Understanding Fubini's Theorem for Calculating Area in 2D - Explained Simply

[twoflower]

Hi all,

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

Let's say I have to compute the area of

<br /> \Omega \subset \mathbb{R}^2\mbox{ , }\Omega\mbox{ is bounded with }y = x^2\mbox{ and }y = x + 2<br />

<br /> \mbox{area(\Omega)} = \iint_{\Omega} 1\ dx\ dy<br />

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

<br /> \int_{-1}^{2} ...<br />

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:

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

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

<br /> \int_{a_1}^{b_1} \left( \int_{a_2}^{b_2} ...<br />

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

And other question - why is it

<br /> \int_{x^2}^{x+2}<br />

and not

<br /> \int_{x+2}^{x^2}\mbox{ ?}<br />

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

Thank you very much for clarifying this...

 

[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.

 

[twoflower]

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.

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 :)

 

[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?