# Thread: setting up a triple integral

1. ## setting up a triple integral

if you are given a function and then given boundaries of that function(ex z is between 3 and 4 and then y and x defined as functions like y=x and y=x^2) how do I set up triple integral for these boundaries?

I know the basic idea and I figure the z limits would be between 3 and 4 but I'm not sure how to go about calculating the x and y limits, the integration is no problem so any advice would be appreciated.

2. If might be easier if you can post a specific example of this type of question, then we can run you through the process using that.

3. I want to express the mass of f(x,y,z) of 12xy +9y^2 + 3z

z=3 and z=4 and the surfaces are y=x and y=x^2

4. Originally Posted by billmasters26
I want to express the mass of f(x,y,z) of 12xy +9y^2 + 3z

z=3 and z=4 and the surfaces are y=x and y=x^2
Here's a general process for you:

You know the z limits, and are only interested in the limits for y and x. So, ignore z, and focus on the x-y plane. Draw your functions on the x-y plane. The region you're integrating over is the region enclosed by these two curves.

Now, pick an axis at random, either x or y, and that's the axis you're going to find constant limits for, and the other axis you'll find limits in terms of other variables. So let's choose x as the axis for which we'll get constant limits. The limits for x, in this case, are simply the x coordinates at which y = x^2 and y = x intersect. So that's 0, and 1. $0 \leq x \leq 1$.

Now, draw a vertical line parallel with the y axis from negative values to positive values, and make sure it cuts through the region. Now, through which function did it ENTER the region? y = x^2, and through which did it leave? y = x.

Hence $x^2 \leq y \leq x$. So your integral looks like this:

$I = \int_3^4 \int_0^1 \int_{x^2}^x (12xy +9y^2 + 3z) \, dy \, dx \, dz$, we always integrate the variable whose limits are function of other variables FIRST.