Integral with infinitesimal under root

1. Jun 15, 2014

Jhenrique

I want compute the following integral:

$$\\ \int f(x,y) \sqrt{dx^2+dy^2}$$
Is correct this pass-by-pass:
$$\\ \sqrt{\left( \int f(x,y) \right)^2} \sqrt{dx^2+dy^2} = \sqrt{\left( \int f(x,y) \right)^2 (dx^2+dy^2)} = \sqrt{\left( \int f(x,y) \right)^2 dx^2 + \left( \int f(x,y) \right)^2 dy^2} = \sqrt{\left( \int f(x,y) dx \right)^2 + \left( \int f(x,y) dy\right)^2}$$
?

2. Jun 15, 2014

micromass

Please define what this means. Or at least, please tell us where you saw this.

3. Jun 15, 2014

DrewD

The only time that I have ever seen notation like this is when $x$ and $y$ can be parameterized. Then This is interpreted as
$\displaystyle\int f(x,y)\sqrt{\left(\frac{dx}{dt}\right)^2dt^2+\left(\frac{dy}{dt}\right)^2dt^2}=\int f(x,y)\sqrt{x'(t)^2+y'(t)^2}dt$
which may be hard to compute but should make sense. Passing the integral under the square root does not make sense. An integral is a limit of sums which can't be passed under the square root.

The integral property $\left|\int fdx\right|\leq\int\left|\ fdx\right|$ should convince you that your first step is not guaranteed to work.

4. Jun 15, 2014

verty

Converting to polar coordinates is an option, that ugly square-root will become $dr$.

5. Jun 15, 2014

DrewD

That's true too. It depends on whether this is a surface or line integral.