# Volume of a tetrahedron using triple integration

1. Jul 25, 2007

### aliaze1

1. The problem statement, all variables and given/known data

Find the volume of the tetrahedron in the first octant bounded by the coordinate planes and the plane passing through (1,0,0), (0,2,0) and (0,0,3)

2. Relevant equations

V=∫∫∫dV
...D

3. The attempt at a solution

I set up the problem as so:

1 -2x+2...-3x+3
∫ ∫........∫......dz dy dx
0 0........0

(the dots are not significant, they are only for spacing)

and integrated.....

the result is -1, but this is wrong because:

1. a volume cannot be negative
2. the book answer is 1

2. Jul 25, 2007

### aliaze1

actually, i realize that the line should be

still, i get the answer as 2.....

3. Jul 26, 2007

### HallsofIvy

Staff Emeritus

How did you get z= -3x+3 for the upper limit? Doesn't z change as y changes?

Again, you should not be thinking of a line. The first integral should be from the xy-plane up to the plane forming the upper boundary.

First, as I presume you have done, draw a picture. The base, in the xy-plane, is the triangle bounded by the x-axis, the y-axis, and the line from (1,0) to (0,2): its equation is y= 2- 2x. Obviously, we can cover the entire triangle by taking x running from 0 to 1 and, for each x, y running from 0 to 2-2x. The plane through (1, 0, 0), (0, 2, 0), and (0, 0, 3) has equation
$$x+ \frac{y}{2}+ \frac{z}{3}= 1$$
or
$$z= 1- 3x-\frac{3y}{2}$$
The volume is given by
$$\int_{x=0}^1\int_{y= 0}^{2-2x}\int_{z=0}^{1-3x-\frac{3y}{2}} dzdydx$$
Integrate that and see what you get.

4. Jul 27, 2007

### huyen_vyvy

honestly, this problem can be made very easy like this:
volume= 1/3height * area of base. Height is z=3, area of base is the area of a right triangle with two sides being 1 and 2. Therefore, V=1.

5. Jul 27, 2007

### HallsofIvy

Staff Emeritus
Yes. In fact, the volume of any tetrahedron with vertices at (a, 0, 0), (0, b, 0), and (0, 0, c) is (1/6)abc but I assumed from the original post that he wanted to do this without using specific formulas.

6. Aug 24, 2007

### aliaze1

thanks for your help everyone, indeed the issue was the Z limit.....and yea i needed to know how to do it without specific formulas (it was on a Calculus III test) but I did use the formulas to confirm my answer