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

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

Short answer

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

Step-by-step solution

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

role="math" localid="1653943372914" $R=\left\{\right(x,y)\mid a\le x\le b\text{and}c\le y\le d\}\text{, that is,}\Omega \subseteq R\text{and}c$ is real number.

Simplify using property

We know property of double integral

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

and

localid="1653944299120" $F(x,y)=(x,y),\text{if}(x,y)\in \Omega$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.

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

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