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For $a>0$, the cardioid $r=a(1+\cos \theta )$ traces its boundary once over $0\le \theta \le 2\pi$. The polar area element givesThe polar arc length integrand is $2a|\cos (\theta /2)|$. Splitting at the cusp parameter $\theta =\pi$ gives perimeter $8a$. The powers of $a$ distinguish the scaling of area and length.

The projection onto the $x$-$y$ plane is the disk $x^2+y^2\le 4$. In polar coordinates, the height is $4-r^2$ and the polar area element is $rdrd\theta$. The volume is thereforeThe radial factor here is required because this integral is being constructed from the geometric volume, unlike the already specified integral in (a).

Each radial segment from $0$ to $1+\cos \theta$ lies inside the cardioid. Integrating the polar area element givesThe three integrals are respectively $2\pi$, $0$ and $\pi$. Thusin square coordinate units, agreeing with the scale-one case of the area and perimeter of a cardioid.

The polar area element is $rdrd\varphi$. Integrating first in radius and using symmetry givesThis is the specialization $a=1$, $c=3/2$ of the polar cap between a cardioid and a circle area formula.

For $a>0$ and $0<c<2a$, consider the region $c\le r\le a(1+\cos \varphi )$ in polar coordinates. It exists precisely for $|\varphi |\le \beta$, where $\beta =\arccos (c/a-1)\in (0,\pi )$. The polar area element $rdrd\varphi$ givesExpanding the square and integrating its even terms gives the title formula. This cap lies inside the cardioid and outside the concentric circle, so reversing the radial bounds would select the wrong region or a signed area.