Find the volume of the solid generated by rotating

Find the volume of the solid generated by rotating the cardioid $\rho = a(1 + \cos\varphi)$ about the polar axis.

We apply the standard formula for the volume swept by a polar curve revolved about the polar axis, integrating over the curve's upper half only (which alone generates the complete solid).

Step 1: Properties of the cardioid

$\rho=a(1+\cos\varphi)$: maximum $\rho=2a$ at $\varphi=0$; cusp $\rho=0$ at $\varphi=\pi$; symmetric about the polar axis.

Step 2: The volume formula

For a polar curve symmetric about the polar axis, revolving just its upper half ($\varphi\in[0,\pi]$, where $\sin\varphi\ge0$) about that axis already sweeps out the complete solid:

\[V = \frac{2\pi}{3}\int_0^{\pi}\rho^3\sin\varphi\,d\varphi\]

(This formula can be checked on a circle $\rho=2R\cos\varphi$, $\varphi\in[0,\frac\pi2]$, which correctly reproduces the sphere volume $\frac{4}{3}\pi R^3$.)

Step 3: Substitute the cardioid

\[V = \frac{2\pi a^3}{3}\int_0^{\pi}(1+\cos\varphi)^3\sin\varphi\,d\varphi\]

Step 4: Evaluate via substitution

Let $u=1+\cos\varphi$, $du=-\sin\varphi\,d\varphi$. When $\varphi=0$: $u=2$; when $\varphi=\pi$: $u=0$.

\[\int_0^{\pi}(1+\cos\varphi)^3\sin\varphi\,d\varphi = \int_0^2 u^3\,du = \left[\frac{u^4}{4}\right]_0^2 = 4\]

Step 5: Calculate the volume

\[V = \frac{2\pi a^3}{3}\cdot4 = \frac{8\pi a^3}{3}\]

\[\boxed{V = \frac{8\pi a^3}{3} \text{ cubic units}}\]

Step 6: Comparison with a sphere

The cardioid extends out to $\rho=2a$; a sphere of that radius would have volume $\frac{4\pi(2a)^3}{3}=\frac{32\pi a^3}{3}$. Our result is exactly $\frac14$ of that, reflecting the cardioid's cusp and much thinner profile.

Summary

\[\boxed{V = \frac{8\pi a^3}{3} \text{ cubic units}}\]

Key insight: Rotated about its axis of symmetry, the heart-shaped cardioid sweeps out an apple-shaped solid whose volume is exactly one-quarter that of the sphere circumscribing its widest extent — a clean consequence of the cusp pinching the solid inward near the origin.