Touching circle area

It is reasonable to assume that the circle to be constructed will have one point of intersection with the given curve at the origin. The radius $r$ circle is then described by the equation $x^{2} + (y - r)^{2} = r^{2},$ Now for the circle to touch (but not intersect) the curve $y = x^{3}$ at a second point we require that not only do the curves pass through this point but that their tangents coincide at this point too.

Working in the first quadrant, and noting that the point of tangency $P(x,y)$ will have a $y$ -coordinate less than $r,$ (since $x^{3}$ is an increasing function), we have that, for the circle, $y = r - \sqrt{r^{2} - x^{2}}.$ So for the two curves to meet, we require that

$x^{3} = r - \sqrt{r^{2} - x^{2}},$ (A),

and since their tangents must coincide, we also require that the derivatives are equal at $P,$ i.e.,

$3x^{2} = \dfrac{x}{\sqrt{r^{2} - x^{2}}} \Longrightarrow \sqrt{r^{2} - x^{2}} = \dfrac{1}{3x},$ (B)

where we divided through by $x$ since we want $P$ other than the origin. Substituting (B) into (A) gives us that

$x^{3} = r - \dfrac{1}{3x} \Longrightarrow r = x^{3} + \dfrac{1}{3x}.$

Substituting this result into (A) and simplifying gives us that

$\dfrac{1}{3x} = \sqrt{x^{6} + \dfrac{1}{9x^{2}} + \dfrac{2x^{2}}{3} - x^{2}}$

$\Longrightarrow \dfrac{1}{9x^{2}} = x^{6} + \dfrac{1}{9x^{2}} + \dfrac{2x^{2}}{3} - x^{2}$

$\Longrightarrow x^{6} = \dfrac{x^{2}}{3} \Longrightarrow x^{4} = \dfrac{1}{3} \Longrightarrow \large x = 3^{-\frac{1}{4}}.$

This in turn gives us that $r = x^{3} + \dfrac{1}{3x} = \large 2*3^{-\frac{3}{4}}.$

So now we just need to evaluate the integral

$\displaystyle \int_{0}^{3^{-\frac{1}{4}}} (r - r\sqrt{1 - \left(\dfrac{x}{r}\right)^{2}} - x^{3}) dx,$

(where of course $x$ is now being used as a variable). The first and third components of the integrand are straightforward and yield a value of $\dfrac{7}{12}.$

The second component can be evaluated using the substitution

$\dfrac{x}{r} = \sin(\theta) \Longrightarrow dx = r\cos(\theta) d\theta.$

The integral of this component then becomes

$\displaystyle\int r^{2}\cos^{2}(\theta) d\theta = \dfrac{r^{2}}{2}(\theta + \sin(\theta)\cos(\theta)) d\theta =$

$\dfrac{2}{3\sqrt{3}}\left(\arcsin\left(\dfrac{x}{r}\right) + \dfrac{x}{r}\sqrt{1 - \left(\dfrac{x}{r}\right)^{2}}\right),$

which when evaluated from $x = 0$ to $\large x = 3^{-\frac{1}{4}}$ comes out to

$\dfrac{2}{3\sqrt{3}}*\left(\arcsin\left(\dfrac{\sqrt{3}}{2}\right)+ \dfrac{\sqrt{3}}{2}*\dfrac{1}{2}\right) = \dfrac{2}{3\sqrt{3}}*\left(\dfrac{\pi}{3} + \dfrac{\sqrt{3}}{4}\right) = \dfrac{2\pi}{9\sqrt{3}} + \dfrac{1}{6}.$

Subtracting this from $\dfrac{7}{12}$ yields a desired area value of

$\dfrac{5}{12} - \dfrac{2\pi}{9\sqrt{3}} = \boxed{0.01360}$ to 4 decimal places.