Double integral over general regions between 2 curves. Using a double integral, how do I find the area enclosed by the curves $y=x^2/2$ and $x=y^2/2?$?
I got as far as $\int_{0}^{2} \int_{y^2/2}^{\sqrt{2y}} dx dy$, but I can't get any further, since I don't understand what do I integrate in that case. is it 1 or y? or even x? Or did I do everything wrong from the start?
 A: You do not need a double integral.

$$\int\limits_{x=0}^1 \sqrt{2 x} - x^2/2\ dx = \frac{1}{6} \left(4 \sqrt{2}-1\right)$$
A: I got something I want to add to @David G. Stork 's answer, which is too long for a comment.
In this case can easily express your equation for $x(y)$ in terms of $y(x)$ which gives:
$$y(x) = \pm \sqrt{2x}$$
Now you would choose the approriate sign, which this case is positive (look at the picture above).
Furthermore you know that the region between a "$1D$"-function $f(x)$ and the x-axis for the intervall $[a,b]$ is given by the integral
$$\int_{a}^{b}f(x) dx$$
In your example you can define: $f(x) = \sqrt{2x}$, and  $g(x) = \frac{x^2}{2}$. For the boundaries of the integral calculate the x-values where your curves meet. Now can calculate the difference between the integral (i.e. the Area A) and use the linearity of the integral:
$$A = \int_0^2 f(x) dx  - \int_0^2 g(x) = \int_0^2 (f(x)-g(x) )dx$$
Which gives:
$$\int_0^2 \big( \sqrt{2x} - x^2/2 \big) dx= \bigg[ \frac{2}{3}\sqrt{2}x^{3/2} - \frac{x^3}{6} \bigg]_{x=0}^{x=2} = \frac{2}{3}\sqrt{2} \ 2^{3/2} - \frac{2^3}{6} = \frac{2}{3} 2^{4/2} - \frac{8}{6} = \frac{8}{3} - \frac{8}{6} = \frac{4}{3}$$
Note: The area under a given curve and the x-Axis might be negative, so beware of probably needed absolute values! (I.e. when the function is negative.)
EDIT: Corrected some Errors.
EDIT: Corrected the upper bound and added the evalution of the integral.
A: I think your integral is correct, then we have that
\begin{align*}
\int_{0}^{2}\int_{y^2/2}^{\sqrt{2y}}dxdy&=\int_{0}^{2}(\sqrt{2y}-y^2/2)dy\\
&=\int_{0}^{2}\sqrt{2y}dy-\int_{0}^{2}y^2/2dy\\
&=\frac{4}{3}
\end{align*}
I will write a bit more about the idea.

Suppose that you are working with a region $R$ of the form
$$R=\{(x,y): a\leqslant x\leqslant b, u_{1}(x)\leqslant y\leqslant u_{2}(x)\}$$
and we want know the area of the region $R$, call it $A(R)$. Using double integrals we need density equals to one, that is,
$$A(R):=\iint_{R}1\, dA$$
But, since we know $R$ we can write it as
$$A(R)=\int_{a}^{b}\int_{u_{1}(x)}^{u_{2}(x)}dydx=\boxed{\int_{a}^{b}(u_{2}(x)-u_{1}(x))dx}$$
And we find one classical formula for the area in calculus of one variable. Similarly,  if we can write the region as $R'$ given by $$R'=\{(x,y): c\leqslant y\leqslant d, v_{1}(y)\leqslant x\leqslant v_{2}(y)\}$$
Then, the area of the region $R'$ call it, $A(R')$ in terms of one integral is given by the other classical formula $$A(R')=\boxed{\int_{c}^{d}(v_{2}(y)-v_{1}(y))dxdy}$$
Returning to your problem, the key part is making a plot of the region because we need to know as the region it is described and we need to decide whether it is convenient to use a description in terms of $R$ or $R'$. We can plot the curves $y=\frac{x^2}{2}$ and $x=\frac{y^2}{2}$ and then noticed that
$$R=\{(x,y): 0\leqslant x\leqslant 2, x^2/2\leqslant y\leqslant \sqrt{2x}\}$$
or in this case we consider
$$R'=\{(x,y): 0\leqslant y\leqslant 2, y^2/2\leqslant x\leqslant \sqrt{2y} \}$$
Of course, since both region they are equivalents, then the area of both regions must be the same. Indeed, we have that
$$A(R)=\int_{0}^{2}(\sqrt{2x}-x^2/2)dx= \frac{4}{3}=\int_{0}^{2}(\sqrt{2y}-y^2/2)dy=A(R')$$
Therefore, the area of the region is $\frac{4}{3}$.
Note: In order to find the region we can find the intersection between both curves, letting $x=(x^2/2)^2/2$ since $y=x^2/2$ then we find $(0,0)$ and $(2,2)$ over the real plane.
