Find the area bounded by the curves: 1^circ rho = a cos

Find the area bounded by the curves: $1^{\circ}\ \rho = a\cos\varphi$, $2^{\circ}\ \rho = a\cos 2\varphi$, $3^{\circ}\ \rho = a\cos 3\varphi$.

We will treat all three curves the same way: find one petal's angular range, integrate it, and multiply by the petal count.

Part 1: $\rho = a\cos\varphi$ (a circle)

For $\rho\geq0$ we need $\cos\varphi\geq0$, so $\varphi\in\left[-\frac\pi2,\frac\pi2\right]$ traces the whole curve (a single circle).

\[A_1 = \frac12\int_{-\pi/2}^{\pi/2} a^2\cos^2\varphi\,d\varphi = \frac{a^2}{4}\int_{-\pi/2}^{\pi/2}(1+\cos2\varphi)\,d\varphi\]

\[= \frac{a^2}{4}\left[\varphi+\frac{\sin2\varphi}{2}\right]_{-\pi/2}^{\pi/2} = \frac{a^2}{4}\cdot\pi\]

\[\boxed{A_1 = \frac{\pi a^2}{4}}\]

Part 2: $\rho = a\cos2\varphi$ (a four-petaled rose)

For $\rho\geq0$, one petal spans $\varphi\in\left[-\frac\pi4,\frac\pi4\right]$; by symmetry there are 4 identical petals.

\[A_{\text{petal}} = \frac12\int_{-\pi/4}^{\pi/4} a^2\cos^22\varphi\,d\varphi = \frac{a^2}{4}\left[\varphi+\frac{\sin4\varphi}{4}\right]_{-\pi/4}^{\pi/4} = \frac{a^2}{4}\cdot\frac\pi2 = \frac{\pi a^2}{8}\]

\[A_2 = 4\cdot\frac{\pi a^2}{8} = \frac{\pi a^2}{2}\]

\[\boxed{A_2 = \frac{\pi a^2}{2}}\]

Part 3: $\rho = a\cos3\varphi$ (a three-petaled rose)

For $\rho\geq0$, one petal spans $\varphi\in\left[-\frac\pi6,\frac\pi6\right]$; by symmetry there are 3 identical petals.

\[A_{\text{petal}} = \frac12\int_{-\pi/6}^{\pi/6} a^2\cos^23\varphi\,d\varphi = \frac{a^2}{4}\left[\varphi+\frac{\sin6\varphi}{6}\right]_{-\pi/6}^{\pi/6} = \frac{a^2}{4}\cdot\frac\pi3 = \frac{\pi a^2}{12}\]

\[A_3 = 3\cdot\frac{\pi a^2}{12} = \frac{\pi a^2}{4}\]

\[\boxed{A_3 = \frac{\pi a^2}{4}}\]

Summary

\[\boxed{A_1=\frac{\pi a^2}{4}\ \text{(circle)}, \qquad A_2=\frac{\pi a^2}{2}\ \text{(four-petaled rose)}, \qquad A_3=\frac{\pi a^2}{4}\ \text{(three-petaled rose)}}\]

Key insight: The circle and the three-petaled rose have exactly the same total area, while the four-petaled rose has twice that area. In general, $\rho=a\cos n\varphi$ traces $n$ petals when $n$ is odd (total area $\frac{\pi a^2}{4}$) and $2n$ petals when $n$ is even (total area $\frac{\pi a^2}{2}$).