Lets say you have a function:

$$f(x) = 3x^4, \quad 0 \leq x \leq 1$$

and we want to revolve it around the x-axis. We can find the volume of the solid created by:

$$\pi \int_{0}^{1} (3x^4)^2 dx$$

However, instead of finding the volume of the solid, is there a way to find the equation of the solid in terms of $x$ and $y$?

For example, I believe if $f(x)=x^2$, the result should be $g(x,y)=x^2+y^2$.

Any help would be greatly appreciated!


1 Answer 1


If we have the function $f$ defined on $D \subseteq \mathbb R$ so that $f(x) \ge 0$ for all $x \in D$, revolving its graph about the $x$-axis gives us the points $p = (x,y,z)$ whose distance from the $x$-axis is equal to $f(x)$. Since $f(x) \ge 0$, this happens iff the squared distance from $p$ to the $x$-axis is equal to $f(x)^2$. Thus, the surface of revolution about the $x$-axis consists precisely of those points $(x,y,z)$ so that $x$ is in $D$ and the equation$$ f(x)^2 = y^2 + z^2 $$ holds.

If we want $z$ as a function of $x$ and $y$, we can rearrange the above equation to $$ z^2 = f(x)^2 - y^2 $$ which tells us that $z$ will be defined whenever $f(x)^2 - y^2 \ge 0$, and it will have up to two values given by $$ z = \pm \sqrt{f(x)^2 - y^2}$$ So for your example of $f(x) = x^2$, we have the equation $y^2 + z^2 = x^4$, which can also be written as $z = \pm \sqrt{x^4 - y^2}$.

  • $\begingroup$ Awesome, thanks! I presume a similar method could be applied to revolving around the y-axis? Sort of finding the distances between the axis of rotation and the surface? $\endgroup$
    – Anders
    Commented Mar 5, 2016 at 19:55
  • $\begingroup$ Yes. That would work for a function $f$ defined on $D \subseteq [0,+\infty)$ by taking $y = f(r_y)$, where $r_y = \sqrt{x^2 + z^2}$ is the distance to the $y$-axis. $\endgroup$ Commented Mar 6, 2016 at 0:24

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