Visualizing surfaces in three dimensions. Graphing in three dimensions, an example So, let's say we have a more complicated function, how do we see it? We will examine the graph of f(x, y) = 1 − x2 − y2. First, we set x = 0; when x = 0, y = 1. Then we evaluate f at (0, 1). This yields f(0) = 1. Next, we set y = 0; when y = 0, x = −1. Then we evaluate f at (−1, 0). This yields f(−1) = −1. Finally, since f(−1) < 0 and f(0) > 0, the surface z = 1 − x2 − y2 is concave upward in the plane consisting of points (x, y) with both x and y greater than −1 and less than or equal to zero. Then z becomes 1 minus y squared. This is a parabola pointing downward and stopping at 1. So we should draw maybe this downward parabola. It starts at 1, and it cuts the y-axis at 1. When y is 1, that gives us 0. So we might have an idea of what it might look like, or maybe not. Let's get more slices—the horizontal ones—and see what happens in the xz-plane, which is a vertical plane that's coming toward us. So in the xz- plane, which we set y equal to 0 and we get z equals 1 minus x squared, it's again a parabola coming downward. So we are going to draw a parabola that goes downward and also forward and backward. So we're starting to have a slightly better idea but we still don't know whether a cross-section of this thing might be round, square, something else. So if we want more confirmation we might want to figure out maybe where this surface intersects the xy- plane? So we hit the xy-plane when z equals 0. That means 1 minus x squared minus y squared should be 0. That becomes x squared plus y squared equals 1. That is a circle of radius 1. That's the unit circle. We can now imagine that we slice the unit circle by a vertical plane, producing a downward parabola that is actually the surface of revolution for a given ratio of diameter to radius. One way to determine the graph is to guess what it looks like, and then plug the values into an equation. Another way is to ask your computer to solve for the points on the graph.