Question

Prove Theorem 13.10 (b). That is, show that if $f(x,y)$and $g(x,y)$are integrable functions on the general region ,then

${\iint }_{\Omega }\left(f\right(x,y)+g(x,y\left)\right)dA={\iint }_{\Omega }f(x,y)dA+{\iint }_{\Omega }g(x,y)dA$

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 ${\iint }_{\Omega }\left[f\right(x,y)+g(x,y\left)\right]dA={\iint }_{\Omega }f(x,y)dA+{\iint }_{\Omega }g(x,y)dA$

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

Using Property of Double Integrals

For any function$f$ that is continuous over region$\Omega$

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

and $F\left(x,y\right)=f(x,y),\text{if}(x,y)\in \Omega$and role="math" localid="1653944271832" $F\left(x,y\right)=0,\text{if}(x,y)\notin \Omega$

Write the double integral on left hand side as double Reimann sum 

Writing as double Riemann sum, we get

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

and

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

Simplify RHS

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

$=\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}F\left(x_i^{*},y_j^{*}\right)\Delta A+\underset{\Delta \to 0}{\lim }\sum _{i=1}^m\sum _{j=1}^{n}G\left(x_i^{*},y_j^{*}\right)\Delta A$

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

Using same property again

${\iint }_{\Omega }\left[f\right(x,y)+g(x,y\left)\right]dA={\iint }_{\Omega }f(x,y)dA+{\iint }_{\Omega }g(x,y)dA$

The equation is true.

Simplification

Changing order of sum

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

Simplify RHS

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

$=\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{j=1}^{m}F\left(x_i^{*},y_j^{*}\right)\Delta A+\underset{\Delta \to 0}{\lim }\sum _{j=1}^n\sum _{j=1}^{m}G\left(x_i^{*},y_j^{*}\right)\Delta A$

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

Using same property again

${\iint }_{\Omega }\left[f\right(x,y)+g(x,y\left)\right]dA={\iint }_{\Omega }f(x,y)dA+{\iint }_{\Omega }g(x,y)dA$

Hence, equation is true.