integrate $\int_0^1 \int_{\sqrt{x}}^1 e^{\frac{y^3}{3}} \, dy \, dx$ 
$$\int_0^1 \int_{\sqrt{x}}^{1}e^{\frac{y^3}{3}} \, dy \, dx$$

So I understand I need to change the integration limits by looking at the domain. I have sketched it

But how do I know if I need to take the area under the curve or above it?
(in the drawing the area filled with vercital or horizontal lines) 
 A: While $x$ varies between $0$ and $1$, $y$ varies between $\sqrt x $ and $1$; hence it is the above region (horizontal lines)
The other region would be $y$ varies between $0$ and $\sqrt x $
A: Change the order of the integration.
$$ \int_0^1 \int_0^{y^2} \exp\left(\dfrac{y^3}{3}\right)\,dx\, dy$$

$$  = \int_0^1 y^2 \exp\left(\dfrac{y^3}{3}\right)\,dy$$

Then you can use a $u$ substitution to solve this integral

 $$\int_0^1 \exp(u) \,du$$

A: Every fixed value of $x$ between $0$ and $1$ corresponds to a vertical line in your sketch.  That vertical line intersects the curve you drew at a height of $\sqrt x$ above the $x$-axis, and intersects the horizontal line above that at a height of $1$ above the $x$-axis; hence as $y$ goes from $\sqrt x$ up to $1$, you're above the curve.
If you do the integral in the opposite order, then you think of a fixed value of $y$ corresponding to a horizontal line.  That horizontal line intersects the left boundary of the region you drew at the $y$-axis, which is the line $x=0$, and intersects the curved right boundary when $x=y^2$.  So on that horizontal line, $x$ goes from $0$ up to $y^2$.  Thus you have
$$
\int_0^{y^2} \cdots\cdots\, dx.
$$
This integral is evaluated on every vertical line corresponding to a $y$ value between $0$ and $1$, so that integral is inside another one:
$$
\int_0^1 \left( \int_0^{y^2} \cdots\cdots \,dx \right) \, dy.
$$
