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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.

Example 1

Find the area of the region bounded above by $$y=e^x$$, below by $$y=1$$, left by $$x=0$$ and right by $$x=1$$.

Solution

To find the area use the formula derived above:

$A= \iint_R dA.$

This double integral can be computed by using Fubini's Theorem:

$\int_{x_0}^{x_1} \int_{g_1 (x)}^{g_2 (x)} f(x,y) \ dy \ dx.$

Where $$x_0 \ \text{and} \ x_1$$ are the left and right bounds and $$g_1 (x) \ \text{and} \ g_2 (x)$$ are the lower and upper bounds, respectively.

Remember $$f(x,y) = 1$$,

$$g_1 (x) = 1$$ $$g_2 (x) = e^x$$ $$x_0 = 0$$ $$x_1 = 1$$

$\text{So, Area} = \int_0^1 \int_1^{e^x} 1 \ dy \ dx.$

First, integrate from $$g_1 (x) \ \text{to} \ g_2 (x)$$ in terms of $$y$$, holding $$x$$ constant:

\begin{align} &= \int_0^1 \left. y \right|_1^{e^x} dx \\ &= \int_0^1 ({e^x} - 1) dx. \end{align}

Then, integrate the remainding single integral from $$x_0 \ \text{to } x_1$$:

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

Example 2

Find the average value of $$f(x,y) = \frac{1}{x^3 y}$$ bounded above by $$y=1$$, below by $$y = e^{-x}$$, to the left by $$x=1$$, and the right by $$x=5$$.

Solution

First compute the integral to find the total value of $$f (x,y)$$ within the region $$R$$:

$\int_{x_0}^{x_1} \int_{g_1 (x)}^{g_2 (x)} f(x,y) \ dy \ dx .$

Where $$dy\; dx$$ is equal to $$dA$$ and $$f(x,y)$$ is equal to the height. So the integral equals a volume, since Area x Height = Volume.

Also, $$x_0 \ \text{and} \ x_1$$ are the left and right bounds and $$g_1 (x) \ \text{and} \ g_2 (x)$$ are the lower and upper bounds, respectively, as shown in example one.

$$f(x,y) = \frac{1}{{x^3}y}$$ $$g_1 (x) = e^{-x}$$ $$g_2 (x) = 1$$ $$x_0 =1$$ $$x_2=5$$

So the total value of $$f(x,y)$$ within the boundaries of $$x$$ and $$y$$ =

$\int_1^5 \int_{e^{-x}}^1 \frac{1}{{x^3}y} \ dy \ dx.$

First integrating with respect to $$y$$, keeping $$x$$ constant:

\begin{align} & = \int_1^5 \left. \frac{\ln|y|}{x^3} \right |_{e^{-x}}^1 \ dx \\ & = \int_1^5 \frac{x}{x^3} \ dx \\ &= \int_1^5 \frac{1}{x^2} \ dx. \end{align}

Then, integrate the single integral between $$x_0 \ \text{and} \ x_1$$ with respect to $$x$$:

\begin{align} & = \left. - \frac{1}{x} \right |_1^5 \\ & = - \frac{1}{5} + 1 \\ & = \frac{4}{5} = 0.8. \end{align}

Then to find the average value of $$f(x,y)$$, we must divide this value by the total area of the region. To find the area of the region:

$\int_{x_0}^{x_1} \int_{g_1 (x)}^{g_2 (x)} f(x,y) \ dy \ dx$

$$f(x,y) = 1$$ $$g_1 (x) = e^{-x}$$ $$g_2 (x) = 1$$ $$x_0 =1$$ $$x_2=5$$

So the area is equal to:

\begin{align} &= \int_1^5 \int_{e^{-x}}^1 1 \ dy \ dx \\ & = \int_1^5 \left. y \right |_{e{-x}}^1 \ dx \\ & = \int_1^5 (1-e{-x}) \ dx \\ & = \left. (x + e^{-x}) \right |_1^5 \\ & = (5+e^{-5}) - (1+e^{-1}) \\ & = 3.63886. \end{align}

Thus, the average value of $$f(x,y) = \frac{1}{{x^3}y}$$ is

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

Example 3

Bob owns an uneven region of land that could be described by the curves $$y=x^3 + 2x$$, $$y=0$$, $$x=0$$, and $$x=7$$. The height above sea level of the land at each point is a function of $$f(x,y)= \frac{1}{x}$$. What is the average height above sea level of Bob's obtained land?

Solution

$\text{Average Height } \ = \ \dfrac{\text{Total Volume}}{\text{Total Area}}$

First, find the total volume:

$\text{Volume} \ = \ \int_{x_0}^{x_1} \int_{g_1 (x)}^{g_2 (x)} f(x,y) \ dy \ dx.$

Where $$dy \(dx$$ is equal to $$dA$$ and $$f(x,y)$$ is equal to the height. So the integral equals a volume, since Area x Height = Volume.

Also, $$x_0 \ \text{and} \ x_1$$ are the left and right bounds and $$g_1 (x) \ \text{and} \ g_2 (x)$$ are the lower and upper bounds, respectively, as shown in example one.

$$f(x,y) = \frac{1}{x}$$ $$g_1 (x) = 0$$ $$g_2 (x) = x^3 + 2x$$ $$x_0 = 0$$ $$x_2=7$$

\begin{align} \text{Volume } &= \int_0^7 \int_0^{x^3 + 2x} \frac{1}{x} \ dy \ dx \\ & = \int_0^7 \left. \frac{y}{x} \right|_0^{x^3 +2x} \ dx \\ & = \int_0^7 \frac{x^3 +2x}{x} - \frac {0}{x} \ dx \\ & = \int_0^7 x^2 +2 \ dx \\ & = \left. {\frac{x^3}{3} + 2x} \right|_0^7 \\ & = \frac{7^3}{3} + 2(7) \\ & = 128.333 \end{align}

Then to find the average value of $$f(x,y)$$, we must divide this value by the total area of the region.

Find the total area:

$\int_{x_0}^{x_1} \int_{g_1 (x)}^{g_2 (x)} f(x,y) \ dy \ dx$

$$f(x,y) = 1$$ $$g_1 (x) = 0$$ $$g_2 (x) = x^3 + 2x$$ $$x_0 = 0$$ $$x_2=7$$

\begin{align} \text{Area } &= \int_0^7 \int_0^{x^3 +2x} 1 \ dy \ dx \\ & = \int_0^7 \left. y \right|_0^{x^3 +2x} \ dx \\ & = \int_0^7 x^3 + 2x \ dx \\ & = \left. \frac{x^4}{4} + x^2 \right|_0^7 \\ & = \frac{7^4}{4} + 7^2 \\ &= 649.25 \end{align}

Thus, the average height above sea level of Bob's land is:

\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

• (UCD)
• Integrated by Justin Marshall.