Right minus left, sideways AB & BC

Slice the region with a thin horizontal strip at height $y$. Its length is the horizontal distance between the boundaries and its thickness is $dy$, so

$$A = \int_c^d \big(x_{\text{right}} - x_{\text{left}}\big)\,dy.$$

Set that beside 8.4's $\int_a^b (y_{\text{top}} - y_{\text{bottom}})\,dx$ and the correspondence is exact:

1. Top becomes right, bottom becomes left. The subtraction still runs from the far boundary to the near one along the direction the strip points.
2. $dx$ becomes $dy$. The strip's thickness is measured along the axis it sweeps.
3. $x$-limits become $y$-limits. They are the heights at which the region starts and stops.

Every check from 8.4 survives the rotation. Solve for all intersections; test a value strictly between two consecutive ones to see which boundary is on the right; a negative result means the subtraction ran backwards.

The mirror example makes the point. In 8.4 the region between $y = x + 2$ and $y = x^{2}$ had area $\frac{9}{2}$. The region between $x = y + 2$ and $x = y^{2}$ is that picture reflected across the line $y = x$, so its area is $\frac{9}{2}$ as well, and the work is the same algebra with the letters exchanged: $y^{2} = y + 2$ gives $y = -1$ and $y = 2$, and testing $y = 0$ puts the line on the right.

A $dy$ integral needs its boundaries written as $x$ in terms of $y$. Leaving one as $y = f(x)$ and integrating it against $dy$ is the error CA6 names, and it produces an expression that is not wrong so much as meaningless.

1. Rewrite each boundary as $x = g(y)$. $y = x - 2$ becomes $x = y + 2$; $y = \sqrt{x}$ becomes $x = y^{2}$, for $y \geq 0$.
2. Watch the branch. $y = \sqrt{x}$ is only the upper half of $x = y^{2}$. The restriction $y \geq 0$ travels with it.
3. Recompute the limits. The old $x$-limits are not the new $y$-limits. Substitute the intersection points back and read off their $y$-coordinates.

That third step is where the arithmetic goes quietly wrong. If two curves meet at $(1, -1)$ and $(4, 2)$, then a $dx$ integral runs from $1$ to $4$ and a $dy$ integral runs from $-1$ to $2$. Both pairs come from the same two points, and using $1$ and $4$ as $y$-limits integrates over a region that does not exist.

The units check has no analogue here, so the guard is structural instead: a $dy$ integral should contain no $x$ anywhere. If one survives into the integrand or into a limit, a boundary was never converted.

Both variables always work. What differs is how many integrals each needs, and the count is decided by whether the boundaries change identity partway across the region.

Take the region enclosed by $x = y^{2}$ and $x = y + 2$. Sweeping horizontally, every strip runs from the parabola on the left to the line on the right, from $y = -1$ to $y = 2$: one integral.

Sweeping vertically, the region is bounded below by the lower half of the parabola throughout, but the upper boundary is the upper half of the parabola for $0 \leq x \leq 1$ and the line for $1 \leq x \leq 4$. That is two integrals, plus the work of splitting the parabola into branches:

$$\int_0^1 \big(\sqrt{x} - (-\sqrt{x})\big)\,dx + \int_1^4 \big(\sqrt{x} - (x - 2)\big)\,dx.$$

Both give $\frac{9}{2}$. The rule of thumb that follows: sweep in the direction in which the boundaries keep their identity. A region with a sideways parabola, or one bounded left and right by two curves and top and bottom by nothing in particular, usually wants $dy$.

Note what is not the criterion. It is not which letter the equations arrive in, and it is not whether the functions look easier to integrate. It is how many times a boundary hands off to a different curve as the strip sweeps.

The diagnosis is identical to 8.4's, and it is worth stating in the rotated language because students who learned the two topics separately often fail to transfer it.

If $\int_c^d (g_1(y) - g_2(y))\,dy$ comes out negative and the two curves do not cross inside $(c, d)$, then $g_2$ was the right-hand boundary and the subtraction ran the wrong way. Swap the order.

A second failure is specific to this orientation. Because a sideways parabola like $x = y^{2}$ has two branches, a student can subtract the correct pair of formulas while pairing the wrong branches, producing a length that is not the width of any strip. Testing a single value of $y$ inside the interval catches it: at $y = 0$ the parabola gives $x = 0$ and the line gives $x = 2$, so the strip runs from $0$ to $2$ and is $2$ units long. Any setup that does not reproduce that number at $y = 0$ is describing a different region.