Science:Math Exam Resources/Courses/MATH105/April 2011/Question 09 (e)/Solution 1

As the hint suggests, we can check each equation with the point (0,0,0), which lies on the surface. The point (0,0,0) does not lie on the surface given by $x^{2}+y^{2}+z^{2}=1$, so we can rule that out.

Since the surface takes negative $z$ values and $z=x^{2}+y^{2}$ only takes non-negative values, we can rule this out as an option as well.

Finally, we notice that the surface is symmetric about the $x,y$ plane (this means that for a given (x,y) we get a value z above the xy-plane and the same value (but negative), -z below the xy-plane. For this reason, we can rule out $z=x^{2}-y^{2}$ since for a given (x,y) we only produce one z value.

Thus the surface is $z^{2}=x^{2}+y^{2}$.

We can check that this solution makes sense by solving the equation for $z$ to get $z=\pm {\sqrt {x^{2}+y^{2}}}$. The two solutions give identical surfaces above and below the $x,y$ plane, the point (0,0,0) does indeed lie on the surface $z^{2}=x^{2}+y^{2}$, and as $x$ and $y$ increase, so does $z$. These all match the given graph.

The level curve at $z=2$ is what we would see if we sliced into the surface horizontally at $z=2$ and looked at it from above. We can find the equation by substituting into the equation for the surface: $2^{2}=x^{2}+y^{2}$. This is simply a circle of radius $2$, the graph as follows: