Most regions are not rectangles. In area between curves you sliced a region vertically, with each slice running from a lower curve to an upper curve. Do the same for a double integral, and let the inner limits be those curves.
Key idea
Type I (vertical slices): $D = \{a \le x \le b,\ g_1(x) \le y \le g_2(x)\}$ gives $$\displaystyle\iint_D f\,dA = \int_a^b\int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx.$$
Type II (horizontal slices): $D = \{c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$ gives $\displaystyle\int_c^d\int_{h_1(y)}^{h_2(y)} f\,dx\,dy$.
Inner limits may depend on the outer variable; outer limits are always constants. The area of $D$ is $\iint_D 1\,dA$.
Type II (horizontal slices): $D = \{c \le y \le d,\ h_1(y) \le x \le h_2(y)\}$ gives $\displaystyle\int_c^d\int_{h_1(y)}^{h_2(y)} f\,dx\,dy$.
Inner limits may depend on the outer variable; outer limits are always constants. The area of $D$ is $\iint_D 1\,dA$.
Why it works
Fix $x$ and look at the vertical line through the region: it enters at the lower boundary $y = g_1(x)$ and leaves at the upper boundary $y = g_2(x)$. The inner integral adds $f$ along that slice only, so it naturally depends on $x$. The outer integral then sweeps the slices across, from the leftmost to the rightmost $x$ in the region; those are fixed numbers, because they describe the whole region, not one slice. After integrating, no variable may be left, and that happens only if the outer limits are constants.
Worked examples
Worked example: Between two curves
Evaluate $\iint_D (x + y)\,dA$ where $D$ is the region between $y = x^2$ and $y = x$.
- The curves meet at $x = 0$ and $x = 1$, and $x \ge x^2$ there: ${}0 \le x \le 1$, $x^2 \le y \le x$.
- Inner: $$\int_{x^2}^{x}(x + y)\,dy = \left[xy + \frac{y^2}{2}\right]_{x^2}^{x} = \frac{3x^2}{2} - x^3 - \frac{x^4}{2}.$$
- Outer: $$\int_0^1 \left(\frac{3x^2}{2} - x^3 - \frac{x^4}{2}\right) dx = \frac{1}{2} - \frac{1}{4} - \frac{1}{10} = \frac{3}{20}.$$
Answer: $\frac{3}{20}$
Worked example: An area
Find the area between $y = x^2$ and $y = 2x$.
- They meet at $x = 0$ and $x = 2$; the line is on top.
- Area $$= \int_0^2\int_{x^2}^{2x} dy\,dx = \int_0^2 (2x - x^2)\,dx = 4 - \frac{8}{3} = \frac{4}{3}.$$
Answer: $\frac{4}{3}$
Worked example: Horizontal slices are easier
Find the area of the region bounded by $x = y^2$ and $x = y + 2$.
- Vertical slices would need two pieces, but horizontally the region runs from the parabola $x = y^2$ (left) to the line $x = y + 2$ (right).
- They meet where $y^2 = y + 2$: $y = -1$ and $y = 2$.
- Area $$= \int_{-1}^{2} (y + 2 - y^2)\,dy = \left[\frac{y^2}{2} + 2y - \frac{y^3}{3}\right]_{-1}^{2} = \frac{10}{3} + \frac{7}{6} = \frac{9}{2}.$$
Answer: $\frac{9}{2}$
Common mistake
Putting variables in the outer limits. An answer like $\int_{x^2}^{x}\int_0^1 \ldots\,dy\,dx$ is upside down: after the inner integral you would integrate in $x$ between limits that contain $x$, which makes no sense. The outer integral describes the whole region with numbers; the inner one describes a single slice with curves. Check: your final answer must be a number.