# Give the equation of the surface

Given $$z = y^2 + 3,$$ give the equation of the surface if rotated around the $z$-axis.

After I plot this out, I get a simple parabola in the $yz$-plane... so flipping it about the $z$-axis is just a parabola opening down instead of opening up.. and thus I have $-z = y^2 + 3$... correct?

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If I read the question correctly you want the equation of a surface of revolution. I don't understand the second part of your question, but you want to imagine rotating the parabola around the z axis and the surface that it sweeps out. –  AnonymousCoward Jun 10 '12 at 22:05
What you have done is reflected the graph of the parabola in the plane given by $z=0$. Rotating the parabola around the $z$-axis should yield a surface, as you say. So picturing the parabola rotated around the $z$-axis, what do you obtain? –  Alex Petzke Jun 10 '12 at 22:06
Think polar coordinates in the $xy$ plane. –  AnonymousCoward Jun 10 '12 at 22:06
In 3space, this is just a cylinder on the zy plane correct? So if I rotate it around the z-axis... nothing changes? –  Nick Jun 10 '12 at 22:17
Think of it this way. If you take $y = 1$ in the $yz$ plane you have the equation of a line. If you consider $(x^2+y^2)^{1/2} = 1$ then you have the equation of a cylinder in 3space. Do you see what I am hinting at? –  AnonymousCoward Jun 10 '12 at 22:32
$z = y^2 +3$ is a parabola opening upwards in the $yz$-plane. For any point along the $y$-axis the vertical distance from $y$ to the rotated surface is $y^2 +3$. Let $(x', y')$ be any point in the $xy$-plane. Then under rotation $(x',y')$ crosses the $y$-axis in the points where $y^2 = x'^2 + y'^2$. But as the height is constant when we rotate around the $z$-axis it follows that the surface is given by $z = x^2 + y^2 +3$.