Find $\int_0^a{f(x)}\, dx$ SMT 2013 Calculus #8: 
The function $f(x)$ is defined for all $x\ge 0$ and is always nonnegative. It has the additional property that if any line is drawn from the origin with any positive slope $m$, it intersects the graph $y=f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin. Let a be the unique real number for which f takes on its maximum value at $x=a$ (you may assume that such an a exists). Find $\int_{0}^{a}f(x) \, dx$.
My work so far:
I tried interpreting the question in the following way:
"if any line is drawn from the origin with any slope $m$, it intersects the graph $y = f(x)$ at precisely one point, which is $\frac{1}{\sqrt{m}}$ units from the origin"
This can be parameterized as $$\begin{cases} y = mx \\ x^2 + y^2 = \frac{1}{m}\end{cases}$$
Then to find the maximum value $x = a$, you simply maximize the function $y = \sqrt{\frac{m}{1+m^2}}$ (which is the same as maximizing the function $\frac{m}{1+m^2}$).
$\left(\frac{m}{1+m^2}\right)' = 0 \implies m = 1 \implies$ the maximum point is $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$, so $ a=\frac{1}{\sqrt{2}}$.
After this point, I don't know how to compute the integral $$\int_0^\frac{1}{\sqrt{2}}{f(x)}\, dx.$$ (I tried using polar coordinates, but then I got confused.)
Edit 2: I have found an algebraic solution! It involved solving $x$ in terms $y$ (too tired to type out the entire solution). But if anyone can explain step-by-step a solution using polar coordinates, that would be greatly appreciated :)
 A: First Approach
If $y=mx$ and $x^2+y^2=\frac1m$, then $x^2+y^2=\frac xy$, which implies
$$
yx^2-x+y^3=0\tag{1}
$$
Taking the implicit derivative of $(1)$:
$$
y'=\frac{1-2xy}{x^2+3y^2}\tag{2}
$$
$y'=0$ means $1=2xy$; then applying $y=mx$, we get
$$
1=2xy=2mx^2\tag{3}
$$
applying $x^2+y^2=\frac1m$ gives
$$
1+2my^2=2mx^2+2my^2=2m\tag{4}
$$
using $y=mx$ yields
$$
m^2=\frac{y^2}{x^2}=2m-1\tag{5}
$$
therefore,
$$
m=1\tag{6}
$$
so we have $a=\frac1{\sqrt2}$.
Applying the quadratic formula to $(1)$:
$$
x=\frac{1-\sqrt{1-4y^4}}{2y}\tag{7}
$$
Integration by parts and substituting $z^2=1-4y^4$ gives
$$
\begin{align}
\int_0^{1/\sqrt2}y\,\mathrm{d}x
&=\frac12-\int_0^{1/\sqrt2}x\,\mathrm{d}y\\
&=\frac12-\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{2y}\,\mathrm{d}y\\
&=\frac12-\frac18\int_0^{1/\sqrt2}\frac{1-\sqrt{1-4y^4}}{4y^4}\,\mathrm{d}4y^4\\
&=\frac12-\frac18\int_0^1\frac{1-z}{1-z^2}2z\,\mathrm{d}z\\
&=\frac12-\frac14\int_0^1\frac{z}{1+z}\,\mathrm{d}z\\
&=\frac12-\frac14\int_0^1\left(1-\frac1{1+z}\right)\,\mathrm{d}z\\
&=\frac14(1+\log(2))\tag{8}
\end{align}
$$

Polar Coordinate Approach
By virtue of the work done above, we know that the curve extends from $m=1$ to $m=\infty$, that is $\theta=\frac\pi4$ to $\theta=\frac\pi2$. There is also the triangle below that is not bounded by the curve.

Note that $r^2=x^2+y^2=\frac1m$ and $\tan(\theta)=\frac yx=m$. Therefore,
$$
r=\frac1{\sqrt{\tan(\theta)}}\tag{9}
$$
The region is given by $(9)$ for $\frac\pi4\le\theta\le\frac\pi2$, and for $0\le\theta\le\frac\pi4$ the region is a right triangle with legs of $\frac1{\sqrt2}$. Thus the area is
$$
\begin{align}
\color{#C000FF}{\frac12\cdot\frac1{\sqrt2}\cdot\frac1{\sqrt2}}+\color{#00A000}{\int_{\pi/4}^{\pi/2}\frac12r^2\,\mathrm{d}\theta}
&=\frac14+\frac12\int_{\pi/4}^{\pi/2}\frac{\mathrm{d}\theta}{\tan(\theta)}\\
&=\frac14+\frac12\left[\vphantom{\int}\log(\sin(\theta))\right]_{\pi/4}^{\pi/2}\\
&=\frac14(1+\log(2))\tag{10}
\end{align}
$$
A: here in one way to solve this problem. the parametric equation of the graph is 
$$x = \dfrac{1}{\sqrt{m(1 + m^2)}}, y = \sqrt{\dfrac{m}{1+m^2}}$$ note that $y$ reaches a maximum at $m = 1.$  
$y \ dx = \sqrt{\dfrac{m}{1+m^2}} \left(  -\dfrac{1}{2}[m(1 + m^2)]^{-3/2}(1 + 3m^2)\right) \ dm =  -\dfrac{1}{2} \dfrac{1 + 3m^2}{m(1 + m^2)^2} \ dm$
so the area is $$ \dfrac{1}{2} \int_1^\infty \dfrac{1 + 3m^2}{m(1 + m^2)^2} \ dm$$
