Inner volume of quadric surface Can anyone help me work out the inner volume of the following quadric surface $x^{2/3} + y^{2/3} + z^{2/3} = 1$?
Edit: Adjusted the title and the line above but have left the rest of the question unchanged. The original question was in fact incorrectly formulated  due to a misunderstanding on my behalf. When plotting the graph in WA the surface only existed in the 0,1 plane on all 3 axes. This is of course incorrect as pointed out by @Rory - who actually provided the correct answer to my original question. However as I was looking for the internal volume I have marked @Kuije's answer as the correct one. Thank you both for your assistance!
Original Question
My reasoning thus far after reading through other questions posted on the site is that:
Given that the surface sits between 0 and 1 on all 3 axes one could work out the volume on any plane as they would all be the same. So R=[0,1]x[0,1].
Thus if I simply to $z=\sqrt{(1-x^{2/3}-y^{2/3})^3}$ then the volume should be defined by the following double integral:
$\iint z \, dA = \int_0^1 \int_0^1 \sqrt{(1-x^{2/3}-y^{2/3})^3} \, dy \,dx$.
However that is as far as I get and I'm really not sure how to go about working out the double integral.
 A: This is more of a clarification of your problem rather than an answer, but you do need to address these issues. Also, this will not fit into a comment.
First, what do you mean by "under the following quadric surface"? The surface surrounds a bounded volume, so it makes sense to speak of "inside the surface", but "under the surface" is not clear. From the rest of your question it seems that you want the part of the volume above the plane $z=0$ and for $x\ge 0,\ y\ge 0$, which is in the first octant and would be one-eighth the volume of the entire region. That is a defensible interpretation of your problem but is not the only one.
If you use Cartesian double integration, the next step is to find the 2D region on the $xy$ plane inside the curve given with $z=0$. However, that region is not $[0,1]\times[0,1]$. For one thing, either $x$ or $y$ can range from $-1$ to $1$. Now if you are told to find the volume only in the first octant, then each $x$ and $y$ are limited to $[0,1]$--but that is not what you said.
Let's assume we want only the first octant. But still, we cannot have both $x=1$ and $y=1$ at the same time. The 2D region you want is $x^{2/3}+y^{2/3}=1$ for $x\ge 0,\ y\ge 0$, but the left hand side is $2$ if $x=y=1$, not $1$. However, you can let $x$ range from $0$ to $1$, but for a given $x$ the limits on $y$ are
$$0\le y\le \sqrt{\left(1-x^{2/3}\right)^3}$$
Therefore, your desired volume in the first octant is
$$\int_0^1 \int_0^{\sqrt{\left(1-x^{2/3}\right)^3}} \sqrt{\left(1-x^{2/3}-y^{2/3}\right)^3} \,dy\,dx$$
Good luck with that!
A: Assuming you want half of the volume inside the solid defined by 
$
x^{2/3}+y^{2/3}+z^{2/3}=1,
$
$$
V=\int_0^1A(z)\; dz,
$$
where $A(z)$ is the area of a slice of the solid at height $z\in [0,1]$. For a fixed $z$, such a slice has equation
$$
\frac{x^{2/3}}{1-z^{2/3}}+\frac{y^{2/3}}{1-z^{2/3}}\le 1
$$
Therefore its boundary can be parametrized as follows:
\begin{cases}
x(t)=\cos^3 t\sqrt{(1-z^{2/3})^3}\\
y(t)=\sin^3 t\sqrt{(1-z^{2/3})^3}\\
\end{cases}
with $t\in [0,2\pi]$. By Green's Theorem, this slice has area
$$
A(z)=\frac{1}{2}\oint_0^{2\pi}x(t)y'(t)-y(t)x'(t)\; dt =
\frac{3}{8}\pi -\frac{9}{8}{z}^{2/3}\pi +\frac{9}{8}{z}^{4/3}\pi -
\frac{3}{8}{z}^{2}\pi
$$
and it follows that
$$
V=\int_0^1A(z)\; dz= \frac{2\pi}{35}
$$
