§12.6 주면과 이차곡면Cylinders and Quadric Surfaces

Solution. The variable $y$ does not appear in $z=x^2$. Substituting $y=k$ leaves the same equation $z=x^2$, independent of $k$, so the trace in every plane $y=k$ is congruent to the parabola $z=x^2$ lying in the $xz$-plane.

The surface is therefore the stack of translated copies of that parabola along the $y$-axis — the cylinder of Definition 3, whose rulings are the lines parallel to the $y$-axis.

Hence the surface is a parabolic cylinder, with rulings parallel to the $y$-axis.

(a) The equation $x^2+y^2=1$ has no $z$ in it. The trace in every plane $z=k$ is the same unit circle $x^2+y^2=1$, regardless of $k$, so the surface is the unit circle in the $xy$-plane stacked along the $z$-axis: a circular cylinder with axis the $z$-axis and rulings parallel to the $z$-axis.

(b) By the same reasoning, $y^2+z^2=1$ has no $x$ in it, so the trace in every plane $x=k$ is the unit circle $y^2+z^2=1$ in the $yz$-plane. This surface is also a circular cylinder, with axis the $x$-axis and rulings parallel to the $x$-axis.

Solution. All three variables occur only to even powers, so reversing the sign of any one of them leaves the equation unchanged: the surface is symmetric in all three coordinate planes.

Substituting $z=k$ gives $x^2+\dfrac{y^2}{9}=1-\dfrac{k^2}{4}$, an ellipse when the right side is positive, i.e. $-2\lt k\lt2$; it is the point $(0,0,\pm2)$ at $k=\pm2$ and empty for $|k|\gt2$. In the same way, $x=k$ gives $\dfrac{y^2}{9}+\dfrac{z^2}{4}=1-k^2$, an ellipse for $-1\lt k\lt1$, and $y=k$ gives $x^2+\dfrac{z^2}{4}=1-\dfrac{k^2}{9}$, an ellipse for $-3\lt k\lt3$.

Since the traces in all three directions are ellipses (apart from the degenerate boundary values), the surface is an ellipsoid. Setting two coordinates to $0$ at a time gives the intercepts $(\pm1,0,0)$, $(0,\pm3,0)$, $(0,0,\pm2)$.

Solution. In the plane $x=k$, $z=y^2+4k^2$: the base parabola $z=y^2$ translated upward as $k$ varies. In the plane $y=k$, $z=4x^2+k^2$, again an upward-opening parabola.

In the plane $z=k$, $4x^2+y^2=k$: an ellipse when $k\gt0$, the single point at the origin when $k=0$, and empty when $k\lt0$.

Since $x$ and $y$ occur only squared while $z$ occurs only to the first power, the equation has the standard form $Ax^2+By^2+Iz=0$. The horizontal traces are ellipses (apart from the degenerate case) and the vertical traces are parabolas, so by the classification of Table 1 the surface is an elliptic paraboloid whose axis is the axis of $z$, the variable appearing to the first power. The horizontal trace is a genuine (non-degenerate) ellipse exactly when $k\gt0$.

Solution. In the plane $x=k$, $z=y^2-k^2$: a parabola opening upward in $y$, shifted downward as $|k|$ grows. In the plane $y=k$, $z=-x^2+k^2$: a parabola opening downward in $x$.

In the plane $z=k$, $y^2-x^2=k$. For $k\gt0$ this is a hyperbola with branches crossing the $y$-axis, for $k\lt0$ a hyperbola with branches crossing the $x$-axis, and at $k=0$ it degenerates to the pair of lines $y=\pm x$.

Near the origin the surface rises when moving along the $y$-axis and falls when moving along the $x$-axis; the opposite concavity in the two directions gives the shape of a saddle there. The surface is a hyperbolic paraboloid, and the origin is its saddle point.

Solution. In the plane $z=k$, $\dfrac{x^2}{4}+y^2=1+\dfrac{k^2}{4}$. The right side is positive for every real $k$, so the horizontal trace is an ellipse for every $k$, growing larger as $|k|$ increases.

The trace in the $xz$-plane ($y=0$) is $\dfrac{x^2}{4}-\dfrac{z^2}{4}=1$, and the trace in the $yz$-plane ($x=0$) is $y^2-\dfrac{z^2}{4}=1$; both are hyperbolas. Setting $z=0$ in each gives the vertices $(\pm2,0,0)$ and $(0,\pm1,0)$, which mark where the surface pinches in most tightly.

Since the horizontal traces are always (non-degenerate) ellipses and the vertical traces are hyperbolas, the classification of Table 1 identifies this as a hyperboloid of one sheet. The variable with the negative sign is $z$, so the axis of symmetry is the $z$-axis.

Solution. Dividing the given equation by $-4$ produces the standard form

Two variables, $x$ and $z$, carry a negative sign, while only $y$ has a positive coefficient, so by the classification of Table 1 this is a hyperboloid of two sheets with axis the $y$-axis.

For the traces in the coordinate planes: setting $z=0$ gives $-x^2+\dfrac{y^2}{4}=1$, a hyperbola, and setting $x=0$ gives $\dfrac{y^2}{4}-\dfrac{z^2}{2}=1$, also a hyperbola. But setting $y=0$ gives $-x^2-\dfrac{z^2}{2}=1$, whose left side is at most $0$ for every real $x,z$, so no point satisfies it — the $xz$-plane carries no trace at all.

In the plane $y=k$, $x^2+\dfrac{z^2}{2}=\dfrac{k^2}{4}-1$. This is an ellipse only when the right side is positive, i.e. $|k|\gt2$; it is a single point at $|k|=2$ and empty for $|k|\lt2$.

Hence the surface is a hyperboloid of two sheets with axis the $y$-axis, its two sheets lying in $y\ge2$ and $y\le-2$, with vertices at $(0,\pm2,0)$.

Solution. Completing the square in $x$, $x^2-6x=(x-3)^2-9$, turns the given equation into $(x-3)^2-9+2z^2-y+10=0$, i.e.

Here $y$ occurs only to the first power while $(x-3)$ and $z$ occur only squared, both with positive coefficients, so by the classification of Table 1 this is an elliptic paraboloid whose axis is parallel to the $y$-axis. The right side vanishes only at $x=3,\ z=0$, where $y=1$, so the vertex is $(3,1,0)$.

The trace in the plane $y=k$ is $(x-3)^2+2z^2=k-1$: an ellipse for $k\gt1$, the single point of the vertex at $k=1$, and empty for $k\lt1$. The trace in the $xy$-plane ($z=0$) is $y=1+(x-3)^2,\ z=0$, a parabola starting at the vertex $(3,1,0)$ and opening in the direction of increasing $y$.