# Finding the equation of a solid in terms of x and y when rotating a 2-d function about a line

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!

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 $z = \pm \sqrt{x^4 - y^2}$.
• 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. Commented Mar 6, 2016 at 0:24