3
$\begingroup$

In many examples of computation of multiple integrals, it is necessary to change the order of integration to achieve the computation. For example, $I=\int_0^1\int_y^1 \cos(x^2)\ dx\ dy$ can be computed using Fubini's theorem as $I=\int_0^1\int_0^x\cos(x^2)\ dy\ dx=\int_0^1 x\cos(x^2)\ dx=\left[\frac{cos(x^2)}{2}\right]_0^1=\frac{\cos(1)}{2}$.

It is generally easy to change the order of integration in double integral, but way more difficult in some cases of triple (or more) integrals.

For example,

$\begin{array}{rcl}\int_0^1\int_0^{1-x^2}\int_0^{1-x}f(x,y,z)\ dy\ dz\ dx & = & \int_0^1\int_0^{1-\sqrt{1-z}}\int_0^{\sqrt{1-z}}f(x,y,z)\ dx\ dy\ dz \\ & + & \int_0^1\int_{1-\sqrt{1-z}}^{1}\int_0^{1-y}f(x,y,z)\ dx\ dy\ dz \end{array}$

Is there a general process to obtain this result? Please note that I know how to tackle this example, I am looking for a generic method/algorithm (hopefully simple enough to explain it to high school students.) Thank you.

$\endgroup$

1 Answer 1

1
$\begingroup$

In the first place, you should disambiguate the integration bounds, indicating to what variable they relate, and show the nesting, like

$\int_{y=0}^{1}[\int_{x=y}^{1}\cos(x^2)\ dx]\ dy=\int_{x=0}^{1}[\int_{y=0}^{x}\cos(x^2)\ dy]\ dx$.

Now, the domain of integration is represented by the inequations $0\le y\le1$ and $y\le x\le 1$, a triangle inscribed in the tile $[0,1]\times[0,1]$.

Choosing a value for $y$, let $Y$, gives you a range for $x$: $Y\le x\le 1$. Similarly, choosing a value for $x$ gives you a range for $y$: $0\le y\le X$.

Your second example is presumably based on the inequations $0\le x\le 1$, $0\le z \le 1-x$ and $0\le y\le 1-x^2$, a truncated parabolic cone inside $[0,1]\times[0,1]\times[0,1]$.

Choosing a value for $z$ gives you a domain for $(x, y)$: $0\le Z\le 1-x$, i.e. $0\le x\le Z$ or $1-Z\le x\le 1$, and $0\le y \le 1-x^2$.

Then choosing a value for $y$ gives you a range for $x$: $x^2\le 1-y$, i.e. $-\sqrt{1-y}\le x\le \sqrt{1-y}$.

Hence: $\int_{z=0}^1[\int_{y=0}^{z}[\int_{x=-\sqrt{1-z}}^{\sqrt{1-z}}dx]dy+\int_{y=1-z}^{1}[\int_{x=-\sqrt{1-z}}^{\sqrt{1-z}}dx]dy]dz$

The general methodology is to write down the inequations describing the integration domain, then to choose the first integration variable and to "slice" the domain by making this variable a constant. You will obtain a sub-domain in $d-1$ dimensions, and you can repeat the process by choosing the next independent integration variable.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .