A cylindrical ungula of base radiusr and height h has volume Its total surface area is the surface area of its curved sidewall is and the surface area of its top is
''Proof''
Consider a cylinder bounded below by plane and above by plane where k is the slope of the slanted roof: Cutting up the volume into slices parallel to the y-axis, then a differential slice, shaped like a triangular prism, has volume where is the area of a right triangle whose vertices are,,, and, and whose base and height are thereby and, respectively. Then the volume of the whole cylindrical ungula is which equals after substituting. A differential surface area of the curved side wall is which area belongs to a nearly flat rectangle bounded by vertices,,, and, and whose width and height are thereby and , respectively. Then the surface area of the wall is where the integral yields, so that the area of the wall is and substituting yields The base of the cylindrical ungula has the surface area of half a circle of radius r:, and the slanted top of the said ungula is a half-ellipse with semi-minor axis of length r and semi-major axis of length, so that its area is and substituting yields Note how the surface area of the side wall is related to the volume: such surface area being, multiplying it by gives the volume of a differential half-shell, whose integral is, the volume. When the slope k equals 1 then such ungula is precisely one eighth of a bicylinder, whose volume is. One eighth of this is.
Conical ungula
A conical ungula of height h, base radius r, and upper flat surface slope k has volume where is the height of the cone from which the ungula has been cut out, and The surface area of the curved sidewall is As a consistency check, consider what happens when the height of the cone goes to infinity, so that the cone becomes a cylinder in the limit: so that which results agree with the cylindrical case.
''Proof''
Let a cone be described by where r and H are constants and z and ρ are variables, with and Let the cone be cut by a plane Substituting this z into the cone's equation, and solving for ρ yields which for a given value of θ is the radial coordinate of the point common to both the plane and the cone that is farthest from the cone's axis along an angle θ from the x-axis. The cylindrical height coordinate of this point is So along the direction of angle θ, a cross-section of the conical ungula looks like the triangle Rotating this triangle by an angle about the z-axis yields another triangle with,, substituted for,, and respectively, where and are functions of instead of. Since is infinitesimal then and also vary infinitesimally from and, so for purposes of considering the volume of the differential trapezoidal pyramid, they may be considered equal. The differential trapezoidal pyramid has a trapezoidal base with a length at the base of, a length at the top of, and altitude, so the trapezoid has area An altitude from the trapezoidal base to the point has length differentially close to The volume of the pyramid is one-third its base area times its altitudinal length, so the volume of the conical ungula is the integral of that: where Substituting the right hand side into the integral and doing some algebraic manipulation yields the formula for volume to be proven. For the sidewall: and the integral on the rightmost-hand-side simplifies to. ∎ As a consistency check, consider what happens when k goes to infinity; then the conical ungula should become a semi-cone. which is half of the volume of a cone. which is half of the surface area of the curved wall of a cone.
When, the "top part" has a parabolic shape and its surface area is When then the top part has an elliptic shape and its surface area is where When then the top part is a section of a hyperbola and its surface area is where where the logarithm is natural, and