Calculate the surface area generated by rotating

Calculate the surface area generated by rotating the cardioid $\rho=a(1+\cos\varphi)$ about the polar axis.

Step 1: Differentiate

\[\rho'=-a\sin\varphi\]

Step 2: Simplify $\rho^2+(\rho')^2$

\[\rho^2+(\rho')^2=a^2(1+\cos\varphi)^2+a^2\sin^2\varphi = a^2[2+2\cos\varphi]=2a^2(1+\cos\varphi)\]

Using $1+\cos\varphi=2\cos^2\frac\varphi2$: \[\sqrt{\rho^2+(\rho')^2}=2a\left|\cos\frac\varphi2\right|=2a\cos\frac\varphi2\quad(\varphi\in[0,\pi])\]

Step 3: Set up the integral

The curve traced by $\varphi\in[0,\pi]$ runs from the far point $(2a,0)$ (on the axis, at $\varphi=0$) to the cusp at the origin (also on the axis, at $\varphi=\pi$), so rotating just this trace about the polar axis already sweeps out the complete cardioid surface once:

\[S=2\pi\int_0^\pi a(1+\cos\varphi)\sin\varphi\cdot2a\cos\frac\varphi2\,d\varphi = 4\pi a^2\int_0^\pi(1+\cos\varphi)\sin\varphi\cos\frac\varphi2\,d\varphi\]

Step 4: Reduce using half-angle identities

With $1+\cos\varphi=2\cos^2\frac\varphi2$ and $\sin\varphi=2\sin\frac\varphi2\cos\frac\varphi2$:

\[(1+\cos\varphi)\sin\varphi\cos\frac\varphi2 = 4\sin\frac\varphi2\cos^4\frac\varphi2\]

\[S=16\pi a^2\int_0^\pi\sin\frac\varphi2\cos^4\frac\varphi2\,d\varphi\]

Step 5: Substitute $u=\cos\frac\varphi2$

$du=-\frac12\sin\frac\varphi2\,d\varphi$; limits $\varphi=0,\pi$ become $u=1,0$:

\[S=16\pi a^2\cdot2\int_0^1u^4\,du = 32\pi a^2\cdot\frac15=\frac{32\pi a^2}5\]

\[\boxed{S=\frac{32\pi a^2}5\approx20.11a^2}\]

Step 6: Sanity check

This mirrors the analogous sphere computation (Problem 118): there, integrating the generating semicircle once over its full extent (from one point on the axis to the other) gave the entire sphere's area with no extra doubling. The cardioid's $\varphi\in[0,\pi]$ trace plays exactly the same role here, confirmed by direct numerical integration of the original polar surface formula.