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How to Describe a 2D Region

We define a region as a bounded part of a graph. To actually work with a region mathematically (integrate over it, find its area, etc.), we need a precise description.

Vertically Simple Regions

Observe the region bounded by $y = x^2$ and $y = 2x$.

Region bounded by y equals x squared and y equals 2x
Figure 1: Vertically simple region bounded by $y = x^2$ and $y = 2x$.

Notice that:

  • In the x-direction, the region spans from $x = 0$ to $x = 2$, two fixed real numbers.
  • In the y-direction, the region is bounded above by $y = 2x$ and below by $y = x^2$, two functions of $x$.

Since the x-axis description uses constants rather than functions, don’t you think it is more simple? We call this a vertically simple region and write it as:

$$R = \{(x, y) \mid 0 \leq x \leq 2, \quad x^2 \leq y \leq 2x\}$$

Horizontally Simple Regions

Conversely, a region is horizontally simple if the y-direction is bounded by two real numbers and the x-direction is bounded by two functions of $y$.


Your Turn

Below is a new region:

$$\text{Left boundary: } x = y^2, \qquad \text{Right boundary: } x = 2y, \qquad y \in [0, 2]$$
Region bounded by x equals y squared and x equals 2y
Figure 2: Horizontally simple region bounded by $x = y^2$ and $x = 2y$.

Question 1: Can you describe this region as a horizontally simple region using set notation?

$$R = \{(x, y) \mid \ldots \leq y \leq \ldots, \quad \ldots \leq x \leq \ldots\}$$
Answer

When $y$ is fixed, $ y^2$ is to the left of $2y$ giving us the boundaries of $x$ as

$$\quad y^2 \leq x \leq 2y$$

$$R = \{(x, y) \mid 0 \leq y \leq 2, \quad y^2 \leq x \leq 2y\}$$

Question 2: Now describe the same region as a vertically simple region.

$$R = \{(x, y) \mid \ldots \leq x \leq \ldots, \quad \ldots \leq y \leq \ldots\}$$
Answer

Invert the boundary curves to write $y$ as a function of $x$:

  • From $x = y^2$, we get $y = \sqrt{x}$.
  • From $x = 2y$, we get $y = \dfrac{x}{2}$.

At $x = 1$, $\sqrt{1} = 1 > \dfrac{1}{2}$, so $y = \sqrt{x}$ is the upper boundary and $y = \dfrac{x}{2}$ is the lower boundary. The $x$-bounds come from setting $y^2 = 2y$, giving $y = 0$ or $y = 2$, so $x$ ranges from $0$ to $4$.

$$R = \left\{(x, y) \;\middle|\; 0 \leq x \leq 4, \quad \frac{x}{2} \leq y \leq \sqrt{x}\right\}$$

This example illustrates our freedom to switch between descriptions. Horizontal and vertical simplicity are not intrinsic properties of a region itself, but of how we choose to describe it. A region is not inherently horizontally simple; our description of it is.

Notice how unnatural it feels to go against the original formulation here. The region was defined with $x$ as a function of $y$, so the horizontally simple description follows directly. Switching to a vertically simple description forces us to invert the boundary curves, adding an extra step that the original description did not require. Although we are always free to switch, one description is often more natural than the other depending on how the region is given.


A Slightly Harder Example

Consider the region bounded by $y^2 = x$ and $y = x^2$.

Region bounded by y squared equals x and y equals x squared
Figure 3: Region bounded by $y^2 = x$ and $y = x^2$.

Notice something interesting here. The boundary $y^2 = x$ can be written as $y = \sqrt{x}$, a function of $x$, but also as $x = y^2$, a function of $y$. Similarly, $y = x^2$ can be written as $x = \sqrt{y}$. So unlike the previous example, both boundaries are naturally expressible either way. You have the freedom to describe this region as vertically simple or horizontally simple with equal ease.

Question: Describe this region in both ways using set notation.

Answer

First find the intersection points by substituting $y = x^2$ into $y^2 = x$:

$$(x^2)^2 = x \implies x^4 = x \implies x(x^3 - 1) = 0$$

So $x = 0$ or $x = 1$, giving intersection points $(0, 0)$ and $(1, 1)$.

As a vertically simple region, $y = \sqrt{x}$ is the upper boundary and $y = x^2$ is the lower boundary, with $x$ ranging from $0$ to $1$:

$$R = \{(x, y) \mid 0 \leq x \leq 1, \quad x^2 \leq y \leq \sqrt{x}\}$$

As a horizontally simple region, rewrite the boundaries as functions of $y$: $y^2 = x$ gives $x = y^2$ and $y = x^2$ gives $x = \sqrt{y}$. With $y$ ranging from $0$ to $1$:

$$R = \{(x, y) \mid 0 \leq y \leq 1, \quad y^2 \leq x \leq \sqrt{y}\}$$

Notice how both descriptions feel equally natural here, precisely because the boundary curves can be expressed as functions of either variable without any extra effort.


Exercise

Consider the square region with corners at $(0, 0)$, $(1, 0)$, $(1, 1)$, and $(0, 1)$. Would you describe it as vertically simple or horizontally simple?

Answer

Both descriptions are equally simple, and this is actually the most natural case of all. Since all four boundaries are constants, neither description requires any functions:

As a vertically simple region:

$$R = \{(x, y) \mid 0 \leq x \leq 1, \quad 0 \leq y \leq 1\}$$

As a horizontally simple region:

$$R = \{(x, y) \mid 0 \leq y \leq 1, \quad 0 \leq x \leq 1\}$$

They are identical in structure. This is a good reminder that vertically and horizontally simple are not competing categories; they describe how we choose to express the bounds, and for simple shapes like a square, the distinction disappears entirely.

Consider the annular region between the two circles $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.

Annular region between circles of radius 1 and radius 2
Figure 4: Annular region bounded by $x^2 + y^2 = 1$ and $x^2 + y^2 = 4$.

Can you describe this region as vertically simple or horizontally simple? If not, what goes wrong? Can you find a way around it?