Radius of a circle Let $f:\mathbb{R} \rightarrow \mathbb{R}$ defined as $f(x)=x^2+1$. Find a real positive number $r$ such that the graph of $f$ intersects the circle about the origin and radius $r$ into two pieces , of which the one contains the origin and has an area equal to the value of the integral $ \displaystyle \mathcal{I}=\int_1^0 \frac{\ln x}{\sqrt{x}}\,{\rm d}x $. 
Path to a solution: It is pretty much obvious that the radius $r$ will be greater than $1$, because if the radius is $1$ then the two graphs attach to each and if $r<1$ then they do not have an intersection. 
That integral evaluates to $4$ with a direct calculation. And here comes the most difficult part of all:
The points at which the two graphs intersect are of the form:
$${\rm A}\left ( -\frac{1}{2}\sqrt{-6+2\sqrt{5+4r^2}}, \frac{1}{2}\left ( -1+\sqrt{5+4r^2} \right ) \right ), \;\; {\rm B}\left ( \frac{1}{2}\sqrt{-6+2\sqrt{5+4r^2}}, \frac{1}{2}\left ( -1+\sqrt{5+4r^2} \right ) \right )$$
Thus, the wanted area is:
$${\rm E}\left ( r \right )=\pi r^2 -\int_{-1/2 \cdot \sqrt{-6+2\sqrt{5+4r^2}}}^{1/2 \cdot \sqrt{-6+2\sqrt{5+4r^2}}} \left [ \sqrt{r^2-x^2}-x^2-1 \right ]\,{\rm d}x$$
All we have to do is to solve ${\rm E}(r)=4$. Basically I cannot solve it (at least not by hand). 
Do you see an alternative?
P.S: Integrating with respect to $y$ does not have have much difference.

For first, by setting $x=t^2$:
$$ \int_{1}^{0}\frac{\log x}{\sqrt{x}} = 4\int_{0}^{1}(-\log t)\,dt = 4.\tag{1} $$
Next, let we assume that the circle and the parabola intersect in two points having ordinate $y>1$ and abscissa $x=\pm\sqrt{y-1}$. The area of the parabolic segment defined by these two points is:
$$ \frac{2}{3}(y-1)\cdot 2\sqrt{y-1}\tag{2} $$
hence, since $r^2=y^2+y-1$, the area of the region inside the circle and below the parabola is given by:
$$A_y= \left(\pi+2\arctan\frac{y}{\sqrt{y-1}}\right)\frac{y^2+y-1}{2}+y\sqrt{y-1}-\frac{4}{3}(y-1)\sqrt{y-1}\tag{3}$$
but I do not see many ways to solve $A_y=4$ without the help of a CAS. 
Numerically, I got that the right radius should be $r\approx 1.13622\ldots$.
