Homework Help: Yet another volume

1. Jan 14, 2006

twoflower

Hi,

I'm having a trouble doing this:

Compute volume of the solid

$$T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}$$

First I need to express bounds for $x$ and $y$, for $z$ I have it already. So because

$$0 \leq z \leq 1 - x - y$$

then

$$0 \leq x \leq 1 - z - y$$

and also

$$0 \leq y \leq 1 - z - x$$

But that's probably not the right approach, because evaluating integral

$$\iiint_{T}\ dx\ dy\ dz = \int_{0}^{1-z-x}\int_{0}^{1-z-y}\int_{0}^{1-x-y}\ dz\ dx\ dy$$

still lefts me with some $z$ and $x$ variables at the end...

Will somebody point me to the right direction?

Thank you.

2. Jan 14, 2006

siddharth

twoflower, maybe this will help.
The boundary of this region is given by the plane
$$x+y+z=1$$.
To find the integral, for some x and y, z varies from 0 to 1-x-y (ie, till the top of the column containing the volume element).
Now keep x constant and integrate wrt to y. This corresponds to adding the columns with base in the x-y plane (from 0 to the line y=1-x). So, y varies from 0 to 1-x.
Finally you can integrate wrt to x, which by similar argument varies from 0 to 1.

3. Jan 14, 2006

twoflower

Btw what does "wrt" stand for? I would guess "with respect to", but it doesn't make sense to me in the previous post..

4. Jan 14, 2006

arildno

The simplest way to do this is first to recognize that the plane z=1-x-y meets the plane z=0 in the line x+y=1 (the line in z=0, that is)

Now, regard the part of T lying in the plane z=0.
This is the triangle bounded by the lines x=0, y=0, x+y=1, that is, we see that 0<=x<=1 and 0<=y<=1-x (alternatively, you could describe the triangle as 0<=y<=1, 0<=x<=1-y)

5. Jan 14, 2006

Staff: Mentor

In general, if you know that you are going to get a definite volume, then you know that the limits on a given integral cannot depend on any of the inner variables of integration. So in your example, the limits for z can be a function of x and y, the limits for x can be a function of y but not z, and the limits for y cannot be a function of x or z.

-Dale

6. Jan 14, 2006

HallsofIvy

You have
$$T = \left\{[x,y,z] \in \mathbb{R}^3; x \geq 0, y \geq 0, 0 \leq z \leq 1 - x - y\right\}$$
and you write

$$V= \iiint_{T}dV = \int\int\int dz\ dx\ dy$$
which indicates that you are going to integrate first with respect to z, then with respect to x, and finally with respect to y.
(That's an arbitrary choice- doing the integrations in any order must give the same answer- although one way might be easier than another.)
You are correct, of course, that the answer is a number which means that the "outside" integral, wrt y, must have only numbers as limits.
How do you find them? Draw a picture! (That should always be the first thing you do with problems like this.) Since you are told that x and y will be larger than or equal to 0, you are looking at the first quadrant in the xy-plane. z= 1- x- y is a plane, of course. It crosses the xy-plane when z= 0= 1-x-y which is the line x+y= 1, through the points (0,1) and (1,0), in the first quadrant. That tells you everything you need to know. Overall, y can be as low as 0 or as high as 1. Your "outside" integral must be $\int_0^1 dy$.
Now look at the "middle" integral- the one with respect for x. Since you will not yet have integrated with respect to y, the limits of integration may depend upon y. For each y what is the range for x? A fixed y is, of course, a horizontal line running from the vertical line x= 0 to the line x+ y= 1 or x= 1- y. For each y, x ranges between 0 and 1- y. The "middle" integral is $\int_0^{1-y}dx$.
Finally, when doing the "inner" integral, we have not yet integrated with respect to x or y so the limits of integration may be functions of both x and y. For each x,y, what is the range for z? The plane z= 1- x- y slants up from the line x+y= 1 in the xy-plane to z= 1 at the z-axis (x=y=0). The crucial point is that the xy-plane (z= 0) is the "floor" and the plane z= 1- x- y is the "roof". For a given (x,y) point, z ranges from z= 0 to z= 1- x-y. That "inner" integral is $\int_0^{1-x-y}dz$

Personally, I think it helps to write "x= ", "y= ", "z= " in the limits of integration themselves. The volume is given by:
$$\int_{y= 0}^{y=1}\int_{x=0}^{x=1-y}\int_{z=0}^{z=1-x-y}dzdxdy$$.

It would be good practice for you to find the integrals in the other 5 orders, and do the integrations to see that they do indeed give the same answer!

Last edited by a moderator: Jan 14, 2006
7. Jan 14, 2006

twoflower

Thank you HallsoftIvy very much for comprehensible explanation! I already managed it as soon as I draw it :) First it seemed ugly to me to draw it but it turned out to be pretty straightforward. And then the expressing 'x' for fixed 'y' was easy.