Chapter 15.P, Problem 1P

If $〚x〛$ denotes the greatest integer in x, evaluate the integral

${\displaystyle \underset{R}{\iint }〚x+y〛}dA$

where $R=\left\{\left(x,y\right)|1\le x\le 3,2\le y\le 5\right\}$.

To determine

To evaluate:

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA$

Answer to Problem 1P

Solution:

$30$

Explanation of Solution

1) Concept:

i. If $R$ is a rectangle of type $R=\left\{\left(x,y\right)\right| a\le x\le b , c\le y\le d \}$, then we can divide region $R$ into $m$ subregions, $R_1, R_2, \dots \dots \dots ., R_m$

Then $\underset{R}{\overset{ }{\iint }}f(x, y)dA==\sum _{i=1}^m\underset{R_i}{\overset{ }{\iint }}f(x, y)dA$

ii. $⟦x⟧$ is the greatest integer in $x$.

2) Given:

The region  $R=\left\{\left(x,y\right)\right| 1\le x\le 3 , 2\le y\le 5 \}$

3) Calculation:

We have to evaluate the integral

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA$

where $R$ is a region

$R=\left\{\left(x,y\right)\right| 1\le x\le 3 , 2\le y\le 5 \}$

The region of $R$ integration is the rectangular region shown below:

Divide the region $R$ into five parts as shown in the figure.

From the figure, see that

$R=R_1\cup R_2\cup R_3\cup R_4\cup R_5$

$R={\bigcup }_{i=1}^5{R}_i$

where, $R_i=\left\{\left(x, y\right)\right| x+y\ge i+2 and x+y\le i+3, 1\le x\le 3 , 2\le y\le 5$

Therefore,

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA=\underset{R_1}{\overset{ }{\iint }}⟦x+y⟧dA+\underset{R_2}{\overset{ }{\iint }}⟦x+y⟧dA+\underset{R_3}{\overset{ }{\iint }}⟦x+y⟧dA+\underset{R_4}{\overset{ }{\iint }}⟦x+y⟧dA+\underset{R_5}{\overset{ }{\iint }}⟦x+y⟧dA$

$=\sum _{i=1}^5\underset{R_i}{\overset{ }{\iint }}⟦x+y⟧dA$

By concept (ii),

$⟦x+y⟧$ is the greatest integer in $x+y;$ hence $⟦x+y⟧$ is constant and equal to the biggest integer strictly less than or equal to  $x+y$.

Therefore, $⟦x+y⟧=i+2$

Therefore,

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA=\sum _{i=1}^5\underset{R_i}{\overset{ }{\iint }}(i+2)dA$

$=\sum _{i=1}^5(i+2)\underset{R_i}{\overset{ }{\iint }}dA$

$=\sum _{i=1}^5\left(i+2\right)\left[A\left(R_i\right)\right]$

Expand the sum:

$=\left(1+2\right)\left[A\left(R_1\right)\right]+\left(2+2\right)\left[A\left(R_2\right)\right]+\left(3+2\right)\left[A\left(R_3\right)\right]+\left(4+2\right)\left[A\left(R_4\right)\right]+\left(5+2\right)\left[A\left(R_5\right)\right]$

$=3\left[A\left(R_1\right)\right]+4\left[A\left(R_2\right)\right]+5\left[A\left(R_3\right)\right]+6\left[A\left(R_4\right)\right]+7\left[A\left(R_5\right)\right]$

Now find the area of each region $R_i, i=1 to 5$

From the figure,

$R_1 and R_5$ are triangles with length $1$ and height $1.$

Therefore,

$A\left(R_1\right)=A\left(R_5\right)=\frac{1}{2}\times 1\times 1=\frac{1}{2}$

Also $R_2$ and ${ R}_4$ lie in triangle $∆abc and ∆pqr$ with length $2$ and height $2.$

$A\left(∆abc\right)=A\left( ∆pqr\right)=\frac{1}{2}\times 2\times 2=2$

From the figure, see that

$A\left(R_2\right)=A\left(∆abc\right)-A\left(R_1\right)$ and

$A\left(R_4\right)=A\left( ∆pqr\right)-A\left(R_5\right)$

Therefore,

$A\left(R_2\right)=2-\frac{1}{2}=\frac{3}{2}$

$A\left(R_4\right)=2-\frac{1}{2}=\frac{3}{2}$

$R_3$ is a rectangle with length $2$ and height $1$

Therefore,

$A\left(R_3\right)=length \times heigth=1\times 2=2$

Therefore,

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA=3\left[\frac{1}{2}\right]+4\left[\frac{3}{2}\right]+5\left[2\right]+6\left[\frac{3}{2}\right]+7\left[\frac{1}{2}\right]$

$=30$

Therefore,

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA=30$

Conclusion:

$\underset{R}{\overset{ }{\iint }}⟦x+y⟧dA=30$