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15.3: Area by Double Integration

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    In this section, we will learn to calculate the area of a bounded region using double integrals, and using these calculations we can find the average value of a function of two variables.

    Areas of Bounded Regions in the Plane

    Using Reimann sums, the volume or surface mass is equal to the sum of the areas at each point \(k\), \( \Delta A_k \), multiplied by height or surface mass density at each point, described by the function, \(f(x,y)\).

    \[ S_n = \sum_{k=1}^n f(x_k,y_k) \Delta A_k = \sum_{k=1}^n \Delta A_k \]

    Using this notation to find the area, we set \(f(x,y)\) (height or surface mass density) equal to 1.

    Volume = Area x Height Surface Mass = Area x Surface Mass Density

    if Height = 1, Volume = Area x 1 if Surface Mass = 1, Surface Mass = Area x 1

    So, Volume = Area So, Surface Mass = Area

    Therefore, we simply sum all the \( \Delta A_k \) values , allowing us to find the area of a boundary. To calculate the area, we sum the areas of infinitely small rectangles within the closed region \( R \). We find the limit of the sum as the length and width in the partition approach zero.

    \[ \lim_{||P|| \rightarrow 0} \sum_{k=1}^n \Delta A_k = \iint_R dA\]

    Therefore, the area of a closed, bounded plane region R is defined as

    \[A= \iint_R dA\]

    Average Value

    Using double integrals to find both the volume and the area, we can find the average value of the function \(f(x,y)\).

    \[ \text{Average Value of} \ f \ \text{over} \ R = \frac{1}{\text{area} \ \text{of} \ R} \iint_R f \ dA \]

    \[ \bar{f} = \frac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA}\]

    The value describes the average height of the calculated volume or the average surface mass of the calculated total mass.

    \[\begin{align} &= \left. ({e^x} - x) \right |_0^1 \\ &= (e-1) \ - \ (1-0) \\ &= (e-2). \end{align}\]

    \[\begin{align} \bar{f} &= \frac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA} \\ & = \frac{0.8}{3.63886} = 0.2198 . \end{align}\]

    \[\begin{align} \bar{f} &= \dfrac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA} \\ & = \dfrac{128.333}{649.25} = 0.1976. \end{align}\]

    Contributors and Attributions

    • (UCD)

    • Integrated by Justin Marshall.