3.2: Area by Double Integration

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Oct 27, 2024
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- 3.1: Double and Iterated Integrals Over Rectangles
- 3.3: Double Integrals Over General Regions

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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, $f(x,y)$ .

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

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.

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

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

$A= \iint_R dA \nonumber$

Average Value

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

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

$\bar{f} = \frac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA} \nonumber$

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

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

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

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

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Integrated by Justin Marshall.

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

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

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

Areas of Bounded Regions in the Plane

Average Value

=(ex−x)|10=(e−1) − (1−0)=(e−2).\begin{align} &= \left. ({e^x} - x) \right |_0^1 \\ &= (e-1) \ - \ (1-0) \\ &= (e-2). \end{align} \nonumber

ˉf=∬Rf(x,y) dA∬R(1) dA=0.83.63886=0.2198.\begin{align} \bar{f} &= \frac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA} \\ & = \frac{0.8}{3.63886} = 0.2198 . \end{align} \nonumber

ˉf=∬Rf(x,y) dA∬R(1) dA=128.333649.25=0.1976.\begin{align} \bar{f} &= \dfrac{\iint_R f(x,y) \ dA}{\iint_R (1) \ dA} \\ & = \dfrac{128.333}{649.25} = 0.1976. \end{align} \nonumber

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

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

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