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What i wanna ask is that how do you tackle the problem oif multiple integral when you are not able to draw the diagram?Also how you would determine the order of integrand. For instance, what is the volume bounded by the surface $x^3+xyz^2, x^3y^2+z^y, z+y+2x=10$ Those function are just creat without any try, so may be it is not solvable, but what i want to say is if we face the function which cannot really draw it, how would you deal with it.

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I would suggest that you first think about the two-dimensional case and how you would tackle it without a diagram. – Phira May 1 '12 at 9:20
up vote 1 down vote accepted

First of all, "$x^3 + xyz^2$" is not a surface. "$x^3 + xyz^2 = c$" for some constant $c$, defines a surface.

Second of all, when you have a volume delimited by functions like this : $f_1(x,y,z) = 0$, $f_2(x,y,z) = 0$, $\dots$, $f_n(x,y,z) = 0$, the idea that works most of the time (and is almost always the only one) is to pull off a change of a variables that results in some integral of this form : $$ \iiint_V f(x,y,z) \, dx \, dy \, dz = \int_{a}^{b} \int_{c(u)}^{d(u)} \int_{e(u,v)}^{g(u,v)} f(u,v,w) \frac{\partial(u,v,w)}{\partial(x,y,z)}\, dw \, dv \, du $$ where $u$, $v$ and $w$ are chosen as functions of $x,y,z$ and this guy $\left( \frac{\partial(u,v,w)}{\partial(x,y,z)} \right)$ is the jacobian of your change of variables. This way you can compute your first integral with respect to $w$ and obtain a resulting function of $u$ and $v$, which you can then integrate with respect to $v$ to obtain a function of $u$ and then complete your integral.

The trick always resides in the change of variables. Here is an example I am fond of.

Compute the integral $$ \iiint_V \frac 1{(x+y+z)^3} \, dx \, dy \, dz $$ where $V = \{ (x,y,z) \in \mathbb R^3 \, | \, x,y,z \ge 0, x+y+z \le 1 \}$.

Now clearly we would want to change the $x+y+z$ to some variable, right? So let us do that : let $w(x,y,z) = x+y+z$, because $x+y+z$ feels like an "height" component for me, so I put it in the third variable (that's rather arbitrary, but it'll help computations). What about $u$ and $v$? Since they don't appear in the integral, let's make them as simple as possible : $u(x,y,z) = x$, and $v(x,y,z) = y$. (Another reason for why I let $u = x$ and $v = y$ is because I only want to change one coordinate, not all three of them, so perhaps the less notationally convenient way to do this is to let "$x = x$", "$y = y$ and $w = x+y+z$ (the new $x$, "$u$", has the same value as $x$, and the new $y$, $v$, has the same value as $y$). Now what happens with the integral? Well if you compute the jacobian : $$ \frac{\partial(u,v,w)}{\partial(x,y,z)} = \left| \begin{bmatrix} 1 & 0 & 0 \\\ 0 & 1 & 0 \\\ 1 & 1 & 1 \end{bmatrix} \right| = 1 $$ (Maybe I computed the transpose of the jacobian or some weird turning around of numbers... I never recall in which order the columns appear! But I have faith that the jacobian is $1$ for geometrical reasons.)

It remains to compute the bounds $a$,$b$,$c(u)$, $d(u)$, $e(u,v)$ and $g(u,v)$ to finish. Just take one variable at a time and look at its maximal range in function of the preceding variables which have bounds determined : the constraints are $x,y,z \ge 0$ and $x + y + z \le 1$. We need to translate those constraints in $u,v,w$. The first one says that $w - v - u, v, w \ge 0$, and the second one says that $u \le 1$. Now play around a little bit and you'll get $0 \le u+v \le w \le 1$, $u \ge 0$, $v \ge 0$. This means that if we start with the variable $u$, $0 \le u \le 1$, then fixing some $u$, we have $0 \le v \le 1-u$, and now fixing $u$ and $v$ gives $u+v \le w \le 1$. (You can check that these bounds are indeed correct by making sure that an element of $\mathbb R^3$ is in $V$ if and only if it satisfies those conditions.)

Hence $$ \iiint_V \frac 1{(x+y+z)^3} \, dV = \int_0^1 \int_0^{1-u} \int_{u+v}^1 \frac 1{w^3} \, dw \, dv \, du. $$ The rest is elementary integrals ; I leave them up to you if you wish to try.

Hope that helps,

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i see you point, but if there involves more surface like 4 or more, then it seems that it is not easy to find the bound. And i think transformation is a good idea but it may be very difficult to think what should be transformed like sometime we may need to transform into the spherical coordinates and sometime just one dummy variable is fine – Mathematics May 1 '12 at 17:50
@Mathematics (...) It's not like multiple integrals are supposed to be easy to compute! I never said it was easy, I'm just saying that in the general case that is the most efficient trick we have ; Phira's answer goes in that direction, that in general, there is no "simple" expression to perform this task... integrating between $a$ and $b$ can already be damn hard in $\mathbb R$ as it is, don't expect things to get any easier in $\mathbb R^3$! – Patrick Da Silva May 1 '12 at 19:16

I doubt that there is a general method to show whether an integral between surfaces can be expressed "simply" (because it is already difficult to decide whether a one-dimensional integral can be expressed with elementary functions or whether $\zeta(5)$ is irrational).

For polynomial surfaces, one way to reduce the problem to one-dimensional integrals is cylindrical algebraic decomposition AFTER you decide on which side of each surface you search for your bounded region (but this can be decided by trial and error to find the non-empty regions).


The only general method is: Calculate the curves of intersection if you can, decide on the relative position of the surfaces within the closed portions of these curves of intersection and then parametrize with the curve and the height above the surfaces in the interior. But it is more instructive to do this first for a 2-dimensional case and only make the picture afterwards to understand the method better.

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