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I'm starting with the simplest problem I can relate to mine: the force imposed on a point mass at the origin by a rectangle that is orthogonal to the $x$- and $y$-axis, stretching from $(x_1, y_1)$ to $(x_2, y_2)$.

I'm using Newton's formula, but currently ignoring the mass and density.

I start off doing something like

$\int^{y_2}_{y_1}\int^{x_2}_{x_1}\frac{1}{x^2 + y^2}dxdy$

Where $x^2+y^2$ equals the squared distance between $(x,y)$ and the origin.

This leads me to the following integral:

$\int^{y_2}_{y_1}(\frac{1}{y}\arctan\frac{x_2}{y}) - (\frac{1}{y}\arctan\frac{x_1}{y})dy$

Is there a better way? I can't find any way to solve this.

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Forces in your case are vectors with an $x$- and a $y$-component. An "area element" ${\rm d}(x,y)$ at the point $(x,y)$ pulls at the origin with the force $$d\vec F=\Bigl({x\over (x^2+y^2)^{3/2}},{y\over (x^2+y^2)^{3/2}}\Bigr)\>{\rm d}(x,y)\ .$$ –  Christian Blatter Sep 20 '11 at 10:21
Thanks! I believe you're right. I think my problem just got harder :-( –  Steinbitglis Sep 20 '11 at 13:27

3 Answers 3

up vote 1 down vote accepted

As far as I can see, the final solution takes the form of: (based on Christian Blatters answer)

$$F_1=-ln({c+\sqrt{a^2+c^2})}+ln({d+\sqrt{a^2+d^2})}+ln({c+\sqrt{b^2+c^2})}-ln({d+\sqrt{b^2+d^2})} \ .$$

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Ooohh... nice, ... rectangular planets :-) –  Steinbitglis Sep 27 '11 at 23:02

I'm elaborating my comment.

Consider the rectangle $R:=[a,b]\times[c,d]$ with $0<a<b$ and $0<c<d$. The $x$-component $F_1$ of the force exerted by the rectangle at the origin is given by $$F_1=\int\nolimits_R{x\over(x^2+y^2)^{3/2}} \> {\rm d}(x,y)=\int_c^d\int_a^b {x\over(x^2+y^2)^{3/2}} \>dx\>dy\ .$$ Here the inner integral has the value $${-1\over (x^2+y^2)^{1/2}}\Biggr|_a^b ={1\over\sqrt{a^2+y^2}}-{1\over\sqrt{b^2+y^2}}\ .$$ Now the outer integral can be expressed in terms of $\ {\rm arsinh}{y\over a}\ $ resp. $\ {\rm arsinh}{y\over b}$ for $y=c$ and $y=d$. I leave the details to you.

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You probably meant: $\int^{y_2}_{y_1}\int^{x_2}_{x_1}\frac{1}{x^2 + y^2}dxdy = \int^{y_2}_{y_1}(\frac{1}{y}\arctan\frac{x_2}{y})dy - \int^{y_2}_{y_1}(\frac{1}{y}\arctan\frac{x_1}{y})dy$

By substitution, you arrive at integrals of type $\int \frac{1}{x}\arctan\frac{x}{a}dx$. There's no solution in closed form, except for this expansion:$$\int \frac{1}{x}\arctan\frac{x}{a}dx=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)^2}(\frac{x}{a})^{2n+1}$$ Which is valid only for $|\frac{x}{a}|<1$.

Only other way is to integrate numerically.

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"There's no solution in closed form" - maybe you meant to say that there's no elementary expression for these, which is more accurate. The function can be expressed in terms of dilogarithms with purely imaginary argument. –  J. M. Sep 20 '11 at 16:33
Yes, I meant that. Sorry for not being 100% accurate, just thought I'd help. –  vedran Sep 20 '11 at 23:09

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