Section 12.6: Cylinders & Quadric Surfaces

SOLUTION By substituting \(z = 0\), we find that the trace in the xy-plane is \(x^2 + y^2/9 = 1\), which we recognize as an equation of an ellipse. In general, the horizontal trace in the plane \(z = k\) is \[x^2 + \frac{y^2}{9} = 1 - \frac{k^2}{4}, \quad z=k\] which is an ellipse, provided that \(k^2 < 4\), that is, \(-2 < k < 2\). Similarly, vertical traces parallel to the yz- and xz-planes are also ellipses: \[\frac{y^2}{9} + \frac{z^2}{4} = 1 - k^2, \quad x=k \quad (\text{if } -1 < k < 1)\] \[x^2 + \frac{z^2}{4} = 1 - \frac{k^2}{9}, \quad y=k \quad (\text{if } -3 < k < 3)\]

Figure 4 shows how drawing some traces indicates the shape of the surface. It’s called an ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to each coordinate plane; this is a reflection of the fact that its equation involves only even powers of \(x\), \(y\), and \(z\).

SOLUTION If we put \(x = 0\), we get \(z = y^2\), so the yz-plane intersects the surface in a parabola. If we put \(x = k\) (a constant), we get \(z = y^2 + 4k^2\). This means that if we slice the graph with any plane parallel to the yz-plane, we obtain a parabola that opens upward. Similarly, if \(y = k\), the trace is \(z = 4x^2 + k^2\), which is again a parabola that opens upward. If we put \(z = k\), we get the horizontal traces \(4x^2 + y^2 = k\), which we recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface \(z = 4x^2 + y^2\) is called an elliptic paraboloid.

SOLUTION The traces in the vertical planes \(x = k\) are the parabolas \(z = y^2 - k^2\), which open upward. The traces in \(y = k\) are the parabolas \(z = -x^2 + k^2\), which open downward. The horizontal traces are \(y^2 - x^2 = k\), a family of hyperbolas. We draw the families of traces in Figure 6, and we show how the traces appear when placed in their correct planes in Figure 7.

In Figure 8 we fit together the traces from Figure 7 to form the surface \(z = y^2 - x^2\), a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in Section 14.7 when we discuss saddle points.

SOLUTION The trace in any horizontal plane \(z = k\) is the ellipse \[\frac{x^2}{4} + y^2 = 1 + \frac{k^2}{4}, \quad z=k\] but the traces in the xz- and yz-planes are the hyperbolas \[\frac{x^2}{4} - \frac{z^2}{4} = 1, \quad y=0 \quad \text{and} \quad y^2 - \frac{z^2}{4} = 1, \quad x=0\] This surface is called a hyperboloid of one sheet and is sketched in Figure 9.