# Triple Integral Problems

• Aug 1st 2006, 11:49 AM
Triple Integral Problems
Question: List five other iterated integrals that equals to -

a) $
\int_0^1 \int_y^1 \int_0^y f(x,y,z) dz dx dy
$

b) $
\int_0^1 \int_0^{x^2} \int_0^y f(x,y,z) dz dx dy
$

To be honest, I don't even know how to start this problem...

Thank you.

KK
• Aug 1st 2006, 01:53 PM
ThePerfectHacker
The problem asks to change the order of interated integration. By Fubini's Theorem order in iterated integration is not important. Thus there are six other representations of an integral each having,
$\left\{ \begin{array}{c}dxdydx\\dxdzdy\\dydxdz\\dydzdx\\dz dxdy\\dzdydx$
That is why there are 5 others cuz one is already given.
• Aug 1st 2006, 02:37 PM
Yes, that is what I thought initially, but I looked up the answer from the back of the book, even the limits of integrals had been changed.

Here is the answer to a), but I have no idea how to get it:

$
\int_0^1 \int_0^x \int_0^y f(x,y,z) dz dy dx
$

$
\int_0^1 \int_z^1 \int_y^1 f(x,y,z) dx dy dz
$

$
\int_0^1 \int_0^y \int_y^1 f(x,y,z) dx dz dy
$

$
\int_0^1 \int_0^x \int_z^x f(x,y,z) dy dz dx
$

$
\int_0^1 \int_z^1 \int_z^x f(x,y,z) dy dx dz
$

Any thoughts?

KK
• Aug 1st 2006, 03:53 PM
galactus
It may help if you list out the respective limits of your original integral, then you can see how they switched things around.

z=0
z=y

y=0
y=x

x=0
x=1

Now, see how they arrived at the second one?.
• Aug 1st 2006, 04:05 PM
Soroban
Hello, KK!

Quote:

List five other iterated integrals that equals to:

. . $a)\;\int_0^1 \int_y^1 \int_0^y f(x,y,z)\, dz\, dx\, dy$

I write in the limits "completely".

. . $\int^{y = 1}_{y = 0}\int^{x=1}_{x=y}\int^{z=y}_{z=0}f(x,y,z)\,dz\,dx \,dy$

I start with the "outside" limits: . $y = 0,\;y = 1$
That is, $y$ ranges from $0$ to $1.$
Code:

            y             |       - - -1+ - - - - -       :::::|:::::::::       :::::|:::::::::       :::::|:::::::::       ------+---------- x             |

The next limits are: . $x = y,\;x = 1$
That is, $x$ ranges between the line $x = y$ and $x = 1$
Code:

            y             |       - - -1+ - - - * -             |    /:|             |  /:::|             | /:::::|       ------/-------+-- x             |      1

Now lay this diagram "on the floor"; the z-axis goes straight up.
Code:

              z               |               |               |       - - - - *---------- y             *::*           *:::::*         *::::::::*       * * * * * * *     /   x

I can't draw this next part . . . hope you can follow.

With no restriction on $z$, we have a triangular prism
. . with the above triangle as a cross-section, extending up and down infinitely.

But the limits are: $z = 0$ and $z = y.$
. . That is, $z$ ranges between the "floor" and the slanted plane $z = y.$

So our triangular prism is cut off at floor-level below
. . and the slanted plane $z = y$ above.

And that is the solid we are dealing with.

Now if we change the order of integration, the limits are changed.

With the first of the answers, we have: $dz\,dy\,dx$

We will try to describe the same solid using this order of limits.

Starting outside: $x$ goes from $x = 0$ to $x = 1$

Then: $y$ goes from $y = 0$ to $y = x$

Then: $z$ goes from $z = 0$ to $z = y$

So the integral is: . $\int_{x=0}^{x=1}\int_{y=0}^{y=x}\int_{x=0}^{z=y}f( x,y,z)\,dz\,dy,\,dx$

. . I hope this helps . . .
• Aug 1st 2006, 06:00 PM
Thank you for your explaination, I worked the problem and I was managed to get the answers of dzdydx, dxdzdy, and dxdydz.

However, for dydxdz, dydzdx, and dydxdz, I still don't understand why are they the same solid as the given one, why is y=x, y=z?

Any ideas?

KK
• Aug 1st 2006, 08:32 PM
JakeD
Quote:

Thank you for your explaination, I worked the problem and I was managed to get the answers of dzdydx, dxdzdy, and dxdydz.

However, for dydxdz, dydzdx, and dydxdz, I still don't understand why are they the same solid as the given one, why is y=x, y=z?

Any ideas?

KK

As another way of solving these, I work with the inequalities describing the region of integration. Start with the original integral

$
\int_0^1 \int_y^1 \int_0^y f(x,y,z) dz dx dy .
$

The inequalities describing the region are

$
0 \le y \le 1
$

$
y \le x \le 1
$

$
0 \le z \le y .
$

To get the $dydxdz$ you asked for, we work with and rearrange the inequalities. We need the outer limit for $dz$ first. From the first and third inequalities, we have

$
0 \le z \le y \le 1
$

so we can write

$
0 \le z \le 1.
$

For $dx$ and $dy$, we have from the second and third inequalities

$
z \le y \le x \le 1,
$

so we have for $dx$

$
z \le x \le 1
$

and for $dy$

$
z \le y \le x.
$

The result is the new set of inequalities

$
0 \le z \le 1
$

$
z \le x \le 1
$

$
z \le y \le x
$

and the integral

$
\int_0^1 \int_z^1 \int_z^x f(x,y,z) dy dx dz .
$

In writing the new inequalities, you can use any implications from the original inequalities. To make sure you haven't left any essential implications out, check that any triple $(x,y,z)$ that satisfies either one set of inequalities also satisfies the other.