Double Integrals over Rectangular Regions

Consider the function $z=f(x,y)=3x^2-y$

over the rectangular region $R=[0,2]\times [0,2]$

([link]).

1. Set up a double integral for finding the value of the signed volume of the solid S that lies above $R$

   and “under” the graph of

   $f.$
2. Divide R into four squares with $m=n=2,$

   and choose the sample point as the upper right corner point of each square

   $(1,1),(2,1),(1,2),$

   and

   $(2,2)$

   ([link]) to approximate the signed volume of the solid S that lies above

   $R$

   and “under” the graph of

   $f.$
3. Divide R into four squares with $m=n=2,$

   and choose the sample point as the midpoint of each square:

   $(1\text{/}2,1\text{/}2),(3\text{/}2,1\text{/}2),(1\text{/}2,3\text{/}2),\text{and}(3\text{/}2,3\text{/}2)$

   to approximate the signed volume.

   ---

   

1. As we can see, the function $z=f(x,y)=3x^2-y^{}$

   is above the plane. To find the signed volume of S, we need to divide the region R into small rectangles

   $R_{ij},$

   each with area

   $\text{Δ}A$

   and with sides

   $\text{Δ}x$

   and

   $\text{Δ}y,$

   and choose

   $(x_{ij}^{*},y_{ij}^{*})$

   as sample points in each

   $R_{ij}.$

   Hence, a double integral is set up as

   ---

   $V={\displaystyle \underset{R}{\iint }\left(3x^2-y\right)dA}=\underset{m,n\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^m{\displaystyle \sum _{j=1}^n\left[3{\left(x_{ij}^{*}\right)}^2-y_{ij}^{*}\right]\text{Δ}A}}.$
2. Approximating the signed volume using a Riemann sum with $m=n=2$

   we have

   $\text{Δ}A=\text{Δ}x\text{Δ}y=1\times 1=1.$

   Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.

   ---

   

   ---

   Hence,

   ---

   $\begin{array}{cc}\hfill V& ={\displaystyle \sum _{i=1}^2{\displaystyle \sum _{j=1}^{2}f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A}}\hfill \\ & ={\displaystyle \sum _{i=1}^2(f(x_{i1}^{*},y_{i1}^{*})+}f(x_{i2}^{*},y_{i2}^{*}))\text{Δ}A\hfill \\ & =f(x_{11}^{*},y_{11}^{*})\text{Δ}A+f(x_{21}^{*},y_{21}^{*})\text{Δ}A+f(x_{12}^{*},y_{12}^{*})\text{Δ}A+f(x_{22}^{*},y_{22}^{*})\text{Δ}A\hfill \\ & =f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1)\hfill \\ & =(3-1)(1)+(12-1)(1)+(3-2)(1)+(12-2)(1)\hfill \\ & =2+11+1+10=24.\hfill \end{array}$
3. Approximating the signed volume using a Riemann sum with $m=n=2,$

   we have

   $\text{Δ}A=\text{Δ}x\text{Δ}y=1\times 1=1.$

   In this case the sample points are (1/2, 1/2), (3/2, 1/2), (1/2, 3/2),

   ---

   and (3/2, 3/2).

   ---

   Hence

   ---

   $\begin{array}{cc}\hfill V& ={\displaystyle \sum _{i=1}^2{\displaystyle \sum _{j=1}^{2}f(x_{ij}^{*},y_{ij}^{*})\text{Δ}A}}\hfill \\ & =f(x_{11}^{*},y_{11}^{*})\text{Δ}A+f(x_{21}^{*},y_{21}^{*})\text{Δ}A+f(x_{12}^{*},y_{12}^{*})\text{Δ}A+f(x_{22}^{*},y_{22}^{*})\text{Δ}A\hfill \\ & =f(1\text{/}2,1\text{/}2)(1)+f(3\text{/}2,1\text{/}2)(1)+f(1\text{/}2,3\text{/}2)(1)+f(3\text{/}2,3\text{/}2)(1)\hfill \\ & =\left(\frac{3}{4}-\frac{1}{4}\right)(1)+\left(\frac{27}{4}-\frac{1}{2}\right)(1)+\left(\frac{3}{4}-\frac{3}{2}\right)(1)+\left(\frac{27}{4}-\frac{3}{2}\right)(1)\hfill \\ & =\frac{2}{4}+\frac{25}{4}+\left(-\frac{3}{4}\right)+\frac{21}{4}=\frac{45}{4}=11.\hfill \end{array}$

Notice that the approximate answers differ due to the choices of the sample points. In either case, we are introducing some error because we are using only a few sample points. Thus, we need to investigate how we can achieve an accurate answer.

Use Fubini’s theorem to compute the double integral ${\displaystyle \underset{R}{\iint }f(x,y)dA}$

where $f(x,y)=x$

and $R=[0,2]\times [0,1].$

Fubini’s theorem offers an easier way to evaluate the double integral by the use of an iterated integral. Note how the boundary values of the region R become the upper and lower limits of integration.

Evaluate the double integral ${\displaystyle \underset{R}{\iint }\left(xy-3x{y}^2\right)}dA$

where $R=\left\{\left(x,y\right)\backslash |0\le x\le 2,1\le y\le 2\right\}.$

This function has two pieces: one piece is $xy$

and the other is $3x{y}^2.$

Also, the second piece has a constant $3.$

Notice how we use properties i and ii to help evaluate the double integral.

Over the region $R=\left\{\left(x,y\right)\backslash |1\le x\le 3,1\le y\le 2\right\},$

we have $2\le x^2+y^2\le 13.$

Find a lower and an upper bound for the integral ${\displaystyle \underset{R}{\iint }\left(x^2+y^2\right)}dA.$

For a lower bound, integrate the constant function 2 over the region $R.$

For an upper bound, integrate the constant function 13 over the region $R.$

Hence, we obtain $4\le {\displaystyle \underset{R}{\iint }\left(x^2+y^2\right)}dA\le 26.$

Evaluate the integral ${\displaystyle \underset{R}{\iint }{e}^y\text{cos}xdA}$

over the region $R=\left\{\left(x,y\right)\backslash |0\le x\le \frac{\pi }{2},0\le y\le 1\right\}.$

This is a great example for property vi because the function $f(x,y)$

is clearly the product of two single-variable functions $e^y$

and $\text{cos}x.$

Thus we can split the integral into two parts and then integrate each one as a single-variable integration problem.

Let’s return to the function $f(x,y)=3x^2-y$

from [link], this time over the rectangular region $R=[0,2]\times [0,3].$

Use Fubini’s theorem to evaluate ${\displaystyle \underset{R}{\iint }f(x,y)dA}$

in two different ways:

1. First integrate with respect to y and then with respect to x;
2. First integrate with respect to x and then with respect to y.

[link] shows how the calculation works in two different ways.

1. First integrate with respect to y and then integrate with respect to x:

   ---

   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f(x,y)dA}& ={\displaystyle {\int }_{x=0}^{x=2}{\displaystyle {\int }_{y=0}^{y=3}(3x^2-y)dydx}}\hfill \\ & ={\displaystyle {\int }_{x=0}^{x=2}\left({\displaystyle \underset{y=0}{\overset{y=3}{\int }}(3x^2-y)dy}\right)}dx={\displaystyle {\int }_{x=0}^{x=2}\left[{3x^{2}y-\frac{y^2}{2}\backslash |}_{y=0}^{y=3}\right]dx}\hfill \\ & ={\displaystyle {\int }_{x=0}^{x=2}\left(9x^2-\frac{9}{2}\right)}dx={3x^3-\frac{9}{2}x\backslash |}_{x=0}^{x=2}=15.\hfill \end{array}$
2. First integrate with respect to x and then integrate with respect to y:

   ---

   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }f(x,y)dA}& ={\displaystyle {\int }_{y=0}^{y=3}{\displaystyle {\int }_{x=0}^{x=2}(3x^2-y)dxdy}}\hfill \\ & ={\displaystyle {\int }_{y=0}^{y=3}\left({\displaystyle {\int }_{x=0}^{x=2}(3x^2-y)dx}\right)}dy={\displaystyle {\int }_{y=0}^{y=3}\left[{x^3-xy\backslash |}_{x=0}^{x=2}\right]dy}\hfill \\ & ={\displaystyle {\int }_{y=0}^{y=3}\left(8-2y\right)dy}={8y-y^2\backslash |}_{y=0}^{y=3}=15.\hfill \end{array}$

With either order of integration, the double integral gives us an answer of 15. We might wish to interpret this answer as a volume in cubic units of the solid $S$

below the function $f(x,y)=3x^2-y$

over the region $R=[0,2]\times [0,3].$

However, remember that the interpretation of a double integral as a (non-signed) volume works only when the integrand $f$

is a nonnegative function over the base region $R.$

Consider the double integral ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}$

over the region $R=\left\{\left(x,y\right)\backslash |0\le x\le 3,0\le y\le 2\right\}$

([link]).

1. Express the double integral in two different ways.
2. Analyze whether evaluating the double integral in one way is easier than the other and why.
3. Evaluate the integral.

   ---

   

1. We can express ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}$

   in the following two ways: first by integrating with respect to

   $y$

   and then with respect to

   $x;$

   second by integrating with respect to

   $x$

   and then with respect to

   $y.$

   ---

   $\begin{array}{}\\ \\ \\ \\ {\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}\hfill & & & \\ ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}\hfill & & & \text{Integrate first with respect to}y.\hfill \\ ={\displaystyle \underset{y=1}{\overset{y=2}{\int }}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}x\text{sin}(xy)dxdy}}\hfill & & & \text{Integrate first with respect to}x.\hfill \end{array}$
2. If we want to integrate with respect to y first and then integrate with respect to $x,$

   we see that we can use the substitution

   $u=xy,$

   which gives

   $du=xdy.$

   Hence the inner integral is simply

   ${\displaystyle \int \text{sin}u}du$

   and we can change the limits to be functions of x,

   ---

   ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\displaystyle \underset{u=x}{\overset{u=2x}{\int }}\text{sin}(u)du}\right]}dx.$

   ---

   However, integrating with respect to

   $x$

   first and then integrating with respect to

   $y$

   requires integration by parts for the inner integral, with

   $u=x$

   and

   $dv=\text{sin}(xy)dx.$

   ---

   Then

   $du=dx$

   and

   $v=-\frac{\text{cos}(xy)}{y},$

   so

   ---

   ${\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}={\displaystyle \underset{y=1}{\overset{y=2}{\int }}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}x\text{sin}(xy)dxdy}}={\displaystyle \underset{y=1}{\overset{y=2}{\int }}\left[-{\frac{x\text{cos}(xy)}{y}\backslash |}_{x=0}^{x=\pi }+\frac{1}{y}{\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\text{cos}(xy)dx}\right]}dy.$

   ---

   Since the evaluation is getting complicated, we will only do the computation that is easier to do, which is clearly the first method.

3. Evaluate the double integral using the easier way.

   ---

   $\begin{array}{cc}\hfill {\displaystyle \underset{R}{\iint }x\text{sin}(xy)dA}& ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}{\displaystyle \underset{y=1}{\overset{y=2}{\int }}x\text{sin}(xy)dydx}}\hfill \\ & ={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\displaystyle \underset{u=x}{\overset{u=2x}{\int }}\text{sin}(u)du}\right]}dx={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left[{\text{−}\text{cos}u\backslash |}_{u=x}^{u=2x}\right]}dx={\displaystyle \underset{x=0}{\overset{x=\pi }{\int }}\left(\text{−}\text{cos}2x+\text{cos}x\right)dx}\hfill \\ & ={-\frac{1}{2}\text{sin}2x+\text{sin}x\backslash |}_{x=0}^{x=\pi }=0.\hfill \end{array}$

Find the area of the region $R=\left\{\left(x,y\right)\backslash |0\le x\le 3,0\le y\le 2\right\}$

by using a double integral, that is, by integrating 1 over the region $R.$

The region is rectangular with length 3 and width 2, so we know that the area is 6. We get the same answer when we use a double integral:

Find the volume $V$

of the solid $S$

that is bounded by the elliptic paraboloid $2x^2+y^2+z=27,$

the planes $x=3$

and $y=3,$

and the three coordinate planes.

First notice the graph of the surface $z=27-2x^2-y^2$

in [link](a) and above the square region $R_1=[\mathrm{-3},3]\times [\mathrm{-3},3].$

However, we need the volume of the solid bounded by the elliptic paraboloid $2x^2+y^2+z=27,$

the planes $x=3$

and $y=3,$

and the three coordinate planes.

Now let’s look at the graph of the surface in [link](b). We determine the volume V by evaluating the double integral over $R_2\text{:}$

The weather map in [link] shows an unusually moist storm system associated with the remnants of Hurricane Karl, which dumped 4–8 inches (100–200 mm) of rain in some parts of the Midwest on September 22–23, 2010. The area of rainfall measured 300 miles east to west and 250 miles north to south. Estimate the average rainfall over the entire area in those two days.

Place the origin at the southwest corner of the map so that all the values can be considered as being in the first quadrant and hence all are positive. Now divide the entire map into six rectangles $\left(m=2\text{and}n=3\right),$

as shown in [link]. Assume $f(x,y)$

denotes the storm rainfall in inches at a point approximately $x$

miles to the east of the origin and y miles to the north of the origin. Let $R$

represent the entire area of $250\times 300=75000$

square miles. Then the area of each subrectangle is

Assume $(x_{ij}^{*},y_{ij}^{*})$

are approximately the midpoints of each subrectangle $R_{ij}.$

Note the color-coded region at each of these points, and estimate the rainfall. The rainfall at each of these points can be estimated as:

At $\left(x_{11},y_{11}\right)$

the rainfall is 0.08.

At $\left(x_{12},y_{12}\right)$

the rainfall is 0.08.

At $\left(x_{13},y_{13}\right)$

the rainfall is 0.01.

At $\left(x_{21},y_{21}\right)$

the rainfall is 1.70.

At $\left(x_{22},y_{22}\right)$

the rainfall is 1.74.

At $\left(x_{23},y_{23}\right)$

the rainfall is 3.00.

According to our definition, the average storm rainfall in the entire area during those two days was

During September 22–23, 2010 this area had an average storm rainfall of approximately 1.10 inches.