Problem 320
Medium Cylinders and Cones

Find the surface obtained by rotating z=x^2, y=0 about a)…

Find the surface obtained by rotating $z=x^2,\ y=0$ about a) the $z$-axis; b) the $y$-axis.

Explanation of Method

Rotating the parabola z = x² about an axis in its plane and about an axis perpendicular to it.

Step 1: a) About $Oz$

The distance from the axis, $|x_0|$, becomes $\sqrt{x^2+y^2}$, and the height $z$ stays the same: $z=x^2+y^2$.

This is a circular paraboloid.

Step 2: b) About $Oy$

The curve lies in the plane $y=0$, which is perpendicular to the $y$-axis. Every point $(x_0,0,x_0^2)$ turns on a circle in that same plane, centered at $O$, with radius $\sqrt{x_0^2+x_0^4}$. This radius takes every value in $[0,\infty)$.

\[\text{b) the whole plane } y=0\ \text{(degenerate)}\]

No proper surface of revolution arises in case b).

x y z $z=x^2$
a) The parabola $z=x^2,\ y=0$ (red) rotated about $Oz$ gives the paraboloid $z=x^2+y^2$.

Result

\[\boxed{\,z=x^2+y^2\,}\]

\[\boxed{\,\text{b) the whole plane } y=0\ \text{(degenerate)}\,}\]

Remark

Correction: the source replaces the curve by $y=x^2,\ z=0$ and answers $y=x^2+z^2$. That is the rotation of a different curve. If the intended axis is $Ox$, which lies in the plane of the curve, the distance $|z_0|=x^2$ becomes $\sqrt{y^2+z^2}$:

\[ y^2+z^2=x^4 . \]