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The equation of a quadric surface after orthogonal diagonalization of its symmetric matrix and appropriate translation. The spectral theorem for real symmetric matrices removes mixed quadratic terms, and completing squares removes linear terms in directions with nonzero eigenvalues. Linear terms in zero-eigenvalue directions can remain and generate elliptic paraboloids or hyperbolic paraboloids. The signs, zero coefficients and constant distinguish ordinary surfaces, degenerate conics in plane sections, and empty real loci.

For the rotational elliptic paraboloid $z=H(1-(x^2+y^2)/R^2)$ above $z=0$, allow inscribed shapes to touch the boundary. A coaxial cylinder of radius $r$ and height $h$ obeys $h+H{r}^2/R^2=H$. Applying a Lagrange multiplier to $\pi r^{2}h$ gives $h=H{r}^2/R^2$, hence $h=H/2$ and $r=R/\sqrt{2}$.

A centred rectangular parallelepiped with horizontal side lengths $a,b$ and height $c$ obeys $c+H(a^2+b^2)/(4R^2)=H$. The Lagrange multiplier equations for $abc$ give $a=b$ and $c=H{a}^2/(2R^2)$, hence $a=b=R,c=H/2$. These yield the displayed maximal volumes; both positive stationary shapes are interior optima among the dimension variables, while their degenerate boundary shapes have zero volume. The cylinder's volume is $\pi /2$ times the box's. If strict containment in the open region is required, these are suprema approached by slightly smaller shapes rather than attained maxima.

Suppose a solid is above the elliptic paraboloid $z=x^2+y^2$ and below a plane whose vertical separation from it is $A-|\mathbf{x}-\mathbf{p}{|}^2$, where $A>0$. A second lower plane exceeds the elliptic paraboloid by $B-|\mathbf{x}-\mathbf{q}{|}^2$, with $B>0$. If $|\mathbf{p}-\mathbf{q}|+\sqrt{B}\le \sqrt{A}$, the disk where the lower plane matters lies within the upper projection. Raising the bottom to this lower plane therefore removes exactly its excess volume, without changing the outer projection.

Translate polar coordinates separately to the two centres. The double integral over a disk of squared radius $A$ isSubtracting the analogous integral with $B$ proves the displayed volume formula. The containment condition is essential: if the two disks overlap only partly, subtract only over their intersection.

The displayed three-dimensional parabolic coordinates use $u,v\ge 0$ and an azimuthal angle $\varphi$. Their coordinate tangent vectors have pairwise zero dot products, with scale factors of orthogonal coordinates $h_u=h_v=\sqrt{u^2+v^2}$ and $h_{\varphi }=uv$. The Jacobian determinant is $uv(u^2+v^2)$. Constant $u$ or $v$ gives a family of confocal circular elliptic paraboloids; the coordinate system degenerates on the axis where $uv=0$. For example, $0\le u\le v\le 1$ fills the cap $(x^2+y^2-1)/2\le z\le 0$, whose rim is the unit circle and whose vertex is $(0,0,-1/2)$.

These orthogonal coordinates extend two confocal families of parabolas parallel to an axis. In the displayed scaling their scale factors of orthogonal coordinates are $h_u=h_v=2\sqrt{u^2+v^2}$ and $h_w=1$, so the Laplacian is $\Delta \varphi =({\varphi }_{uu}+{\varphi }_{vv})/[4(u^2+v^2)]+{\varphi }_{ww}$ away from $u=v=0$. With $u\ge 0$, signed $v$ distinguishes the upper and lower half-planes; the two signs at $u=0$ describe the same negative $x$ half-axis. Fixed positive $u^2$ or $v^2$ describes a parabolic cylinder. This differs from the rotational three-dimensional parabolic coordinates, whose coordinate surfaces are elliptic paraboloids.

An invertible $4\times 4$ matrix $M$, modulo scalar multiples, acts linearly on homogeneous coordinates. In an affine chart its projective transformation has the formwhere the denominator must be nonzero for a finite image. The important nested groups are Euclidean motions, Euclidean similarities, invertible affine maps, and general projective transformations. Their defining preserved structures become progressively weaker.

A Euclidean motion has $x^{\prime }=Rx+b$ with $R^{T}R=I$. It preserves all distances and angles, hence lengths, areas and volumes. Proper motions have $\det R=1$ and preserve orientation as well; allowing reflections preserves unsigned measurements but can reverse signed ones. A Euclidean similarity has $x^{\prime }=sRx+b$ with $s>0$. It preserves angles and all ratios of lengths; lengths, areas and volumes scale by $s,s^2,s^3$. Thus rigid shape is preserved up to one overall scale.

An invertible affine map has $c=0$ and preserves the plane at infinity. It preserves incidence, parallelism, collinear ratios, affine combinations, and ratios of volumes. Absolute volume scales by $|\det A|$; the volume-preserving affine subgroup has $|\det A|=1$. Lengths and angles are not affine invariants. A general projective transformation preserves incidence, collinearity, concurrence, tangency, and the cross-ratio of four collinear points, but can move the plane at infinity. The cross-ratio, rather than a three-point affine ratio, survives arbitrary projective changes. For a homogeneous quadric hypersurface $X^{T}QX=0$, its transformed matrix is $M^{-T}Q{M}^{-1}$, so its rank and projective degeneracy are preserved.

For surface classes, rather than a fixed numerical radius or curvature, the consequences are:

For example, the affine stretch $(x,y,z)\mapsto (2x,y,z)$ changes a unit sphere to $x^{\prime 2}/4+y^{\prime 2}+z^{\prime 2}=1$ and a unit circular cylinder to $x^{\prime 2}/4+y^{\prime 2}=1$. A general conical surface written $P+tD(s)$ transforms affinely into $T(P)+tAD(s)$, so every generator still passes through a finite apex. A nonsingular elliptic paraboloid or hyperbolic paraboloid has quadratic part of rank two and a nonzero linear component in its null direction. An invertible affine map preserves that rank, inertia and nonzero null-direction coupling, so translation and linear coordinate changes again put it in elliptic or hyperbolic paraboloid form.

Under a genuinely projective change, these Euclidean classifications depend on which plane is designated as infinity. For the map $(x,y,z)\mapsto (x,y,z)/(1+\alpha z)$, a circular cylinder $x^2+y^2=1$ becomesa cone whose apex is finite. The inverse map sends that apex to infinity and produces the cylinder. Thus a cone is projectively invariant as a family of concurrent generators if an ideal apex is allowed, but finite-apex cones and cylinders are not separately invariant. Likewise a projective image of a sphere or either type of paraboloid remains a nonsingular quadric hypersurface, but need not retain its affine or metric type. The distinction between general cones and circular cones matters: circularity is a metric property and is not preserved by arbitrary affine changes.

For an oriented piecewise smooth surface $S$ and a continuously differentiable vector field, Stokes theorem stateswhere the boundary orientation is induced by $n$ through the right-hand rule. Choose the upward orientation of the elliptic paraboloid. Its band has cylindrical radius $1/2\le r\le 1$, and a parametrization isIts oriented surface element and scalar area element areThe requested sketch shows the open band, not a capped solid.

For the specified vector field, direct differentiation gives $\nabla \times B=(0,0,3x^2+3y^2)$. HenceTo check this by Stokes theorem, parametrize a counterclockwise circle of radius $\rho$ at height ${\rho }^2$ by $r_{\rho }(\phi )=(\rho \cos \phi ,\rho \sin \phi ,{\rho }^2)$. Along it $dz=0$, andUsing ${\int }_0^{2\pi }{\sin }^4\phi d\phi ={\int }_0^{2\pi }{\cos }^4\phi d\phi =3\pi /4$ gives $3\pi {\rho }^4/2$. The induced outer circle is counterclockwise as viewed from above, and the inner circle is clockwise. ThereforeThis verifies the direct surface integral and the Stokes flux through an annular paraboloid. Reversing the chosen surface orientation reverses the final sign.

In spherical coordinates, $z=r\cos \theta$ and $x^2+y^2=r^2{\sin }^2\theta$, so the elliptic paraboloid has equationAway from the origin this is $r=\cos \theta /{\sin }^2\theta$, with $0<\theta <\pi /2$. The undivided equation also includes the origin, where angular coordinates are not unique.

In cylindrical coordinates, $x^2+y^2={\rho }^2$, so the elliptic paraboloid is

In the standard spherical coordinate system, $\theta$ is measured from the positive $z$ axis. The radial unit vector in spherical coordinates isThe coordinate forms of the elliptic paraboloid are given in the labelled parts below.

For the other surface, $r^2\cos 2\theta =r^2({\cos }^2\theta -{\sin }^2\theta )$ converts toFor $a>0$, this is a two-sheeted hyperboloid, with two disconnected sheets starting at $z=\pm \sqrt{a}$ and opening along the $z$ axis. For $a<0$, it is a one-sheet hyperboloid:Its waist is the circle $z=0$, $x^2+y^2=|a|$. Both are surfaces of revolution about the $z$ axis and approach the cone $z^2=x^2+y^2$ far from the origin.

The figure shows $a=1$ and $a=-1$; scaling all coordinates by $\sqrt{|a|}$ gives the other magnitudes.

For an oriented smooth surface $S$ with piecewise smooth, consistently oriented boundary, Stokes theorem givesfor a continuously differentiable vector field defined on a neighborhood of $S$. Choose the upward normal vector here. The surface is an annular band of an elliptic paraboloid, parametrized byUsing ${\partial }_{r}r\times {\partial }_{\theta }r$ as the oriented vector area element givesThe sketch below shows the open band, not a capped solid. Its upper circle has radius one and its lower circle radius one third.

Differentiating the given vector field yieldsThe surface integral is thereforeFor the boundary line integral, the upward normal vector induces counterclockwise traversal of the outer circle and clockwise traversal of the inner circle, as viewed from above. On a circle of radius $R$, parametrized counterclockwise, $z$ is constant andSince each fourth power integrates to $3\pi /4$, the two-circle line integral isconfirming Stokes theorem and the Stokes flux through an annular paraboloid. Reversing the chosen orientation changes both integrals to $-40\pi /27$.

Start with the solid between the elliptic paraboloid $z=x^2+y^2$ and the upper plane $z=10-2y$. Its projection is the disksince their vertical separation is $11-x^2-(y+1{)}^2$. The second lower bound, $z\ge 2-2x$, removes the part between the elliptic paraboloid and this lower plane whereverCall this second disk $E$. Its centre is distance $\sqrt{2}$ from the centre of $D$, andbecause $5+2\sqrt{6}<11$. Thus $E$ lies wholly inside $D$. In particular the lower plane never exceeds the upper plane on $E$, so the subtraction describes exactly the requested solid.

The nested-disk subtraction for paraboloid volumes therefore gives a difference of two double integrals:Translate polar coordinates to the centre of each disk. A general integral of this form isUsing $A=11$ and $A=3$ gives

The quadric surface $x^2+y^2+z^2=1$ is the unit sphere centered at the origin. Each section $z=c$, for $|c|<1$, is a circle of radius $\sqrt{1-c^2}$.

The quadric surface $z=x^2+y^2$ is a circular elliptic paraboloid, opening in the positive $z$ direction with vertex at the origin. Its horizontal sections $z=c>0$ are circles of radius $\sqrt{c}$, and its vertical sections through the $z$ axis are parabolas.

For $a{x}^2+y^2=0$, $z$ is unrestricted. If $a>0$, both nonnegative terms must vanish, so the locus is the $z$ axis, a straight line. If $a=0$, it is the plane $y=0$, represented by the repeated factor $y^2$. If $a<0$, factorization givesso the locus is two distinct planes, $y=\pm \sqrt{-a}x$, meeting along the $z$ axis. These are degenerate quadrics, not ordinary curved cylinders.

The rotation axis is $L=\{(x,1,0):x\in \mathbb{R}\}$. The distance of the generating function graph from this axis is $\sqrt{x}$, so a parametrization isThe surface of revolution therefore has equation $(y-1{)}^2+z^2=x$: it is a circular elliptic paraboloid opening along the positive $x$ direction, with its vertex and its rim at $x=1$ excluded. Translating the axis does not change the surface area formula, soThe cancellation makes the improper integral finite at the vertex. Neither the missing vertex nor the missing boundary circle changes the surface area, and no cap is included.

In cylindrical coordinates, $x=r\cos \theta$ and $y=r\sin \theta$, so $x^2+y^2=r^2$. Thus the circular elliptic paraboloid has equationAt height $h\ge 0$, the rim radius is $\sqrt{ah}$.

Put $R=|b|$ so that the radial range respects the stated allowance $b\ne 0$. With the azimuth measured from the starting meridian, the cylindrical coordinates of the curved surface satisfyAt $r=0$ all azimuths describe the same apex. The surface is a sector of an elliptic paraboloid.

At height $z>0$, the elliptic paraboloid has a circular cross-section $0\le r\le \sqrt{z}$, of area $\pi z$. The cylindrical coordinates volume element therefore reduces the volume integral toFor the specified $f$, the derivative is $e^{-z}(z^{-1/2}+\frac{1}{2}{z}^{-3/2})$. Thus $z{f}^{\prime }(z)$ is integrable at zero and decays exponentially at infinity. Apply integration by parts first on a truncated interval and then take the endpoint limits:Both boundary terms vanish because $zf(z)=-\sqrt{z}{e}^{-z}$. With $z=u^2$, integration by substitution gives the Gaussian integral:The integrand is singular at the apex but the resulting improper integral is convergent.

Use cylindrical coordinates $x=r\cos \theta$, $y=r\sin \theta$. The Jacobian determinant givesThe inequalities imply $-d<z<0$ and $0\le r<a\sqrt{1+z/d}$. Thus the volume isEquivalently, integrating vertically first gives $V={\int }_0^{2\pi }{\int }_0^a{\int }_{d(r^2/a^2-1)}^{0}rdzdrd\theta$. The region is bounded by an elliptic paraboloid and its top disc; its volume is half that of the cylinder with base radius $a$ and height $d$. Including or excluding the boundary surfaces does not change the volume.

Use cylindrical radius $\rho =\sqrt{x^2+y^2}$. The upper cone is $z=1-\rho$ and the lower elliptic paraboloid is $z={\rho }^2-1$, both with $0\le \rho \le 1$. They meet along the unit circle in $z=0$. The enclosed solid isThus the coordinate ranges on the complete surface areThese separate ranges do not mean every point of their Cartesian product lies on the surface. A vertical cross-section has a straight upper cone profile and a curved lower paraboloid profile.

The upper piece is a downward-opening circular elliptic paraboloid cap, running from the circle $r=1,z=1$ to the apex $r=0,z=2$. The middle piece is the side of a circular cylinder of radius one and height one. The lower piece is the unit disk in the horizontal plane. Their union is the closed boundary ofThe four panels show the three separate pieces and their union; there is no additional fourth component. In a meridional section, the boundary consists of the base segment, two vertical sides, and the parabolic cap. The outward normals point downward on the base, radially outward on the cylindrical side, and upward/outward on the cap.