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A circular sector is the portion of a disc between two radii and the connecting circle arc. If its radius is $R$ and its central angle is $\theta$ radians, its area isIndeed, polar coordinates give $A={\int }_0^{\theta }{\int }_0^{R}rdrd\varphi$. The full disc corresponds to $\theta =2\pi$, a semicircle to $\theta =\pi$, and a quarter-disc to $\theta =\pi /2$. In a solid with horizontal sector cross-sections, radius $R(z)$ and angle $\theta (z)$, its volume is $\frac{1}{2}\int \theta (z)R(z{)}^{2}dz$. Holding the angles and heights fixed while scaling every radius by a factor $s$ scales that volume by $s^2$.

A circular segment is the part of a disk bounded by a chord and one of the two circular arcs joining its endpoints. For radius $R$ and the chosen arc's central angle $0\le \varphi \le 2\pi$ in radians, its area is $R^2(\varphi -\sin \varphi )/2$. Subtract the signed central triangle of area $R^2\sin \varphi /2$ from the circular sector of area $R^2\varphi /2$. Angles below $\pi$ give the smaller segment and angles above $\pi$ the larger one.

Let $A$ be the top vertex, $O$ the center, and $B,C$ the two other points on the circle. The diameter bisects the angle at $A$, so the two right triangles obtained by drawing the diameter to the opposite point show that $B,C$ have coordinates $(\mp \sin \theta ,-\cos \theta )$ when $O=(0,0)$ and $A=(0,1)$. The chord $BC$ cuts off a lower circular segment of central angle $2\theta$. The triangle $ABC$ has area $\sin \theta (1+\cos \theta )$, while the segment has area $\theta -\sin \theta \cos \theta$, by subtracting the central triangle from its circular sector. Adding giveswith $\theta$ measured in radians. This is the unit-radius case of the area enclosed by two chords and a circular arc.

The two labelled regions in the original diagram are complementary circular sectors. Their areas add to $\pi r^2$ and satisfy $A=2B$, so $B=\pi r^2/3$. The sector with central angle $\theta$ has area $r^2\theta /2$. Hence

At each height, the removed circular sector has the same angle and $4/5$ of the original radius. Its area and hence its volume are therefore multiplied by $(4/5{)}^2=16/25$. Equating the removed volume to that of the reformed solid sphere givesConsequently

At height $z$, the circular sector has radius $r(z)=R{e}^{-z/H}$ and angle $\Theta (z)=2\pi -\pi z/H$. Its area is $\Theta (z)r(z{)}^2/2$, so slicing by horizontal cross-sections gives the volume integralUse the substitution rule $t=z/H$. ThenAn integration by parts gives the antiderivative $(t/4-3/8)e^{-2t}$, whose endpoint difference is $(3-e^{-2})/8$. Therefore

A semicircular disk is one half of a disk cut along a diameter, including its interior. It is a circular sector of angle $\pi$, with area $\pi a^2/2$. Source polar coordinates use $0\le r\le a$, $0\le \theta \le \pi$, and area element $rdrd\theta$. Its boundary consists of a semicircle and a diameter; the semicircle alone is a curve and has no planar interior.