Finding volume of a shape using double integral I'm trying to find the volume of a given shape:
$$ V= \begin{cases} \sqrt{x}+\sqrt{y}+\sqrt{z} \leq 1\\x\geq 0,\ y\geq 0,\ z\geq0\end{cases} $$
using double integral. Unfortunately I don't know how to start, namely:
$$ z = (1 - \sqrt{y} - \sqrt{x})^{2} $$
and now what should I do?
Wolfram can't even plot this function, I'm unable to imagine how it looks like...
Would it be simpler with a triple integral?
 A: For fun, here's a generalization.  The shape of the region is the part of a superellipsoid that resides in the first octant.  A superellipsoid is (obviously, I suppose) a generalization of an ellipsoid, which in turn is the 3D version of an ellipse.
A superellipsoid satisfies the inequality $$\left(\left|\frac{x}{A}\right|^r +\left|\frac{y}{B}\right|^r\right)^{t/r} + \left|\frac{z}{C}\right|^t \leq 1.$$
The volume of a superellipsoid is (according to the Wikipedia page) $$V = \frac{2}{3} ABC \frac{4}{rt} \frac{\Gamma(1/r)^2 \Gamma(2/t) \Gamma(1/t)}{\Gamma(2/r) \Gamma(3/t)},$$
where $\Gamma(x)$ is the gamma function. 
Since we just want the volume in the first octant, we need to divide this by $8$.  For the OP's question, we have $r = t = 1/2$ and $A = B = C = 1$.  Also, $\Gamma(n) = (n-1)!$ for integers $n \geq 1$.  
Thus the required volume is
$$\begin{align}
V &= \frac{1}{8} \frac{2}{3} \frac{4}{(1/2)(1/2)} \frac{6}{(6) (120)} \\
&= \frac{4}{360} \\
&= \frac{1}{90}.
\end{align}$$
A: The transformation
$$\phi:\quad (u,v,w)\mapsto(x,y,z):=(u^2,v^2,w^2)$$
maps the simplex
$$S:=\{(u,v,w)\ |\ u\geq0,\ v\geq 0,\ w\geq 0,\  u+v+w\leq1\}$$
one-one onto your body $V$, and the Jacobian of $\phi$ computes to
$$J_\phi(u,v,w)=8uvw\geq0\ .$$
It follows that volume of $V$ is given by
$$\eqalign{{\rm vol}(V)&=\int_S 1\ J_\phi(u,v,w)\ {\rm d}(u,v,w)\cr&=\int_0^1 \int_0^{1-u}\int _0^{1-u-v} 8uvw\ dw\ dv\ du\cr &=
\int_0^1\int_0^{1-u} 4u v(1-u-v)^2\ dv\ du\cr
&=\ldots=
{1\over90}\ .\cr}$$
A: The volume can be calculated as
$$\int_0^1 \int_0^{(1-\sqrt{x})^2} (1-\sqrt{x}-\sqrt{y})^2 dy dx.$$
Maple was able to find the inner integral (function of $x$), but unable to finish the evaluation by integrating that from 0 to 1. However a numeric approximation gave 
$0.011111111...$ which looks like it is $1/90$.
EDIT: If the substitutions $x=r^2, y=s^2, z=t^2$ are made, noting that $dx=2r \ dr$, $dy=2s\ ds$, and $dz=2t\ dt$, the volume element in the $r,s,t$ sysztem is $8rst\ dt \ ds \ dr$, and the volume becomes
$$\int_0^1 \int_0^{1-r} \int_0^{1-r-s}8rst\ dt\ ds\ dr=\frac{1}{90},$$
where the integration in the $r,s,t$ system is not complicated by squareroots.
Christian Blatter did exactly this substitution in his answer, and I just noticed that after doing the above edit. 
A: I'm not sure if you need help setting up the double integral or not, but here's the solid as well as the integral and answer for the volume:

A: Wolfram can plot this and do the nessesary integrals.
The indefinite double integral of the z value with respect to $x$ and $y$ is:
$V = c_1+y*c_2+4/9*x^{3/2}*(2*y^.5-3)y+y*x^2/2+xy(3y-8y^.5+6)/6$
The first integral with respect to x and boundary condition 0 to $(1-2y^.5+y)$ is:
$1/6 (-1+\sqrt{y})^3 (-9+8 \sqrt{(-1+\sqrt{y})^2}+9 \sqrt{y})$
Integrating that from $y = 0$ to $1$ is: $1/90$
