7
$\begingroup$

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?

  • 0
    Have you tried change of variables?2012-12-18

6 Answers 6

4

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 $\eqalign{{\rm vol}(V)&=\int_S 1\ J_\phi(u,v,w)\ {\rm d}(u,v,w)\cr&=\int_0^1 \int_0^{1-u}\int _0^{1-u-v} 8uvw\ dw\ dv\ du\cr &= \int_0^1\int_0^{1-u} 4u v(1-u-v)^2\ dv\ du\cr &=\ldots= {1\over90}\ .\cr}$

  • 0
    +1: I had just made that substitution in an EDIT to my answer, and then saw your answer here.2012-12-18
3

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.

  • 0
    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.2012-12-18
3

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 $\begin{align} V &= \frac{1}{8} \frac{2}{3} \frac{4}{(1/2)(1/2)} \frac{6}{(6) (120)} \\ &= \frac{4}{360} \\ &= \frac{1}{90}. \end{align}$

  • 1
    @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.2012-12-18
2

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:

Mathematica graphics

  • 0
    *Mathematica*. It has a built in function to plot solutions of systems of inequalities like this.2012-12-18
2

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: $1/90$

0

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

  • 0
    The questio$n$ was not about the eigth of a sphere.2014-03-06