Calculus II - Quadric Surfaces

Section 12.4 : Quadric Surfaces

3. Sketch the following quadric surface.

\[z = \frac{{{x^2}}}{4} + \frac{{{y^2}}}{4} - 6\]

Show Solution

This is an elliptic paraboloid that is centered on the $z$-axis. Because the $x$ and $y$ terms are positive we know that it will open upwards. The “-6” tells us that the surface will start at $z = - 6$. We can also say that because the coefficients of the $x$ and $y$ terms are identical the cross sections of the surface will be circles.

Here are a couple of sketches of the region. We’ve given them with the more traditional axes as well as “boxed” axes to help visualize the surface.

$A three dimensional sketch of the surface $z = \frac{{{x^2}}}{4} + \frac{{{y^2}}}{4} - 6$ drawn with the traditional axes. It is a circular paraboloid centered on the z-axis that opens upward with its lowest point at $z = - 6$, so it looks like a tall bowl.$

$A three dimensional sketch of the same circular paraboloid $z = \frac{{{x^2}}}{4} + \frac{{{y^2}}}{4} - 6$, this time drawn with “boxed” axes to make the surface easier to visualize. It opens upward from its lowest point at $z = - 6$.$