I want to find the function that satisfies the following DE:
$$\pi y(x)^2=\sqrt{1+(y'(x))^2}$$
This comes from the fact that the left-hand side gives the volume of radiation around the x-axis and the right-hand side gives the length of the curve. I also know that the initial condition is given by $y(a)=b$ where $a$ and $b$ are real values.
The questions I have are:
- What is the function $y(x)$?
- In what range of $x$ is the function $y(x)$ real-valued?
“The” function?
Take any function $f_1$ whose domain is an interval $[a,b],$ subject only to the restrictions that $(f_1(x))^2$ has a finite integral on the domain of $f_1$ and the length of the curve $y=f_1(x)$ is finite.
Now for any positive real number $t,$ define $f_t(x)=tf(x/t)$ on the interval $[ta,tb].$ That is, $f_t$ is $f_1$ “scaled up” in both the $x$ and $y$ dimensions.
For any choice of $f_t,$ the curve is $t$ times as long as the curve for $f_1$ but the volume of the solid of rotation is $t^3$ times as large. So if $f_1$ has a volume smaller than the curve's length, increase $t$ until the length and volume match. If the volume is larger than the length, decrease $t.$
You can make many very different functions that satisfy your requirements in this way.
A simple example is the function $y(x)=1/\sqrt\pi$ on any finite interval.