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?

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It will look like a wonky eighth of a sphere. If we were looking at $x^2+y^2+z^2=1$, then it would be the sphere, however this will look like a strange spiky sphere. Suggestion: Try using the divergence theorem. This will allow you to write the volume as a double integral. With $F=x\boldsymbol{i}+y\boldsymbol{j}+z\boldsymbol{k}$, we have that $\nabla\cdot F=3$, and so $$\iiint_{V}\nabla\cdot FdV=3V.$$ By the divergence theorem, we also have that this equals $$\iint_{S}F\cdot n dS.$$ Unfortunately, I think this last integral is more difficult to evaluate than the triple integral. –  Eric Naslund Dec 18 '12 at 18:29
Have you tried change of variables? –  23rd Dec 18 '12 at 19:07

by using curve expert program this is an approximation formula in the first quadrant:

v=A*B*C*(a*b+c*n^d)/(b+n^d)

for n={.3,....,1}

a=-7.996*10^-4

b=9.2*10^-1

c=3.198*10^-1

d=4.7

for n={1,....,100}

a=-1.77*10^-1

b=2.43

c=1

d=1.83

so for n = .5 ,volume approximated =0.012

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Does this address the question....? The exact volume ($1/90$) was given by earlier respondents. :) –  user86418 Mar 6 at 13:12
The question was not about the eigth of a sphere. –  flonk Mar 6 at 13:41

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.

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Mathematica evaluates same double integral exactly as $1/90$. –  Mike Spivey Dec 18 '12 at 19:11
Mike: I have only an old version of maple -- maybe maple has improved since. Thanks for the exact answer, as I was only fairly sure of 1/90 to so many decimals. –  coffeemath Dec 18 '12 at 19:43

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: $1/90$

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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}

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Very cool, Mike! –  JohnD Dec 18 '12 at 19:44
@JohnD: Thanks! The figured the OP's equation was too nice not to have a name. It took me a while to track it down. –  Mike Spivey Dec 18 '12 at 19:47

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}

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+1: Nice, Christian. I imagine that's how the general formula for the volume of a superellipsoid is obtained. –  Mike Spivey Dec 18 '12 at 20:02