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Mathematics LibreTexts

15.3: Area by Double Integration

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.