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I need to calculate the area of an object that is limited by this line:

$$\phi=r\!\cdot\!\arctan(r)\;,\quad\phi\in\left[0,\frac{\pi}{\sqrt3}\right]$$

I am using this formula:

$$S=\frac12\int_{r_1}^{r_2}r^2\phi'\mathrm dr$$

When I plugged in values I got:

$$S=\frac12\left(\int r^2\!\cdot\!\arctan(r)\,\mathrm dr+\int\frac{r^3}{1+r^2}\,\mathrm dr\right)$$

which I can calculate. However, I don't know how to find the boundaries of integration $r_1$ and $r_2$ if I know $\phi_1$ and $\phi_2$.

Is this the correct approach? Thanks in advance.

Jean Marie
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ALiCe P.
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1 Answers1

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A preliminary comment : It is much less usual to have curves defined by equations $\phi=F(r)$ than curves of the form $r=G(\phi)$. When $F$ has a known explicit inverse formula, we can revert to the second form. But here, this is not the case. Therefore we have to use another "trick".

  1. Your last formula is correct due to

$$\phi'(r)=\arctan(r)+\frac{1}{1+r^2}$$

which is positive for all $r \ge 0$. Therefore function $\phi$ is increasing on any interval $[0,r_0]$.

  1. The fact that $\phi$ is an increasing function is the key for finding the bounds. Let us recall the following property : a continuous increasing function on a closed bounded interval $[a,b]$ is a bijection from $[a,b]$ onto $[f(a),f(b)]$.

Here, it is sufficient to verify that $$[\phi(a),\phi(b)]=[\phi(0),\phi(\sqrt{3})].$$

Indeed, one can check that

  • $\phi(0)=0 \arctan(0)=0.$

  • $\phi(\sqrt{3})=\sqrt{3}\arctan(\sqrt{3})=\sqrt{3}\tfrac{\pi}{3}=\tfrac{\pi}{\sqrt{3}}$

Therefore, using (implicitly) the reciprocal function $\phi^{-1}$ the bounds for variable $r$ are $r_1=0$ and $r_2=\sqrt{3}.$

enter image description here

Straight line segment has length $\sqrt{3}$ and polar angle $\tfrac{\pi}{\sqrt{3}} \approx \tfrac{4\pi}{7}.$

Matlab program for this figure :

 clear all;close all;hold on;axis equal
 k=0;
 for r=0:0.01:(pi/sqrt(3))
    k=k+1;phi=r*atan(r);
    T(1,k)=r*cos(phi);T(2,k)=r*sin(phi);
 end;
 plot(T(1,:),T(2,:),'r'); % curve
 plot([0,T(1,end)],[0,T(2,end)],'r'); % straight line
Jean Marie
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