13.7 Integrals of Boxes

Putting together polyrulers and boxes, we can measure area!

Definition.

Suppose we have a degree two polyruler and a degree two box

\[ \begin{align} \bar{\diff} r &: \PolyRuler_{(x, y)}^{2} & q &: \BoxType_{(x, y)}^{2} \end{align} \]

where \( \bar{\diff} r \) is the differential of a ruler \( r : \Ruler_{(x, y)} \). The integral of \( q \)

\[ \begin{align} \Bigl\langle{\bar{\diff} r}\mathrel{\textstyle{\int}}{q}\Bigr\rangle &: \Number \\ \Bigl\langle{\bar{\diff} r}\mathrel{\textstyle{\int}}{q}\Bigr\rangle &= \Bigl\langle{r}\mathrel{\textstyle{\int}}{\boundary q}\Bigr\rangle \end{align} \]

is computed by trading the differential for a boundary.

The area polyruler for the \( (x, y) \)-coordinate plane is the polyruler

\[ \begin{align} \bar{\standard}_{x} \mult \bar{\standard}_{y} &: \PolyRuler_{(x, y)}^{2}. \end{align} \]

We get a simple formula when we calculate the area of a box.

Theorem.

The area integral of a box is given as follows

\[ \begin{align} \biggl\langle{\bar{\standard}_{x} \mult \bar{\standard}_{y}}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle &= (x_2 - x_1) \mult (y_2 - y_1) \end{align} \]

where

\[ \begin{align} x_1 &: \Number, & x_2 &: \Number, & y_1 &: \Number, & y_2 &: \Number \end{align} \]

are all constants.

Why?

We can write the area polyruler \( \bar{\standard}_{x} \mult \bar{\standard}_{y} \) as the differential of the ruler \( r = x \mult \bar{\standard}_{y} \). Indeed,

\[ \begin{align} \bar{\diff} r &= \bar{\diff} \bigl(x \mult \bar{\standard}_{y}\bigr) = \bar{\diff} x \mult \bar{\standard}_{y} = \bar{\standard}_{x} \mult \bar{\standard}_{y}. \end{align} \]

Let's use this to calculate the area \( A \).

\[ \begin{align} A &= \biggl\langle{\bar{\standard}_{x} \mult \bar{\standard}_{y}}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle \\ &= \biggl\langle{\bar{\diff} \bigl(x \mult \bar{\standard}_{y}\bigr)}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle \\ &= \biggl\langle{x \mult \bar{\standard}_{y}}\mathrel{\displaystyle{\int}}{\boundary \bigl([x_1, x_2]_{x} \mult [y_1, y_2]_{y}\bigr)}\biggr\rangle \\ &= \biggl\langle{x \mult \bar{\standard}_{y}}\mathrel{\displaystyle{\int}}{\boundary [x_1, x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle - \biggl\langle{x \mult \bar{\standard}_{y}}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult \boundary [y_1, y_2]_{y}}\biggr\rangle \\ &= I_2 - I_1 \end{align} \]

So far our calculation has produced the difference of two integrals \( I_2 - I_1 \). Let's first work with \( I_1 \).

\[ \begin{align} I_1 &= \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult \boundary [y_1, y_2]_{y}}\biggr\rangle \\ &= \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult [y_2]_{y}}\biggr\rangle - \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x} \mult [y_1]_{y}}\biggr\rangle \\ &= \biggl\langle{x \mult \bar{\diff} y_2}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x}}\biggr\rangle - \biggl\langle{x \mult \bar{\diff} y_1}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x}}\biggr\rangle \\ &= \biggl\langle{0}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x}}\biggr\rangle - \biggl\langle{0}\mathrel{\displaystyle{\int}}{[x_1, x_2]_{x}}\biggr\rangle \\ &= 0 \end{align} \]

We find that \( I_1 \) is zero because \( y_1 \) and \( y_2 \) are constants. Let's pick up our calculation of the area with \( I_2 \).

\[ \begin{align} A &= \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{\boundary [x_1, x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle \\ &= \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[x_2]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle - \biggl\langle{x \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[x_1]_{x} \mult [y_1, y_2]_{y}}\biggr\rangle \\ &= \biggl\langle{x_2 \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[y_1, y_2]_{y}}\biggr\rangle - \biggl\langle{x_1 \mult \bar{\diff} y}\mathrel{\displaystyle{\int}}{[y_1, y_2]_{y}}\biggr\rangle \\ &= x_2 \mult \biggl\langle{\bar{\diff} y}\mathrel{\displaystyle{\int}}{[y_1, y_2]_{y}}\biggr\rangle - x_1 \mult \biggl\langle{\bar{\diff} y}\mathrel{\displaystyle{\int}}{[y_1, y_2]_{y}}\biggr\rangle \\ &= x_2 \mult \biggl\langle{y}\mathrel{\displaystyle{\int}}{\boundary [y_1, y_2]_{y}}\biggr\rangle - x_1 \mult \biggl\langle{y}\mathrel{\displaystyle{\int}}{\boundary [y_1, y_2]_{y}}\biggr\rangle \\ &= x_2 \mult (y_2 - y_1) - x_1 \mult (y_2 - y_1) \\ &= (x_2 - x_1) \mult (y_2 - y_1) \end{align} \]

Phew! We found our answer.