How can I find the volume of a solid defined only by inequalities? For example, in this case I have: $$0\le z \le y \le x \le 1$$
Can someone please explain to me step-by-step on how I can do this. This is a very new concept to me.
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maybe you should start to "draw" $0\leq y \leq x\leq 1$ in the plane to get an idea on how things behave – Surb Jul 20 '15 at 18:20
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@Surb I can't grasp this conceptually – Nina Jul 20 '15 at 18:21
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I don't know any general technique (and I remember that I also struggled a lot on interpreting volumes at the beginning). This is indeed a tetrahedron. But have a look in the plane first it is a triangle. Just pick some value of $x$ and look what you can do with $y$, and try to "slice the triangle" (e.g. choose $x=0,0.1,0.2,...,0.9,1$) you will see the triangle appearing. In space you can proceed somehow the same way to get an intuition of the shape. – Surb Jul 20 '15 at 18:25
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That volume is exactly the probability that $$ Z\leq Y\leq X $$ occurs, with $X,Y,Z$ being three indepent random variables, uniformly distributed over $[0,1]$. Quite trivially any arrangement of $X,Y,Z$ has the same probability, hence the wanted volume equals $\dfrac{1}{3!}=\displaystyle\color{red}{\dfrac{1}{6}}.$
Without exploiting symmetry:
$$ V = \int_{0}^{1}\int_{0}^{x}\int_{0}^{y}1\,dz\,dy\,dx = \int_{0}^{1}\int_{0}^{x}y\,dy\,dx=\int_{0}^{1}\frac{x^2}{2}\,dx=\frac{1}{6}.$$
wythagoras
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Jack D'Aurizio
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1Wow I never looked at it this way. But can you also show the solution in a more 'calculus' way? – Nina Jul 20 '15 at 18:28
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