Double Integral of Pythagoras over rectangular region

[ctchervenkov]

Take any given point on the perimeter of a (A x B) rectange and then draw a line from that point to another point on one of the three remaining sides of the rectangle. What is the average length of the line?
Well, the answer to that question involves integrals like this:

\int_0^A \int_0^B \sqrt{x^2 + y^2} dy dx

This thing gets a bit interesting given that it's basically a polar type integral but with rectangular boundaries. Any ideas?

Thanks!

Chris

 

[HallsofIvy]

That looks like a fairly standard "trig substitution" integral. To integrate with respect to y initially, let y= x tan(\theta) so that \sqrt{x^2+ y^2}= \sqrt{x^2+ x^2tan^2(\theta)}= x\sqrt{1+ tan^2(\theta)}= x sec(\theta) while dy= x sec^2(\theta)d\theta. The first integral becomes \int_0^{arctan(B/x)} (x sec(\theta))(x sec^2(\theta)= x^2\int_0^{arctan(B/x)} sec^3(\theta)d\theta

 

[Saitama]

Take any given point on the perimeter of a (A x B) rectange and then draw a line from that point to another point on one of the three remaining sides of the rectangle. What is the average length of the line?
Well, the answer to that question involves integrals like this:

\int_0^A \int_0^B \sqrt{x^2 + y^2} dy dx

This thing gets a bit interesting given that it's basically a polar type integral but with rectangular boundaries. Any ideas?

Thanks!

Chris

Hi ctchervenkov! Welcome to PF!

Polar works here. You will need to split the region into two.
$$\int_0^{\tan^{-1}\frac{B}{A}} \int_0^{\frac{A}{\cos\theta}} r^2\,dr\,d\theta+\int_{\tan^{-1}\frac{B}{A}}^{\pi/2} \int_{0}^{\frac{B}{\sin\theta}} r^2\,dr\,d\theta$$
Not sure if this can be simplified further.

 

[ctchervenkov]

to Pranav-Arora: I had tried that, but the things got a bit ugly...

to HallsofIvy: Had not tried that particular substitution I don't think. Thanks. Will see if that simplifies things...