Triple Integral - Volume of Tetrahedron

[dkotschessaa]

Homework Statement

Actually, the problem was addressed in a prior post:

https://www.physicsforums.com/showthread.php?t=178250

Which is closed.

Homework Equations

I would like to know how HallsofIvy (or anyone) arrived at the formula for the tetrahedron given the vertices (1,0,0), (0,2,0), (0,0,3).

Ultimately I am to find the volume of this tetrahedron using triple integrals.

But I'm not worried about the integral as much as the setup:

The equation I get is 3 -3x -3/2y
not 1 -3x -3/2y

The Attempt at a Solution

I've taken two vectors from these points, taken their cross product, and created an equation of a plane. I am still getting my answer, and consequently an integral that doesn't seem right!-Dave K

 

[DryRun]

I would like to know how HallsofIvy (or anyone) arrived at the formula for the tetrahedron given the vertices (1,0,0), (0,2,0), (0,0,3).

You should label each of the vertices. Let A=(1,0,0), B=(0,2,0), C=(0,0,3). There are 2 methods (that i know of):

First method:
The Cartesian equation of plane is: $ax +by+cz=d$

Just plug in the coordinates of the 3 points and solve the system of 3 linear equations. You should get x, y and z in terms of d. Then, divide throughout by d to get the final equation of the plane.

Second method (what you're expected to use):

Find two vectors that lie in the plane. Do the cross product to get the normal vector, $\vec n$.

Then, use the formula: $\vec r.\vec n=\vec a.\vec n$ where $\vec a$ is any point found in the plane.

Using the second method, you will get the Cartesian equation of the plane: $6x+3y+2z=6$

There is indeed a mistake in that post: https://www.physicsforums.com/showpost.php?p=1387411&postcount=3

$$z=3-3x-\frac{3y}{2}$$

 

[dkotschessaa]

Oh thank goodness.

I wasn't trusting my own answer, and it (the correct answer) makes the integration a little bit uglier.

Thanks man.

Regards,

-Dave K