1. Stuck on this problem

I've included a picture stating the problem. How do I approach this?

I recognize that $x^2 + y^2$ is the equation of a circle, but that's about all I have.

2. Originally Posted by Ares_D1
I've included a picture stating the problem. How do I approach this?

I recognize that $x^2 + y^2$ is the equation of a circle, but that's about all I have.
When you rotate the curve $z = e^{-x^2}$ about the z axis you get the surface $z = e^{-r^2}$ (in cylindrical polar coords). Since $x = r \cos \theta$, $y = r \sin \theta$ then $r^2 = x^2 + y^2$, giving your answer.

3. Originally Posted by Danny
When you rotate the curve $z = e^{-x^2}$ about the z axis you get the surface $z = e^{-r^2}$ (in cylindrical polar coords). Since $x = r \cos \theta$, $y = r \sin \theta$ then $r^2 = x^2 + y^2$, giving your answer.
Thanks for the help Danny, but I'm not sure I follow that last bit. Could you explain further how $x = r \cos \theta$ and $y = r \sin \theta$, and how, from that, you end up with $r^2 = x^2 + y^2$?

4. Originally Posted by Ares_D1
Thanks for the help Danny, but I'm not sure I follow that last bit. Could you explain further how $x = r \cos \theta$ and $y = r \sin \theta$, and how, from that, you end up with $r^2 = x^2 + y^2$?
$x = r \cos \theta$ and $y = r \sin \theta$ are polar coordinates - an alternate way to describe a point in 2D. Since

$
\sin ^2 \theta + \cos ^2 \theta = 1, \text{then }\;x^2 +y^2 = r^2\left(\cos ^2 \theta + \sin ^2 \theta \right) = r^2
$

5. By the way, the statement " $z= e^{x^2}$ lies in the xy-plane" is incorrect. Since you are then to rotate around the z-axis, I imagine it was supposed to be "lies in the xz-plane".