cosine, but \(\displaystyle -\cos^2 \theta\) doesn't. We need to use the double-angle identity, \(\displaystyle \cos^2 \theta = \frac{1}{2}(1 + \cos 2\theta)\), to convert it to an integrable form.

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} - 4\cos\theta - \frac{1}{2}(1 + \cos 2\theta) d\theta\)

Now we multiply to get the basic integral form we need.

\(\displaystyle \frac{1}{2} \int_{\pi/2}^{3\pi/2} - 4\cos\theta - \frac{1}{2} - \frac{1}{2}\cos 2\theta d\theta\)

At this point we can integrate all three terms. For the \(\displaystyle \frac{1}{2}\cos 2\theta\), remember to account for the \(\displaystyle 2\) inside the cosine. The result is as follows.

\(\displaystyle \frac{1}{2}[-4 \sin \theta - \frac{1}{2}\theta - \frac{1}{4}\sin 2\theta] \vert_{\pi/2}^{3\pi/2}\)

Now plug in the upper and lower bounds and simplify.

\(\displaystyle \frac{1}{2}[(-4 \sin (\frac{3\pi}{2}) - \frac{1}{2}(\frac{3\pi}{2}) - \frac{1}{4}\sin 2(\frac{3\pi}{2}))-(-4\sin(\frac{\pi}{2}) - \frac{1}{2}(\frac{\pi}{2}) - \frac{1}{4}\sin 2(\frac{\pi}{2}))]\)

\(\displaystyle \frac{1}{2}[(-4 (-1) - \frac{3\pi}{4} - \frac{1}{4}\sin 3\pi)-(-4(1) - \frac{\pi}{4} - \frac{1}{4}\sin \pi)]\)

Since \(\displaystyle \sin 3\pi\) and \(\displaystyle \sin \pi\) both equal zero, we can drop those terms. Then we distribute the negative through the second group.

\(\displaystyle \frac{1}{2}[4 - \frac{3\pi}{4} +4 + \frac{\pi}{4}]\)

Now simplify.

\(\displaystyle \frac{1}{2}[8 - \frac{\pi}{2}]\)

The final answer is \(\displaystyle 4 - \frac{\pi}{4}\)