Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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

Let's say I have to compute the area of

[tex]

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

[/tex]

[tex]

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

[/tex]

When I draw the situation, I easily find leftmost and rightmost point of this area, which have x-coordinates-1and2, 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

[tex]

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

[/tex]

But I'm not sure with the inner one. I know the bounds can't be0and4, because that would give me the area of the whole rectangle.

So, right solution is this:

[tex]

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

[/tex]

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

[tex]

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

[/tex]

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

And other question - why is it

[tex]

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

[/tex]

and not

[tex]

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

[/tex]

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

Thank you very much for clarifying this...

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# Homework Help: Fubini's theorem help

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