Volume of a tetrahedron using triple integration

[aliaze1]

Homework Statement

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)

Homework Equations

V=∫∫∫dV
...D

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

 

[aliaze1]

actually, i realize that the line should be

-1½x+3 instead of -3x+3

still, i get the answer as 2...

 

[HallsofIvy]

Homework Statement

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)

Homework Equations

V=∫∫∫dV
...D

The Attempt at a Solution

I set up the problem as so:


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

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

(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

actually, i realize that the line should be

-1½x+3 instead of -3x+3

still, i get the answer as 2...

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.

 

[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.

 

[HallsofIvy]

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.

 

[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