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Area integral: examples

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Area integral example.jpg

We want to evaluate the area integral

$$\displaystyle\int\displaystyle\int_{G_1 \cup G_2} f dA,$$

where $G_1$ and $G_2$ are given in the picture. Let

$$G = \{ (x, y): -2 {\leq} x {\leq} 4, -4 {\leq} y {\leq} 4\} {\rm \hspace{3pt} and}$$

$$G_3 = G - ( G_1 {\cup} G_2)$$

and the function f be defined as

$$\begin{array}{} f(x,y) &= 2 {\rm \hspace{3pt} if \hspace{3pt}} (x,y) \in G_1 \\ &= -1 {\rm \hspace{3pt} if \hspace{3pt}} (x,y) \in G_2 \\ &= 0 {\rm \hspace{3pt} if \hspace{3pt}} (x,y) \in G_3. \end{array}$$

Then, exploiting the properties of Riemann integral, we have:

$$\begin{array}{} \displaystyle\int\displaystyle\int_G f dA &= \displaystyle\int\displaystyle\int_{G_1} f dA + \displaystyle\int\displaystyle\int_{G_2} f dA + \displaystyle\int\displaystyle\int_{G_3} f dA {\rm \hspace{3pt} (by \hspace{3pt} additivity)} \\ &= \displaystyle\int\displaystyle\int_{G_1} 2 dA + \displaystyle\int\displaystyle\int_{G_2} (-1) dA + \displaystyle\int\displaystyle\int_{G_3} 0 dA {\rm \hspace{3pt} (by \hspace{3pt} substitution)} \\ &= 2 \displaystyle\int\displaystyle\int_{G_1} dA + (-1) \displaystyle\int\displaystyle\int_{G_2} dA + 0 \displaystyle\int\displaystyle\int_{G_3} dA {\rm \hspace{3pt} (by \hspace{3pt} homogeneity)} \\ &= 2 area(G_1) + (-1) area(G_2) + 0 area(G_3) {\rm \hspace{3pt} (by \hspace{3pt} normalization)} \\ &= 2 ( 2 \cdot 3 ) + ( -1 ) ( 1 \cdot 2 ) + 0 \\ &= 12 - 2 \\ &= 10. \end{array}$$