# Help setting up triple integrals from 4 points

• Jul 10th 2012, 07:02 PM
NickE
Help setting up triple integrals from 4 points
The topic pretty much says it all. I have to evaluate the triple integral of xyz for a tetrahedron with vertices (0, 2, 0), (1, 0, 0), (4, 1, 0) and (1, 0, 1) then I have to do it again with vertices (0, 2, 0), (1, 0, 0), (4, 1, 0) and (1, 1, 1). I shouldn't have a problem solving them once I get them set up, my problem is that my professor did not explain how to set up a triple integral from coordinates. Can someone please give me step by step tutorial on how to go about setting up these integrals?
• Jul 11th 2012, 03:44 AM
tom@ballooncalculus
Re: Help setting up triple integrals from 4 points
Sure. What is your narrative for a single definite integral of y = f(x) with respect to x? Mine is...

I walk along the x-axis from the lower limit towards the upper, and at each point I turn left and note down the distance away (along the x-y plane in the y direction) of the curve y = f(x). The integral, e.g, for the 'curve' (though straight line) $\tfrac{9}{4}x$
...

$\int_0^1\ \tfrac{9}{4}x\ dx$

... is the total of all these values. Hopefully you can intuit this total as the area under the curve. (Do imagine visually, e.g. sketch, all of the curves I refer to.)

Or we could get the same result from observing (at each point on the x-axis journey looking in the y direction) the difference between the distances to each of the two curves $2 - \tfrac{1}{4}x$ and $2 - 2x$ ...

$\int_0^1\ 2 - \tfrac{1}{4}x\ - (2 - 2x)\ dx$

But let's turn either of these into a double integral of z = 1 (with respect to x and y) ...

I walk along the axis from the lower limit towards the upper, and at each point I turn left and this time go on a detour in the x-y plane in the y direction, from the x-axis to as far as the curve for f(x). At each point along the detour I note down a number 1. Then the integral of all those ones...

$\int_0^1 \int_0^{\tfrac{9}{4}x}\ 1\ dy\ dx$

... or else, if each detour in the y direction goes from the curve for $2 - 2x$ up (in the y direction) to the curve for $2 - \tfrac{1}{4}x$, then this integral of all the ones...

$\int_0^1 \int_{2 - 2x}^{2 - \tfrac{1}{4}x}\ 1\ dy\ dx$

... will be the same total as before.

Now, every time we noted down 1, i.e. at each point on our detour along the x-y plane in the y direction, we might have been looking up above us in the z direction (the x-y plane being horizontal ground) and observing the distance above us of the plane z = 1.

What if, above us at each of these points, we saw not the plane z = 1 but the plane $z = \tfrac{1}{7} (8 - x - 4y)$ ...

$\int_0^1 \int_{2 - 2x}^{2 - \tfrac{1}{4}x}\ \tfrac{1}{7} (8 - x - 4y)\ dy\ dx$

And here you can probably see that...

$\int_0^1 \int_{0}^{\tfrac{9}{4}x}\ \tfrac{1}{7} (8 - x - 4y)\ dy\ dx$

... would be want to produce a different volume.

Again, we could turn this into a triple integral by going on a third detour, in the z direction, and observing, at each point on the detour, a distance of 1 in some 'fourth' dimension (that's difficult to visualise as a new spatial dimension, but might be density or temperature etc.)...

$\int_0^1\ \int_{2 - 2x}^{2 - \tfrac{1}{4}x}\ \int_0^{\tfrac{1}{7} (8 - x - 4y)}\ 1\ dz\ dy\ dx$

And what you have in the first section (west of the plane x = 1) of your first tetrahedron is the same volume, but the density to note down at each point is xyz instead of 1...

$\int_0^1\ \int_{2 - 2x}^{2 - \tfrac{1}{4}x}\ \int_0^{\tfrac{1}{7} (8 - x - 4y)}\ xyz\ dz\ dy\ dx$

That's the basic approach.