Let’s say we have $f(x)=\sqrt{x}$ in range $\,x\in[0,5]\,$ and we revolved around the $x$-axis what would the function $z=z(x,y)$ of this new surface? How can I find this surface?
I have the volume of this surface:
$$\int_{0}^{5}\pi xdx = V$$
in 2D the rate of change of the area under the curve is $f(x)$ will this be the same in 2D with the volume, how can I find the equation $\,z=z(x,y)\,?$
I was trying:
$$A = \int\pi xdx$$ $$\frac{dA}{dx} = \pi x$$
If you are interested in the equation of the surface, then read my answer.
A point $P(x,y,z)$ belongs to the surface if and only if the distance from $P$ to $x$-axis is equal to $f(x)=\sqrt x$.
$\mathrm{dist}(P,x\mathrm{-axis})=\sqrt{y^2+z^2}\,.$
Consequently a point $P(x,y,z)$ belongs to the surface if and only if $\,\sqrt{y^2+z^2}=\sqrt x\;.$
Hence the equation of the surface is
$y^2+z^2=x\quad$ (in implicit form)
or
$z=\pm\sqrt{x-y^2}\quad$ (in explicit form).
But if you are interested in calculating the value of the area of the surface, you have to use the following formula:
$\mathrm{Area}=2\pi\displaystyle\int_0^5f(x)\sqrt{1+f’^2(x)}\,\mathrm dx\,.$