Area enclosed between $2$ curves 
Question: The area enclosed by the curves $$ y^2 = x \ \text{and} \ y^2 = \ 3x - 1, \ \ \ \ \  \text{where} \  \ 0\leq \  x\leq \frac{1}{2} \ $$

My work:
To begin with I re-arranged the curves in terms of $y$ to the power $1$
$$y_{1} = \sqrt{x}$$
and
$$y_{2} = \sqrt{3x-1}$$
and then proceeded to take the integral for both over the interval $0\leq \  x\leq \frac{1}{2}$
$$f(x_{1}) = \frac{1}{3\sqrt{2}}$$
and
$$f(x_{2}) = \frac{5^{\frac{3}{2}}}{9\sqrt{2}} - \frac{2}{9}$$
And therefore the area between the curves is $\displaystyle\int f(x_{1}) - f(x_{2})$ right?
Is this the correct method? Because the answer that was provided to me was $\frac{2\sqrt{2}}{9}$.
 A: Say $A_1$ is area between parabola $y^2 = x$ and line $x = \frac{1}{2}$, which you write as $f(x_1)$ and $A_2$ is the area between parabola $y^2 = 3x-1$ and line $x = \frac{1}{2}$
You need to multiple $A_1$ by $2$ as the answer you have is only upper / lower half of the area.
Coming to $A_2$, your mistake is that you are integrating between $x = 0$ and $x = \frac{1}{2}$. The vertex of the second parabola is $(\frac{1}{3}, 0)$. So, you should integrate the second one between $x = \frac{1}{3}$ and $\frac{1}{2}$.
So the second integral should be,
$ \displaystyle 2 \int_{1/3}^{1/2} \int_{0}^{ \sqrt {3x-1}} dy ~dx$
That gives you $A = A_1 - A_2 = \frac{2}{3 \sqrt2} - \frac{2}{9\sqrt2} = \frac{2 \sqrt2}{9}$

Also note that you can directly integrate if you go with respect to $dx$ first. Keep the equations in the form $y^2 = x$ and $y^2 = 3x - 1 \implies x = \frac{y^2+1}{3} ~$.
At intersection $y^2 = x = 3x - 1 \implies x = \frac{1}{2}, y = \pm \frac{1}{\sqrt2}$
So the integral to find area between the curves is,
$\displaystyle \int_{- 1/\sqrt2}^{1/\sqrt2} \int_{y^2}^{ (y^2+1) / 3} ~ dx ~dy$
A: I would integrate with respect to y.
$x = y^2\\
x = \frac {y^2}{3} + \frac 13$
The curves intersect at $(\frac 12, \frac {\sqrt {2}}{2})$ and $(\frac 12, -\frac{\sqrt {2}}{2})$
$\int_{-\frac {\sqrt {2}}{2}}^{\frac {\sqrt 2}{2}} \frac {y^2}{3} + \frac 13 - y^2\ dy$
