1) Concept:
i. If R is a rectangle of type R=x,y a≤x≤b , c≤y≤d }, then we can divide region R into m subregions, R1, R2, ………., Rm
Then ∬R f(x, y)dA==∑i=1m∬Ri f(x, y)dA
ii. x is the greatest integer in x.
2) Given:
The region R=x,y 1≤x≤3 , 2≤y≤5 }
3) Calculation:
We have to evaluate the integral
∬R x+ydA
where R is a region
R=x,y 1≤x≤3 , 2≤y≤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=R1∪R2∪R3∪R4∪R5
R=⋃i=15Ri
where, Ri=x, y x+y≥i+2 and x+y≤i+3, 1≤x≤3 , 2≤y≤5
Therefore,
∬R x+ydA=∬R1 x+ydA+∬R2 x+ydA+∬R3 x+ydA+∬R4 x+ydA+∬R5 x+ydA
=∑i=15∬Ri x+ydA
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,
∬R x+ydA=∑i=15∬Ri (i+2)dA
=∑i=15(i+2)∬Ri dA
=∑i=15i+2ARi
Expand the sum:
=1+2AR1+2+2AR2+3+2AR3+4+2AR4+5+2AR5
=3AR1+4AR2+5AR3+6AR4+7AR5
Now find the area of each region Ri, i=1 to 5
From the figure,
R1 and R5 are triangles with length 1 and height 1.
Therefore,
AR1=AR5=12×1×1=12
Also R2 and R4 lie in triangle ∆abc and ∆pqr with length 2 and height 2.
A∆abc=A ∆pqr=12×2×2=2
From the figure, see that
AR2=A∆abc-AR1 and
AR4=A ∆pqr-AR5
Therefore,
AR2=2-12=32
AR4=2-12=32
R3 is a rectangle with length 2 and height 1
Therefore,
AR3=length ×heigth=1×2=2
Therefore,
∬R x+ydA=312+432+52+632+712
=30
Therefore,
∬R x+ydA=30
Conclusion:
∬R x+ydA=30