The question is: Find the area (R) bounded by the following parabolas:
(1). $y=x^2$
(2). $4y=x^2$
(3). $y^2=2x$
(4). $y^2=3x$
I am looking for a solution with double integrals.
I tried to do it with jacobian found out u={1,4} and v={2,3} but unlike this (Finding Integral over a Region Bounded by 4 curves) one i cannot evaluate. Because it is asking the area.
Edit: Alternative Solution
After determining u and v values draw a u-v graph and find the area. (You will get a small rectangle.) Let's call it A. Then determine the jacobian (j) and since j is "R/A" you will get the answer.
Finding the area of a region is equivalent to integrate the constant function $1$ over that region. I will assume you are trying to find the area of following region:
$$\begin{array}{rcl} &1 \le \frac{x^2}{y} \le 4 & \text{(below (1), above (2))}\\ \text{and}& 2 \le \frac{y^2}{x} \le 3 & \text{(left of (3), right of (4))} \end{array}$$
Change variable to $u = \frac{x^2}{y}, v = \frac{y^2}{x}$, the region becomes $$ 1 \le u \le 4\quad\text{ and }\quad 2 \le v \le 3$$ Since $x = u^{\frac23}v^{\frac13}$ and $y = u^{\frac13}v^{\frac23}$, the Jacobian $J$ equals to $$J = \left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left|\begin{matrix} \frac{2x}{3u} & \frac{x}{3v}\\ \frac{y}{3u} & \frac{2y}{3v} \end{matrix}\right| = \frac{2^2-1}{3^2}\frac{xy}{uv} = \frac13 $$ As a result, the area of the region is $$R = \int_2^3 \int_1^4 J dudv = \frac13\int_2^3 \int_1^4 dudv = \frac13(3-2)(4-3) = 1$$