Question 10.6.4: Sketching a Paraboloid Draw a graph of the quadric surface x...

To get an idea of what the graph looks like, first draw its traces in the three coordinate planes. In the y z \text {-plane, we have } x=0 \text { and so, } y^2=z (a parabola). In the x z \text {-plane, we have } y=0 \text { and so, } x^2=z \text { (a parabola). In the } x y \text {-plane, we have } z=0 and so, x^2+y^2=0 \text { (a point-the origin). } We sketch the traces in Figure 10.57a. Finally,
since the trace in the x y \text {-plane } is just a point, we consider the traces in the planes z=k (for k > 0). Notice that these are the circles x^2+y^2=k, \text { where for larger values of } z (i.e., larger values of k), we get circles of larger radius. We sketch the surface in Figure 10.57b. Such surfaces are called paraboloids and since the traces in planes parallel to the x y \text {-plane } are circles, this is called a circular paraboloid.
Graphing utilities with three-dimensional capabilities generally produce a graph like Figure 10.57c for z=x^2+y^2. Notice that the parabolic traces are visible, but not the circular cross sections we drew in Figure 10.57b. The four peaks visible in Figure 10.57c are due to the rectangular domain used for the plot (in this case, −5 ≤ x ≤ 5 and −5 ≤ y ≤ 5).
We can improve this by restricting the range of the  z \text {-values. With } 0 \leq z \leq 15 \text {, }
you can clearly see the circular cross section in the plane z=15 in Figure 10.57d.

As in example 6.3, a parametric surface plot is even better. Here, we have x=s \cos t, y=s \sin t \text { and } z=s^2 \text {, with } 0 \leq s \leq 5 \text { and } 0 \leq t \leq 2 \pi \text {. } Figure 10.57e clearly shows the circular cross sections in the planes z=k, \text { for } k>0.