Finding the volume using triple integrals I am curious how I would describe the region $$y^2=x$$ bounded by the lines $y=1$, $x=1$, $x=0$ and $y=0$ revolved about the $y$ axis in 3 dimensions in order to find the volume in terms of triple integrals. (The solid looks like a cone.) This can easily be done with 1 integral using the "disk" or "shell method, however suppose I don't know those methods and I wanted to find the volume from a very general idea (i.e. triple integrals). What would be the limits of integration in the $z$ plane? Would the limits in the $x$ and $y$ planes simply be from $0$ to $1$?
 A: Let me clarify that the volume to be found is of

the solid obtained by revolving around the $y$-axis the region bounded by $y=1$, $x=0$ and $y=\sqrt x$ (hence a cone-like shape).

Cross-sections of the solid perpendicular to the $y$-axis are circles. If the cutting plane is $y=a$ then the circle is described by the equation
$$x^2+z^2=(a^2)^2$$
The outermost layer of our triple integral in Cartesian coordinates will be with respect to $y$, which has range $[0,1]$:
$$\int_0^1\dots dy$$
For a given $y$, since the circular cross-section has radius $y^2$ it makes sense to have the middle integral with respect to $x$ and ranging over $[-y^2,y^2]$, the circle's diameter:
$$\int_0^1\int_{-y^2}^{y^2}\dots dx\ dy$$
Given some $x$ and $y$ we must have $-\sqrt{y^4-x^2}\le z\le\sqrt{y^4-x^2}$ if the point $(x,y,z)$ is to lie in the solid. The last integral with respect to $z$ has these limits, and the integrand is just 1 since we are finding volume:
$$V=\int_0^1\int_{-y^2}^{y^2}\int
_{-\sqrt{y^4-x^2}}^{\sqrt{y^4-x^2}}1\ dz\ dx\ dy$$
Evaluating these integrals from the inside out we have
$$V=\int_0^1\int_{-y^2}^{y^2}2\sqrt{y^4-x^2}\ dx\ dy$$
$$V=\int_0^1\pi y^4\ dy=\frac\pi5$$
This answer matches the volumes obtained from the shell method ($2\pi\int_0^1x(1-\sqrt x)\ dx$) or the disc method ($\pi\int_0^1y^4\ dy$).
Just for reference, here is a triple integral for the volume of the same solid in cylindrical coordinates:
$$V=\iiint_R r\ dr\ d\varphi\ dy\\=\int_0^1\int_0^{2\pi}\int_0^{y^2}r\ dr\ d\varphi\ dy=\frac\pi5$$
A: $$
\int_0 ^{2\pi}\int_{0}^{R=1}\int_{0}^{\sqrt{r}} r dz dr d\theta
$$
