Q 67. - Calculus

Question

Prove Theorem 13.10 (a). That is, show that if $f(x, y)$ is an integrable function on the general region  and $c \in R$, then

${\iint }_{\Omega }\alpha f(x,y)dA=\alpha {\iint }_{\Omega }f(x,y)dA$

Step-by-Step Solution

Answer

To prove this, write the double integral on left hand side as double Reimann sum. 

Step 1: Given Information

It is given that

${\iint }_{\Omega }cf(x,y)dA=c{\iint }_{\Omega }f(x,y)dA$

Region is subset of rectangular region defined by

$R=\left\{\right(x,y)\mid a\le x\le b\text{ and }c\le y\le d\}\text{, that is, }\Omega \subseteq R\text{ and }c$ is real number.

Step 2: Simplify using property

We know property of double integral

${\iint }_{\Omega }f(x,y)dA={\iint }_{R}F(x,y)dA$

and

$F(x,y)=(x,y),\text{ if }(x,y)\in \Omega$ and $0, \text{ if }(x,y)\notin \Omega$

write the double integral on left hand side as double Reimann sum. 

${\iint }_{R}cF(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}cF\left(x_i^{*},y_j^{*}\right)\Delta A$ and

$\Delta =\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}$

Simplify RHS

${\iint }_{R}cF(x,y)dA=c\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}F\left(x_i^{*},y_j^{*}\right)\Delta A$

$=c{\iint }_{R}F(x,y)dA$

From same property ${\iint }_{\Omega }cf(x,y)dA=c{\iint }_{\Omega }f(x,y)dA$

Equation is true.

Step 3 Simplification

Changing order of sum

${\iint }_{R}cF(x,y)dA=\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{i=1}^{m}cF\left(x_i^{*},y_j^{*}\right)\Delta A$

${\iint }_{R}cF(x,y)dA=c\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{i=1}^{m}F\left(x_i^{*},y_j^{*}\right)\Delta A$

$=c{\iint }_{R}F(x,y)dA$

From property stated above

${\iint }_{\Omega }cf(x,y)dA=c{\iint }_{\Omega }f(x,y)dA$