Q 68. - Calculus

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

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

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

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

Step 2: 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 $F\left(x,y\right)=0, \text{ if }(x,y)\notin \Omega$

Step 3: 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.

Step 4: 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.