Use single variable calculus methods to find the area of the region Use single variable calculus methods to find the area of the region in the first quadrant bounded by the curves $y^2=3x$, $y^2=4x$, $x^2=3y$, $x^2=4y$
Could someone please show me how to go about this?
 A: First, express all the functions as functions of either $x$ or of $y$. Since we are interested in the first quadrant only, we can take the positive roots where necessary. Since you're likely more comfortable using formulas expressed as functions of $x$, express each formula as $y = f(x)$.
$y^2=3x\implies y = \sqrt{3x}$,
$y^2=4x \implies y = \sqrt{4x} = 2 \sqrt x$, 
$x^2=3y\implies y = \frac 13 x^2$, 
$x^2=4y \implies y = \frac 14 x^2$.
Next, graph the functions, together, in one graph!
GRAPH ADDED: My guess is that you are being asked to find the area of the "parallelogram-like" figure in shown in the upper-left portion of the graph below, each side belonging to one and only one of the curves. 

Find the exact points of intersection to help determine bounds of integration (you will likely need to sum different integrals to find the total area, as the upper and lower curves will change). 
You'll need to find the $x$-values for each of the four points of intersection, so the area will be determined by the sum of four integrals with bounds from $x_0$ to $x_1$, $x_1$ to $x_2$, $x_2$ to $x_3$, and $x_3$ to $x_4$, respectively. You'll also need to determine which functions to use as your upper and lower curve on each interval over which you are integrating.
A: Let $u=\frac{x^2}{y}$ and $v=\frac{y^2}{x}$. Then your transformed region in $uv$-plane is $3\leq u\leq 4$ and $3\leq v\leq 4$ with the Jacobian $$\frac{\partial(u, v)}{\partial(x, y)}=3.$$ You can proceed from here.
