maths.free › Multivariable Calculus › 5. Multiple Integration › Double Integrals over Rectangular Regions

Double Integrals over Rectangular Regions

Recognize when a function of two variables is integrable over a rectangular region.

Volumes and Double Integrals

We begin by considering the space above a rectangular region R. Consider a continuous function \(f(x,y)\ge 0\) of two variables defined on the closed rectangle R:

\[R=[a,b]\ \times \ [c,d]=\{(x,y)\in {ℝ}^{2}|a\le x\le b,c\le y\le d\}\]

Here \([a,b]\ \times \ [c,d]\) denotes the Cartesian product of the two closed intervals \([a,b]\) and \([c,d].\) It consists of rectangular pairs \((x,y)\) such that \(a\le x\le b\) and \(c\le y\le d.\) The graph of \(f\) represents a surface above the \(xy\)-plane with equation \(z=f(x,y)\) where \(z\) is the height of the surface at the point \((x,y).\) Let \(S\) be the solid that lies above \(R\) and under the graph of \(f\) (). The base of the solid is the rectangle \(R\) in the \(xy\)-plane. We want to find the volume \(V\) of the solid \(S.\)

We divide the region \(R\) into small rectangles \({R}_{ij},\) each with area \(\text{\Delta }A\) and with sides \(\text{\Delta }x\) and \(\text{\Delta }y\) (). We do this by dividing the interval \([a,b]\) into \(m\) subintervals and dividing the interval \([c,d]\) into \(n\) subintervals. Hence \(\text{\Delta }x=\frac{b-a}{m},\) \(\text{\Delta }y=\frac{d-c}{n},\) and \(\text{\Delta }A=\text{\Delta }x\text{\Delta }y.\)

The volume of a thin rectangular box above \({R}_{ij}\) is \(f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A,\) where \(({x}_{ij}^{*},{y}_{ij}^{*})\) is an arbitrary sample point in each \({R}_{ij}\) as shown in the following figure.

Using the same idea for all the subrectangles, we obtain an approximate volume of the solid \(S\) as \(V\approx \sum _{i=1}^{m}\sum _{j=1}^{n}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A.\) This sum is known as a double Riemann sum and can be used to approximate the value of the volume of the solid. Here the double sum means that for each subrectangle we evaluate the function at the chosen point, multiply by the area of each rectangle, and then add all the results.

As we have seen in the single-variable case, we obtain a better approximation to the actual volume if m and n become larger.

\[V=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A\ \text{or}\ V=\underset{\text{\Delta }x,\text{\Delta }y\to 0}{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A.\]

Note that the sum approaches a limit in either case and the limit is the volume of the solid with the base R. Now we are ready to define the double integral.

Condensed — the full section is in OpenStax Calculus Volume 3.

Properties of Double Integrals

The properties of double integrals are very helpful when computing them or otherwise working with them. We list here six properties of double integrals. Properties 1 and 2 are referred to as the linearity of the integral, property 3 is the additivity of the integral, property 4 is the monotonicity of the integral, and property 5 is used to find the bounds of the integral. Property 6 is used if \(f(x,y)\) is a product of two functions \(g(x)\) and \(h(y).\)

These properties are used in the evaluation of double integrals, as we will see later. We will become skilled in using these properties once we become familiar with the computational tools of double integrals. So let’s get to that now.

Iterated Integrals

So far, we have seen how to set up a double integral and how to obtain an approximate value for it. We can also imagine that evaluating double integrals by using the definition can be a very lengthy process if we choose larger values for \(m\) and \(n.\) Therefore, we need a practical and convenient technique for computing double integrals. In other words, we need to learn how to compute double integrals without employing the definition that uses limits and double sums.

The basic idea is that the evaluation becomes easier if we can break a double integral into single integrals by integrating first with respect to one variable and then with respect to the other. The key tool we need is called an iterated integral.

The notation \(\int _{a}^{b}[\int _{c}^{d}f(x,y)dy]dx\) means that we integrate \(f(x,y)\) with respect to y while holding x constant. Similarly, the notation \(\int _{c}^{d}[\int _{a}^{b}f(x,y)dx]dy\) means that we integrate \(f(x,y)\) with respect to x while holding y constant. The fact that double integrals can be split into iterated integrals is expressed in Fubini’s theorem. Think of this theorem as an essential tool for evaluating double integrals.

The double integration in this example is simple enough to use Fubini’s theorem directly, allowing us to convert a double integral into an iterated integral. Consequently, we are now ready to convert all double integrals to iterated integrals and demonstrate how the properties listed earlier can help us evaluate double integrals when the function \(f(x,y)\) is more complex. Note that the order of integration can be changed (see ).

Condensed — the full section is in OpenStax Calculus Volume 3.

Applications of Double Integrals

Double integrals are very useful for finding the area of a region bounded by curves of functions. We describe this situation in more detail in the next section. However, if the region is a rectangular shape, we can find its area by integrating the constant function \(f(x,y)=1\) over the region \(R.\)

This definition makes sense because using \(f(x,y)=1\) and evaluating the integral make it a product of length and width. Let’s check this formula with an example and see how this works.

Example

Try it.

Find the area of the region \(R=\{(x,y)|0\le x\le 3,0\le y\le 2\}\) by using a double integral, that is, by integrating 1 over the region \(R.\)

Solution

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:

\[A(R)=\int _{0}^{2}\ \int _{0}^{3}1dx\ dy=\int _{0}^{2}[{x|}_{0}^{3}]dy=\int _{0}^{2}3dy=3\int _{0}^{2}dy=3{y|}_{0}^{2}=3(2)=6.\]

We have already seen how double integrals can be used to find the volume of a solid bounded above by a function \(f(x,y)\) over a region \(R\) provided \(f(x,y)\ge 0\) for all \((x,y)\) in \(R.\) Here is another example to illustrate this concept.

Example

Try it.

Find the volume \(V\) of the solid \(S\) that is bounded by the elliptic paraboloid \(2{x}^{2}+{y}^{2}+z=27,\) the planes \(x=3\) and \(y=3,\) and the three coordinate planes.

Solution

First notice the graph of the surface \(z=27-2{x}^{2}-{y}^{2}\) in (a) and above the square region \({R}_{1}=[-3,3]\ \times \ [-3,3].\) However, we need the volume of the solid bounded by the elliptic paraboloid \(2{x}^{2}+{y}^{2}+z=27,\) the planes \(x=3\) and \(y=3,\) and the three coordinate planes.

Now let’s look at the graph of the surface in (b). We determine the volume V by evaluating the double integral over \({R}_{2}\text{:}\)

\[\begin{array}{lllll}V & =\underset{R}{∬}z\ dA=\underset{R}{∬}(27-2{x}^{2}-{y}^{2})dA & & & \\ & =\int _{y=0}^{y=3}\ \int _{x=0}^{x=3}(27-2{x}^{2}-{y}^{2})dx\ dy & & & \text{Convert to iterated integral.} \\ & =\int _{y=0}^{y=3}{[27x-\frac{2}{3}{x}^{3}-{y}^{2}x]|}_{x=0}^{x=3}dy & & & \text{Integrate with respect to}\ x. \\ & =\int _{y=0}^{y=3}(63-3{y}^{2})dy=63y-{{y}^{3}|}_{y=0}^{y=3}=162. & & & \end{array}\]

Recall that we defined the average value of a function of one variable on an interval \([a,b]\) as

\[{f}_{\text{ave}}=\frac{1}{b-a}\int _{a}^{b}f(x)dx.\]

Condensed — the full section is in OpenStax Calculus Volume 3.

Key Concepts

  • We can use a double Riemann sum to approximate the volume of a solid bounded above by a function of two variables over a rectangular region. By taking the limit, this becomes a double integral representing the volume of the solid.
  • Properties of double integral are useful to simplify computation and find bounds on their values.
  • We can use Fubini’s theorem to write and evaluate a double integral as an iterated integral.
  • Double integrals are used to calculate the area of a region, the volume under a surface, and the average value of a function of two variables over a rectangular region.

Key Equations

Double integral\(\underset{R}{∬}f(x,y)dA=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A\)
Iterated integral\({\int }_{a}^{b}{\int }_{c}^{d}f(x,y)dx\ dy={\int }_{a}^{b}[{\int }_{c}^{d}f(x,y)dy]dx\)
or
\({\int }_{c}^{d}{\int }_{b}^{a}f(x,y)dx\ dy={\int }_{c}^{d}[{\int }_{a}^{b}f(x,y)dx]dy\)
Average value of a function of two variables\({f}_{\text{ave}}=\frac{1}{\text{Area}\ R}\underset{R}{∬}f(x,y)dx\ dy\)

Double Integrals over Rectangular Regions

In the following exercises, use the midpoint rule with \(m=4\) and \(n=2\) to estimate the volume of the solid bounded by the surface \(z=f(x,y),\) the vertical planes \(x=1,\) \(x=2,\) \(y=1,\) and \(y=2,\) and the horizontal plane \(z=0.\)

In the following exercises, estimate \({∬}_{R}f\left(x,y\right)dA\) by using a Riemann sum with \(m=n=2\) and the sample points to be the lower left corners of the subrectangles of the partition.

In the following exercises, calculate the integrals by interchanging the order of integration.

In the following exercises, evaluate the iterated integrals by choosing the order of integration.

In the following exercises, find the average value of the function over the given rectangles.

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

In the following exercises, use property v. of double integrals and the answer from the preceding exercise to show that the following inequalities are true.

  1. Determine the explicit form of the function f.
  2. Find the volume of the solid under the surface \(z=f(x,y)\) and above the region R.
  3. Find the average value of the function f on R.
  4. Use a computer algebra system (CAS) to plot \(z=f(x,y)\) and \(z={f}_{\text{ave}}\) in the same system of coordinates.
  1. Find the volume of the solids \({S}_{n}\) under the surfaces \(z={f}_{n}(x,y)\) and above the region R.
  2. Determine the limit of the volumes of the solids \({S}_{n}\) as n increases without bound.

Condensed — the full section is in OpenStax Calculus Volume 3.

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

  1. Consider the function \(z=f(x,y)=3{x}^{2}-y\) over the rectangular region \(R=[0,2]\ \times \ [0,2]\) ().

    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 \((2,2)\) () 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.
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
    1. As we can see, the function \(z=f(x,y)=3{x}^{2}-{y}^{}\) is both above and below the plane. To find the signed volume of S, we need to divide the region R into small rectangles \({R}_{ij},\) each with area \(\text{\Delta }A\) and with sides \(\text{\Delta }x\) and \(\text{\Delta }y,\) and choose \(({x}_{ij}^{*},{y}_{ij}^{*})\) as sample points in each \({R}_{ij}.\) Hence, a double integral is set up as
      \[V=\underset{R}{∬}(3{x}^{2}-y)dA=\underset{m,n\to \infty }{\text{lim}}\sum _{i=1}^{m}\sum _{j=1}^{n}[3{({x}_{ij}^{*})}^{2}-{y}_{ij}^{*}]\text{\Delta }A.\]
    2. Approximating the signed volume using a Riemann sum with \(m=n=2\) we have \(\text{\Delta }A=\text{\Delta }x\text{\Delta }y=1\ \times \ 1=1.\) Also, the sample points are (1, 1), (2, 1), (1, 2), and (2, 2) as shown in the following figure.

      Hence,
      \[\begin{array}{ll}V & =\sum _{i=1}^{2}\sum _{j=1}^{2}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A \\ & =\sum _{i=1}^{2}(f({x}_{i1}^{*},{y}_{i1}^{*})+f({x}_{i2}^{*},{y}_{i2}^{*}))\text{\Delta }A \\ & =f({x}_{11}^{*},{y}_{11}^{*})\text{\Delta }A+f({x}_{21}^{*},{y}_{21}^{*})\text{\Delta }A+f({x}_{12}^{*},{y}_{12}^{*})\text{\Delta }A+f({x}_{22}^{*},{y}_{22}^{*})\text{\Delta }A \\ & =f(1,1)(1)+f(2,1)(1)+f(1,2)(1)+f(2,2)(1) \\ & =\left(3-1\right)(1)+(12-1)(1)+(3-2)(1)+(12-2)(1) \\ & =2+11+1+10=24\end{array}\]
    3. Approximating the signed volume using a Riemann sum with \(m=n=2,\) we have \(\text{\Delta }A=\text{\Delta }x\text{\Delta }y=1\ \times \ 1=1.\) 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
      \[\begin{array}{ll}V & =\sum _{i=1}^{2}\sum _{j=1}^{2}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A \\ & =f({x}_{11}^{*},{y}_{11}^{*})\text{\Delta }A+f({x}_{21}^{*},{y}_{21}^{*})\text{\Delta }A+f({x}_{12}^{*},{y}_{12}^{*})\text{\Delta }A+f({x}_{22}^{*},{y}_{22}^{*})\text{\Delta }A \\ & =f(\frac{1}{2},\frac{1}{2})(1)+f(\frac{3}{2},\frac{1}{2})(1)+f(\frac{1}{2},\frac{3}{2})(1)+f(\frac{3}{2},\frac{3}{2})(1) \\ & =(\frac{3}{4}-\frac{1}{2})(1)+(\frac{27}{4}-\frac{1}{2})(1)+(\frac{3}{4}-\frac{3}{2})(1)+(\frac{27}{4}-\frac{3}{2})(1) \\ & =\frac{1}{4}+\frac{25}{4}+(-\frac{3}{4})+\frac{21}{4}=\frac{44}{4}=11\end{array}\]

  2. Use the same function \(z=f(x,y)=3{x}^{2}-{y}^{}\) over the rectangular region \(R=[0,2]\ \times \ [0,2].\)

    Divide R into the same four squares with \(m=n=2,\) and choose the sample points as the upper left corner point of each square \((0,1),(1,1),(0,2),\) and \((1,2)\) () to approximate the signed volume of the solid S that lies above \(R\) and “under” the graph of \(f.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(V=\sum _{i=1}^{2}\sum _{j=1}^{2}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A=0\)

  3. Use Fubini’s theorem to compute the double integral \(\underset{R}{∬}f(x,y)dA\) where \(f(x,y)=x\) and \(R=[0,2]\ \times \ [0,1].\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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.

    \[\begin{array}{ll}\underset{R}{∬}f(x,y)dA & =\underset{R}{∬}f(x,y)dx\ dy \\ & ={\int }_{y=0}^{y=1}{\int }_{x=0}^{x=2}x\ dx\ dy \\ & ={\int }_{y=0}^{y=1}[{\frac{{x}^{2}}{2}|}_{x=0}^{x=2}]dy \\ & ={\int }_{y=0}^{y=1}2dy={2y|}_{y=0}^{y=1}=2.\end{array}\]

  4. Evaluate the double integral \(\underset{R}{∬}(xy-3x{y}^{2})dA\) where \(R=\{(x,y)|0\le x\le 2,1\le y\le 2\}.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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.

    \[\begin{array}{llll} \\ \\ \\ \\ \underset{R}{∬}(xy-3x{y}^{2})dA & & & \\ =\underset{R}{∬}xy\ dA+\underset{R}{∬}(-3x{y}^{2})dA & & & \text{Property i: Integral of a sum is the sum of the integrals.} \\ ={\int }_{y=1}^{y=2}{\int }_{x=0}^{x=2}xy\ dx\ dy-{\int }_{y=1}^{y=2}{\int }_{x=0}^{x=2}3x{y}^{2}dx\ dy & & & \text{Convert double integrals to iterated integrals.} \\ ={\int }_{y=1}^{y=2}{(\frac{{x}^{2}}{2}y)|}_{x=0}^{x=2}dy-3{\int }_{y=1}^{y=2}{(\frac{{x}^{2}}{2}{y}^{2})|}_{x=0}^{x=2}dy & & & \text{Integrate with respect to}\ x,\text{holding}\ y\ \text{constant.} \\ ={\int }_{y=1}^{y=2}2y\ dy-{\int }_{y=1}^{y=2}6{y}^{2}dy & & & \text{Property ii: Placing the constant before the integral.} \\ =2{\int }_{1}^{2}y\ dy-6{\int }_{1}^{2}{y}^{2}dy & & & \text{Integrate with respect to}\ y. \\ ={2\frac{{y}^{2}}{2}|}_{1}^{2}-{6\frac{{y}^{3}}{3}|}_{1}^{2} & & & \\ ={{y}^{2}|}_{1}^{2}-{2{y}^{3}|}_{1}^{2} & & & \\ =(4-1)-2(8-1) & & & \\ =3-2(7)=3-14=-11.\end{array}\]

  5. Over the region \(R=\{(x,y)|1\le x\le 3,1\le y\le 2\},\) we have \(2\le {x}^{2}+{y}^{2}\le 13.\) Find a lower and an upper bound for the integral \(\underset{R}{∬}({x}^{2}+{y}^{2})dA.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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 \(R.\)

    \[\begin{array}{lll}{\int }_{1}^{2}{\int }_{1}^{3}2dx\ dy & = & {\int }_{1}^{2}[{2x|}_{1}^{3}]dy={\int }_{1}^{2}2(2)dy={4y|}_{1}^{2}=4(2-1)=4 \\ {\int }_{1}^{2}{\int }_{1}^{3}13dx\ dy & = & {\int }_{1}^{2}[{13x|}_{1}^{3}]dy={\int }_{1}^{2}13(2)dy={26y|}_{1}^{2}=26(2-1)=26.\end{array}\]

    Hence, we obtain \(4\le \underset{R}{∬}({x}^{2}+{y}^{2})dA\le 26.\)

  6. Evaluate the integral \(\underset{R}{∬}{e}^{y}\text{cos}\ x\ dA\) over the region \(R=\{(x,y)|0\le x\le \frac{\pi }{2},0\le y\le 1\}.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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.

    \[\begin{array}{ll}\underset{R}{∬}{e}^{y}\text{cos}\ x\ dA & ={\int }_{0}^{1}{\int }_{0}^{\pi \text{/}2}{e}^{y}\text{cos}\ x\ dx\ dy \\ & =({\int }_{0}^{1}{e}^{y}dy)({\int }_{0}^{\pi \text{/}2}\text{cos}\ x\ dx) \\ & =({{e}^{y}|}_{0}^{1})({\text{sin}\ x|}_{0}^{\pi \text{/}2}) \\ & =e-1.\end{array}\]
    1. Use the properties of the double integral and Fubini’s theorem to evaluate the integral
      \[{\int }_{0}^{1}{\int }_{-1}^{3}(3-x+4y)dy\ dx.\]
    2. Show that \(0\le \underset{R}{∬}\text{sin}\ \pi x\ \text{cos}\ \pi y\ dA\le \frac{1}{32}\) where \(R=\left[0,\frac{1}{4}\right]\times \left[\frac{1}{4},\frac{1}{2}\right].\)
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    a. 26 b. Answers may vary.

  8. Let’s return to the function \(f(x,y)=3{x}^{2}-y\) from , this time over the rectangular region \(R=[0,2]\ \times \ [0,3].\) Use Fubini’s theorem to evaluate \(\underset{R}{∬}f(x,y)dA\) 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.
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    shows how the calculation works in two different ways.

    1. First integrate with respect to y and then integrate with respect to x:
      \[\begin{array}{ll}\underset{R}{∬}f(x,y)dA & ={\int }_{x=0}^{x=2}{\int }_{y=0}^{y=3}(3{x}^{2}-y)dy\ dx \\ & ={\int }_{x=0}^{x=2}(\int _{y=0}^{y=3}(3{x}^{2}-y)dy)dx={\int }_{x=0}^{x=2}[{3{x}^{2}y-\frac{{y}^{2}}{2}|}_{y=0}^{y=3}]dx \\ & ={\int }_{x=0}^{x=2}(9{x}^{2}-\frac{9}{2})dx={3{x}^{3}-\frac{9}{2}x|}_{x=0}^{x=2}=15.\end{array}\]
    2. First integrate with respect to x and then integrate with respect to y:
      \[\begin{array}{ll}\underset{R}{∬}f(x,y)dA & ={\int }_{y=0}^{y=3}{\int }_{x=0}^{x=2}(3{x}^{2}-y)dx\ dy \\ & ={\int }_{y=0}^{y=3}({\int }_{x=0}^{x=2}(3{x}^{2}-y)dx)dy={\int }_{y=0}^{y=3}[{{x}^{3}-xy|}_{x=0}^{x=2}]dy \\ & ={\int }_{y=0}^{y=3}(8-2y)dy={8y-{y}^{2}|}_{y=0}^{y=3}=15.\end{array}\]

  9. Evaluate \({\int }_{y=-3}^{y=2}{\int }_{x=3}^{x=5}(2-3{x}^{2}+{y}^{2})dx\ dy.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(-\frac{1340}{3}\)

  10. Consider the double integral \(\underset{R}{∬}x\ \text{sin}(xy)dA\) over the region \(R=\{(x,y)|0\le x\le \pi ,1\le y\le 2\}\) ().

    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.
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ
    1. We can express \(\underset{R}{∬}x\ \text{sin}(xy)dA\) 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 \(y.\)
      \[\begin{array}{llll} \\ \\ \\ \\ \underset{R}{∬}x\ \text{sin}(xy)dA & & & \\ =\int _{x=0}^{x=\pi }\ \int _{y=1}^{y=2}x\ \text{sin}(xy)dy\ dx & & & \text{Integrate first with respect to}\ y. \\ =\int _{y=1}^{y=2}\ \int _{x=0}^{x=\pi }x\ \text{sin}(xy)dx\ dy & & & \text{Integrate first with respect to}\ x.\end{array}\]
    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 \(du=x\ dy.\) Hence the inner integral is simply \(\int \text{sin}\ u\ du\) and we can change the limits to be functions of x,
      \[\underset{R}{∬}x\ \text{sin}(xy)dA=\int _{x=0}^{x=\pi }\ \int _{y=1}^{y=2}x\ \text{sin}(xy)dy\ dx=\int _{x=0}^{x=\pi }[\int _{u=x}^{u=2x}\text{sin}(u)du]dx.\]
      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
      \[\underset{R}{∬}x\ \text{sin}(xy)dA=\int _{y=1}^{y=2}\ \int _{x=0}^{x=\pi }x\ \text{sin}(xy)dx\ dy=\int _{y=1}^{y=2}[-{\frac{x\ \text{cos}(xy)}{y}|}_{x=0}^{x=\pi }+\frac{1}{y}\int _{x=0}^{x=\pi }\text{cos}(xy)dx]dy.\]
      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.
      \[\begin{array}{ll}\underset{R}{∬}x\ \text{sin}(xy)dA & =\int _{x=0}^{x=\pi }\ \int _{y=1}^{y=2}x\ \text{sin}(xy)dy\ dx \\ & =\int _{x=0}^{x=\pi }[\int _{u=x}^{u=2x}\text{sin}(u)du]dx=\int _{x=0}^{x=\pi }[{\text{-}\text{cos}\ u|}_{u=x}^{u=2x}]dx=\int _{x=0}^{x=\pi }(\text{-}\text{cos}\ 2x+\text{cos}\ x)dx \\ & ={-\frac{1}{2}\text{sin}\ 2x+\text{sin}\ x|}_{x=0}^{x=\pi }=0.\end{array}\]

  11. Evaluate the integral \(\underset{R}{∬}x{e}^{xy}dA\) where \(R=[0,1]\ \times \ [0,\text{ln}\ 5].\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(\frac{4-\text{ln}\ 5}{\text{ln}\ 5}\)

  12. Find the area of the region \(R=\{(x,y)|0\le x\le 3,0\le y\le 2\}\) by using a double integral, that is, by integrating 1 over the region \(R.\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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:

    \[A(R)=\int _{0}^{2}\ \int _{0}^{3}1dx\ dy=\int _{0}^{2}[{x|}_{0}^{3}]dy=\int _{0}^{2}3dy=3\int _{0}^{2}dy=3{y|}_{0}^{2}=3(2)=6.\]

  13. Find the volume \(V\) of the solid \(S\) that is bounded by the elliptic paraboloid \(2{x}^{2}+{y}^{2}+z=27,\) the planes \(x=3\) and \(y=3,\) and the three coordinate planes.

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    First notice the graph of the surface \(z=27-2{x}^{2}-{y}^{2}\) in (a) and above the square region \({R}_{1}=[-3,3]\ \times \ [-3,3].\) However, we need the volume of the solid bounded by the elliptic paraboloid \(2{x}^{2}+{y}^{2}+z=27,\) the planes \(x=3\) and \(y=3,\) and the three coordinate planes.

    Now let’s look at the graph of the surface in (b). We determine the volume V by evaluating the double integral over \({R}_{2}\text{:}\)

    \[\begin{array}{lllll}V & =\underset{R}{∬}z\ dA=\underset{R}{∬}(27-2{x}^{2}-{y}^{2})dA & & & \\ & =\int _{y=0}^{y=3}\ \int _{x=0}^{x=3}(27-2{x}^{2}-{y}^{2})dx\ dy & & & \text{Convert to iterated integral.} \\ & =\int _{y=0}^{y=3}{[27x-\frac{2}{3}{x}^{3}-{y}^{2}x]|}_{x=0}^{x=3}dy & & & \text{Integrate with respect to}\ x. \\ & =\int _{y=0}^{y=3}(63-3{y}^{2})dy=63y-{{y}^{3}|}_{y=0}^{y=3}=162. & & & \end{array}\]

  14. Find the volume of the solid bounded above by the graph of \(f(x,y)=xy\ \text{sin}({x}^{2}y)\) and below by the \(xy\)-plane on the rectangular region \(R=[0,1]\ \times \ [0,\pi ].\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(\frac{\pi }{2}\)

  15. The weather map in 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.

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    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 \((m=3\ \text{and}\ n=2),\) as shown in . 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

    \[\text{\Delta }A=\frac{1}{6}(75,000)=12,500.\]

    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 \((x{*}_{11},y{*}_{11})\) the rainfall is 0.08.

    At \((x{*}_{21},y{*}_{21})\) the rainfall is 0.08.

    At \((x{*}_{31},y{*}_{31})\) the rainfall is 0.01.

    At \((x{*}_{12},y{*}_{12})\) the rainfall is 1.70.

    At \((x{*}_{22},y{*}_{22})\) the rainfall is 1.74.

    At \((x{*}_{32},y{*}_{32})\) the rainfall is 3.00.

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

    \[\begin{array}{ll}{f}_{\text{ave}} & =\frac{1}{\text{Area}\ R}\underset{R}{∬}f(x,y)dx\ dy=\frac{1}{75000}\underset{R}{∬}f(x,y)dx\ dy \\ & \approx \frac{1}{75,000}\sum _{i=1}^{3}\sum _{j=1}^{2}f({x}_{ij}^{*},{y}_{ij}^{*})\text{\Delta }A \\ & \approx \frac{1}{75,000}[f({x}_{11}^{*},{y}_{11}^{*})\text{\Delta }A+f({x}_{21}^{*},{y}_{21}^{*})\text{\Delta }A \\ & \ +f({x}_{31}^{*},{y}_{31}^{*})\text{\Delta }A+f({x}_{12}^{*},{y}_{12}^{*})\text{\Delta }A+f({x}_{22}^{*},{y}_{22}^{*})\text{\Delta }A+f({x}_{32}^{*},{y}_{32}^{*})\text{\Delta }A] \\ & \approx \frac{1}{75,000}[0.08+0.08+0.01+1.70+1.74+3.00]\text{\Delta }A \\ & \approx \frac{1}{75,000}[0.08+0.08+0.01+1.70+1.74+3.00]12500 \\ & \approx \frac{5}{30}[0.08+0.08+0.01+1.70+1.74+3.00] \\ & \approx 1.10.\end{array}\]

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

  16. A contour map is shown for a function \(f(x,y)\) on the rectangle \(R=[-3,6]\ \times \ [-1,4].\)

    1. Use the midpoint rule with \(m=3\) and \(n=2\) to estimate the value of \(\underset{R}{∬}f(x,y)dA.\)
    2. Estimate the average value of the function \(f(x,y).\)
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    Answers to both parts a. and b. may vary.

  17. \(f(x,y)=4x+2y+8xy\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    27.

  18. \(f(x,y)=16{x}^{2}+\frac{y}{2}\)

  19. \(f(x,y)=\text{sin}\ x-\text{cos}\ y,\) \(R=[0,\pi ]\ \times \ [0,\pi ]\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    0.

  20. \(f(x,y)=\text{cos}\ x+\text{cos}\ y,\) \(R=[0,\pi ]\ \times \ [0,\frac{\pi }{2}]\)

  21. Use the midpoint rule with \(m=n=2\) to estimate \(\underset{R}{∬}f(x,y)dA,\) where the values of the function f on \(R=[8,10]\ \times \ [9,11]\) are given in the following table.

    y
    x99.51010.511
    89.856.755.6
    8.59.44.585.43.4
    98.74.665.53.4
    9.56.764.55.46.7
    106.86.45.55.76.8
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    21.3.

  22. The values of the function f on the rectangle \(R=[0,2]\ \times \ [7,9]\) are given in the following table. Estimate the double integral \(\underset{R}{∬}f(x,y)dA\) by using a Riemann sum with \(m=n=2.\) Select the sample points to be the upper right corners of the subsquares of R.

    \({y}_{0}=7\)\({y}_{1}=8\)\({y}_{2}=9\)
    \({x}_{0}=0\)10.2210.219.85
    \({x}_{1}=1\)6.739.759.63
    \({x}_{2}=2\)5.627.838.21

  23. The depth of a children’s 4-ft by 4-ft swimming pool, measured at 1-ft intervals, is given in the following table.

    1. Estimate the volume of water in the swimming pool by using a Riemann sum with \(m=n=2.\) Select the sample points using the midpoint rule on \(R=[0,4]\ \times \ [0,4].\)
    2. Approximate the average depth of the swimming pool.
      y
      x01234
      011.522.53
      111.522.53
      211.51.52.53
      3111.522.5
      41111.52
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    a. 28 \({\text{ft}}^{3}\) b. 1.75 ft.

  24. The depth of a 3-ft by 3-ft hole in the ground, measured at 1-ft intervals, is given in the following table.

    1. Estimate the volume of the hole by using a Riemann sum with \(m=n=3\) and the sample points to be the upper left corners of the subsquares of R.
    2. Approximate the average depth of the hole.
      y
      x0123
      066.56.46
      16.577.56.5
      26.56.76.56
      366.555.6

  25. The level curves \(f(x,y)=k\) of the function f are given in the following graph, where k is a constant.

    1. Apply the midpoint rule with \(m=n=2\) to estimate the double integral \(\underset{R}{∬}f(x,y)dA,\) where \(R=[0.2,1]\ \times \ [0,0.8].\)
    2. Estimate the average value of the function f on R.
    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    a. \(0.112\) b. \({f}_{\text{ave}}≃0.175;\) here \(f(0.4,0.2)≃0.1,\) \(f(0.2,0.6)≃-0.2,\) \(f(0.8,0.2)≃0.6,\) and \(f(0.8,0.6)≃0.2.\)

  26. The level curves \(f(x,y)=k\) of the function f are given in the following graph, where k is a constant.

    1. Apply the midpoint rule with \(m=n=2\) to estimate the double integral \(\underset{R}{∬}f(x,y)dA,\) where \(R=[0.1,0.5]\ \times \ [0.1,0.5].\)
    2. Estimate the average value of the function f on R.

  27. The solid lying under the surface \(z=\sqrt{4-{y}^{2}}\) and above the rectangular region \(R=[0,2]\ \times \ [0,2]\) is illustrated in the following graph. Evaluate the double integral \(\underset{R}{∬}f(x,y)dA,\) where \(f(x,y)=\sqrt{4-{y}^{2}},\) by finding the volume of the corresponding solid.

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(2\pi .\)

  28. The solid lying under the plane \(z=y+4\) and above the rectangular region \(R=[0,2]\ \times \ [0,4]\) is illustrated in the following graph. Evaluate the double integral \(\underset{R}{∬}f(x,y)dA,\) where \(f(x,y)=y+4,\) by finding the volume of the corresponding solid.

  29. \(\int _{-1}^{1}(\int _{-2}^{2}(2x+3y+5)dx)dy\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    40.

  30. \(\int _{0}^{2}(\int _{0}^{1}(x+2{e}^{y}-3)dx)dy\)

  31. \(\int _{1}^{27}(\int _{1}^{2}(\sqrt[3]{x}+\sqrt[3]{y})dy)dx\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(\frac{81}{2}+39\sqrt[3]{2}.\)

  32. \(\int _{1}^{16}(\int _{1}^{8}(\sqrt[4]{x}+2\sqrt[3]{y})dy)dx\)

  33. \(\int _{\text{ln}\ 2}^{\text{ln}\ 3}(\int _{0}^{l}{e}^{x+y}dy)dx\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(e-1.\)

  34. \(\int _{0}^{2}(\int _{0}^{1}{3}^{x+y}dy)dx\)

  35. \(\int _{1}^{6}(\int _{2}^{9}\frac{\sqrt{y}}{{x}^{2}}dy)dx\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \(15-\frac{10\sqrt{2}}{9}.\)

  36. \(\int _{1}^{9}(\int _{4}^{2}\frac{\sqrt{x}}{{y}^{2}}dy)dx\)

  37. \(\int _{0}^{\pi }\ \int _{0}^{\pi \text{/}2}\text{sin}(2x)\text{cos}(3y)dx\ dy\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    0.

  38. \(\int _{\pi \text{/}12}^{\pi \text{/}8}\ \int _{\pi \text{/}4}^{\pi \text{/}3}[\text{cot}\ x+\text{tan}(2y)]dx\ dy\)

  39. \(\int _{1}^{e}\ \int _{1}^{e}[\frac{1}{x}\text{sin}(\text{ln}\ x)+\frac{1}{y}\text{cos}(\text{ln}\ y)]dx\ dy\)

    ಉತ್ತರವನ್ನು ತಿಳಿಸಿ

    \((e-1)(1+\text{sin}\ 1-\text{cos}\ 1).\)

  40. \(\int _{1}^{e}\ \int _{1}^{e}\frac{\text{sin}(\text{ln}\ x)\text{cos}(\text{ln}\ y)}{xy}dx\ dy\)

Symbols used here

\sum_{k=1}^{n} a_k
summation
Add a_k for k = 1 up to n.
\int f(x)\,dx,\ \int_a^b
integral
Antiderivative (indefinite) or signed area from a to b (definite).
\infty
infinity
Not a number: "grows without bound" in limits and intervals.
\pi
pi
Ratio of a circle's circumference to its diameter, 3.14159…
x \in A,\ A \subseteq B
element of, subset
x belongs to A; every element of A is in B.
A \cup B,\ A \cap B,\ A \setminus B
union, intersection, difference
In either; in both; in A but not B.
f'(x),\ \frac{dy}{dx}
derivative
Instantaneous rate of change; slope of the graph.
i
imaginary unit
i² = −1.
\approx
approximately equal
Equal to the precision shown, not exactly.
\leq,\ \geq
less/greater than or equal
Inequalities that allow equality; < and > exclude it.
\sup,\ \inf
supremum, infimum
Least upper bound, greatest lower bound.
\frac{\partial f}{\partial x}
partial derivative
Derivative with respect to x, holding the other variables fixed.
\nabla f
gradient (nabla, del)
Vector of partial derivatives; points uphill.
\iint,\ \oint
double / contour integral
Integral over a region of the plane; integral around a closed curve.
\mathbf{v},\ \vec{v}
vector
A quantity with magnitude and direction; a column of numbers.
\mathbf{u} \cdot \mathbf{v},\ \|\mathbf{v}\|
dot product, norm
Σ u_i v_i; the length of v, √(v·v).

How to: Double Integrals over Rectangular Regions

  1. Recognize when a function of two variables is integrable over a rectangular region.
  2. Recognize and use some of the properties of double integrals.
  3. Evaluate a double integral over a rectangular region by writing it as an iterated integral.
  4. Use a double integral to calculate the area of a region, volume under a surface, or average value of a function over a plane region.
  5. Set up a double integral for finding the value of the signed volume of the solid
  6. Divide
  7. Divide
  8. As we can see, the function

Questions people ask

What is a partial derivative?

The ordinary derivative with respect to one variable while every other variable is frozen — the slope of the surface in one coordinate direction.

What does the gradient point at?

Uphill: the direction of steepest increase, with length equal to that steepest slope. It is perpendicular to the level curves.

ನಿಮ್ಮದೇ ಆದದ್ದನ್ನು ಪ್ರಯತ್ನಿಸಿ

Parts of this page are adapted from OpenStax Calculus Volume 3 (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

ಇನ್ನಷ್ಟು Multivariable Calculus

Functions of several variablesPartial derivatives and the gradientOptimisation in several variablesDouble and triple integralsVector fields, line integrals and the big theorems