<s>[INST] 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 $\frac{2\sqrt{2}}{9}$.
 [/INST] 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 $(\frac 12, -\frac{\sqrt {2}}{2})$
$\int_{-\frac {\sqrt {2}}{2}}^{\frac {\sqrt 2}{2}} \frac {y^2}{3} + \frac 13 - y^2\ dy$
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