Find one part of the area bounded by the curves: rho =

Find one part of the area bounded by the curves: $\rho = a(1 + \cos\varphi)$ and $\rho = \frac{3a}{2}$.

We will find the crossing angles and integrate the difference of the squared radii over the lobe where the cardioid pokes outside the circle.

Step 1: Find intersection points

Setting $a(1+\cos\varphi)=\frac{3a}{2}$: $\cos\varphi=\frac12 \implies \varphi=\pm\frac\pi3$.

Step 2: Determine which curve is outer

At $\varphi=0$: cardioid gives $\rho=2a$, circle gives $\rho=\frac{3a}2$. Since $2a>\frac{3a}2$, the cardioid is outside the circle near $\varphi=0$ — this is the "lobe" of the cardioid poking outside the circle.

Step 3: Set up the integral using symmetry

By symmetry about the polar axis:

\[A = \int_0^{\pi/3} \left[a^2(1+\cos\varphi)^2 - \frac{9a^2}{4}\right] d\varphi\]

Step 4: Expand and simplify

\[(1+\cos\varphi)^2 = \frac32+2\cos\varphi+\frac{\cos2\varphi}{2}\]

\[A = a^2\int_0^{\pi/3}\left[2\cos\varphi+\frac{\cos2\varphi}{2}-\frac34\right]d\varphi\]

Step 5: Integrate

\[A = a^2\left[2\sin\varphi+\frac{\sin2\varphi}{4}-\frac{3\varphi}{4}\right]_0^{\pi/3}\]

At $\varphi=\frac\pi3$: $2\cdot\frac{\sqrt3}{2}+\frac{\sqrt3/2}{4}-\frac{\pi}{4} = \sqrt3+\frac{\sqrt3}{8}-\frac\pi4 = \frac{9\sqrt3}{8}-\frac\pi4$. At $\varphi=0$: all terms vanish.

\[A = a^2\left(\frac{9\sqrt3}{8}-\frac\pi4\right) = \frac{a^2}{8}(9\sqrt3-2\pi)\]

\[\boxed{A = \frac{a^2}{8}(9\sqrt{3} - 2\pi) \text{ square units}}\]

Step 6: Numerical value

\[A \approx a^2(1.949-0.785) \approx 1.164\,a^2 \text{ square units}\]

Summary

\[\boxed{A = \frac{a^2}{8}(9\sqrt3-2\pi) \approx 1.164a^2 \text{ square units}}\]

Key insight: The lens-shaped region where the cardioid bulges outside the circle mixes an algebraic term ($9\sqrt3$) from the cosine terms with a transcendental term ($2\pi$) from the constant part of the expansion.