# Fubini's theorem help

1. Jan 11, 2006

### twoflower

Hi all,

I don't fully understand the usage of Fubini's theorem.

Let's say I have to compute the area of

$$\Omega \subset \mathbb{R}^2\mbox{ , }\Omega\mbox{ is bounded with }y = x^2\mbox{ and }y = x + 2$$

$$\mbox{area(\Omega)} = \iint_{\Omega} 1\ dx\ dy$$

When I draw the situation, I easily find leftmost and rightmost point of this area, which have x-coordinates -1 and 2, respectively.

Well, Fubini tells us that the integral above can be evaluated as (in this case) two nested one-dimensional integrals. But I have problems with the lower and upper bounds.

The outer integral is clear, it will be

$$\int_{-1}^{2} ...$$

But I'm not sure with the inner one. I know the bounds can't be 0 and 4, because that would give me the area of the whole rectangle.

So, right solution is this:

$$\iint_{\Omega} 1\ dx\ dy = \int_{-1}^{2}\left( \int_{x^2}^{x+2} 1\ dy\right)\ dx = ... = \frac{9}{2}$$

I don't fully get it...why are the bounds of the inner integral $x^2$ and $x+2$? I know those are the bounding curves, but in Fubini's theorem there is just

$$\int_{a_1}^{b_1} \left( \int_{a_2}^{b_2} ...$$

Which I can't connect with the bounds in my example.

And other question - why is it

$$\int_{x^2}^{x+2}$$

and not

$$\int_{x+2}^{x^2}\mbox{ ?}$$

Is it because on the interval $(-1, 2)$ the function $x+2$ is greater than $x^2$?

Thank you very much for clarifying this...

Last edited: Jan 11, 2006
2. Jan 11, 2006

### Galileo

You integrate from x^2 to x+2, because x^2 is the lower bound and x+2 the upper bound, as you can easily see from a drawing.

A way to visualize it is this. What does the inner integral mean? In this case, it is the length of the vertical line between the bounding curves as a function of x. Right? It's simply $(x+2)-x^2$ in this case.
A little more generally, you can use double integrals like $\int_a^b \int_{g(x)}^{h(x)} f(x,y)dydx$ to find the volume of the graph under f(x,y) in which case the inner integral will be the area under the graph obtained by intersecting the graph of f(x,y) with a vertical plane parallel to the yz-plane (or xz-plane, depending on the variable) as a function of x. The inner integral is just a function of x, and it's an area like a standard integral. Then you integrate that along y to get the total volume. If you can visualize that procedure I`m sure you see why the bounds are what they are.

Last edited: Jan 11, 2006
3. Jan 11, 2006

### twoflower

Thank you Galileo! I think it's more clear to me now. This was very easy problem, I hope I will understand the three-dimensional as well :)

4. Jan 11, 2006

### benorin

I'm just curious: does Fubini's Theorem extend to the case of multiple integrals taken w.r.t. different measures?

Like, say the Lebesgue and counting measures?