Active Calculus · 11: Multiple Integrals

11.1: Double Riemann Sums and Double Integrals over Rectangles

Motivating Questions
  • What is a double Riemann sum?
  • How is the double integral of a continuous function f=f(x,y)f = f(x,y) defined?
  • What are two things the double integral of a function can tell us?

In single-variable calculus, recall that we approximated the area under the graph of a positive function ff on an interval [a,b][a,b] by adding areas of rectangles whose heights are determined by the curve. The general process involved subdividing the interval [a,b][a,b] into smaller subintervals, constructing rectangles on each of these smaller intervals to approximate the region under the curve on that subinterval, then summing the areas of these rectangles to approximate the area under the curve. We will extend this process in this section to its three-dimensional analogs, double Riemann sums and double integrals over rectangles.

Preview Activity 11.1.111.1.1

In this activity we introduce the concept of a double Riemann sum.

  1. Review the concept of the Riemann sum from single-variable calculus. Then, explain how we define the definite integral ∫abf(x) dx\int_a^b f(x) \ dx of a continuous function of a single variable xx on an interval [a,b].[a,b]\text{.} Include a sketch of a continuous function on an interval [a,b][a,b] with appropriate labeling in order to illustrate your definition.
  2. In our upcoming study of integral calculus for multivariable functions, we will first extend the idea of the single-variable definite integral to functions of two variables over rectangular domains. To do so, we will need to understand how to partition a rectangle into subrectangles. Let RR be rectangular domain R={(x,y):0≤x≤6,2≤y≤4}R = \{(x,y) : 0 \leq x \leq 6, 2 \leq y \leq 4\} (we can also represent this domain with the notation [0,6]×[2,4][0,6] \times [2,4]), as pictured in Figure 11.1.111.1.1.
    fig_11_1_rect_domain.svg

    Figure 11.1.111.1.1. Rectangular domain RR with subrectangles.

    To form a partition of the full rectangular region, R,R\text{,} we will partition both intervals [0,6][0,6] and [2,4];[2,4]\text{;} in particular, we choose to partition the interval [0,6][0,6] into three uniformly sized subintervals and the interval [2,4][2,4] into two evenly sized subintervals as shown in Figure 11.1.111.1.1. In the following questions, we discuss how to identify the endpoints of each subinterval and the resulting subrectangles.
    1. Let 0=x0<x1<x2<x3=60=x_0 \lt x_1 \lt x_2 \lt x_3=6 be the endpoints of the subintervals of [0,6][0,6] after partitioning. What is the length Δx\Delta x of each subinterval [xi−1,xi][x_{i-1},x_i] for ii from 1 to 3?
    2. Explicitly identify x0,x_0\text{,} x1,x_1\text{,} x2,x_2\text{,} and x3.x_3\text{.} On Figure 11.1.111.1.1 or your own version of the diagram, label these endpoints.
    3. Let 2=y0<y1<y2=42=y_0 \lt y_1 \lt y_2=4 be the endpoints of the subintervals of [2,4][2,4] after partitioning. What is the length Δy\Delta y of each subinterval [yj−1,yj][y_{j-1},y_j] for jj from 1 to 2? Identify y0,y_0\text{,} y1,y_1\text{,} and y2y_2 and label these endpoints on Figure 11.1.111.1.1.
    4. Let RijR_{ij} denote the subrectangle [xi−1,xi]×[yj−1,yj].[x_{i-1},x_i] \times [y_{j-1},y_j]\text{.} Appropriately label each subrectangle in your drawing of Figure 11.1.111.1.1. How does the total number of subrectangles depend on the partitions of the intervals [0,6][0,6] and [2,4]?[2,4]\text{?}
    5. What is area ΔA\Delta A of each subrectangle?

Double Riemann Sums over Rectangles

For the definite integral in single-variable calculus, we considered a continuous function over a closed, bounded interval [a,b].[a,b]\text{.} In multivariable calculus, we will eventually develop the idea of a definite integral over a closed, bounded region (such as the interior of a circle). We begin with a simpler situation by thinking only about rectangular domains, and will address more complicated domains in Section 11.3.

Let f=f(x,y)f = f(x,y) be a continuous function defined on a rectangular domain R={(x,y):a≤x≤b,c≤y≤d}.R = \{(x,y) : a \leq x \leq b, c \leq y \leq d\}\text{.} As we saw in Preview Activity 11.1.111.1.1, the domain is a rectangle RR and we want to partition RR into subrectangles. We do this by partitioning each of the intervals [a,b][a,b] and [c,d][c,d] into subintervals and using those subintervals to create a partition of RR into subrectangles. In the first activity, we address the quantities and notations we will use in order to define double Riemann sums and double integrals.

Activity 11.1.211.1.2

Let f(x,y)=100−x2−y2f(x,y) = 100 - x^2-y^2 be defined on the rectangular domain R=[a,b]×[c,d].R = [a,b] \times [c,d]\text{.} Partition the interval [a,b][a,b] into four uniformly sized subintervals and the interval [c,d][c,d] into three evenly sized subintervals as shown in Figure 11.1.211.1.2. As we did in Preview Activity 11.1.111.1.1, we will need a method for identifying the endpoints of each subinterval and the resulting subrectangles.

fig_11_1_rect_general.svg

Figure 11.1.211.1.2. Rectangular domain with subrectangles.
  1. Let a=x0<x1<x2<x3<x4=ba=x_0 \lt x_1 \lt x_2 \lt x_3 \lt x_4 =b be the endpoints of the subintervals of [a,b][a,b] after partitioning. Label these endpoints in Figure 11.1.211.1.2.
  2. What is the length Δx\Delta x of each subinterval [xi−1,xi]?[x_{i-1},x_i]\text{?} Your answer should be in terms of aa and b.b\text{.}
  3. Let c=y0<y1<y2<y3=dc=y_0 \lt y_1 \lt y_2 \lt y_3 =d be the endpoints of the subintervals of [c,d][c,d] after partitioning. Label these endpoints in Figure 11.1.211.1.2.
  4. What is the length Δy\Delta y of each subinterval [yj−1,yj]?[y_{j-1},y_j]\text{?} Your answer should be in terms of cc and d.d\text{.}
  5. The partitions of the intervals [a,b][a,b] and [c,d][c,d] partition the rectangle RR into subrectangles. How many subrectangles are there?
  6. Let RijR_{ij} denote the subrectangle [xi−1,xi]×[yj−1,yj].[x_{i-1},x_i] \times [y_{j-1},y_j]\text{.} Label each subrectangle in Figure 11.1.211.1.2.
  7. What is area ΔA\Delta A of each subrectangle?
  8. Now let [a,b]=[0,8][a,b] = [0,8] and [c,d]=[2,6].[c,d] = [2,6]\text{.} Let (x11∗,y11∗)(x_{11}^*,y_{11}^*) be the point in the upper right corner of the subrectangle R11.R_{11}\text{.} Identify and correctly label this point in Figure 11.1.211.1.2. Calculate the product

    f(x11∗,y11∗)ΔA. f(x_{11}^*,y_{11}^*) \Delta A.

    Explain, geometrically, what this product represents.

  9. For each ii and j,j\text{,} choose a point (xij∗,yij∗)(x_{ij}^*,y_{ij}^*) in the subrectangle Ri,j.R_{i,j}\text{.} Identify and correctly label these points in Figure 11.1.211.1.2. Explain what the product
    f(xij∗,yij∗)ΔA f(x_{ij}^*,y_{ij}^*) \Delta A

    represents.

  10. If we were to add all the values f(xij∗,yij∗)ΔAf(x_{ij}^*,y_{ij}^*) \Delta A for each ii and j,j\text{,} what does the resulting number approximate about the surface defined by ff on the domain R?R\text{?} (You don't actually need to add these values.)
  11. Write a double sum using summation notation that expresses the arbitrary sum from part (jj).

Double Riemann Sums and Double Integrals

Now we use the process from the most recent activity to formally define double Riemann sums and double integrals.

Definition 11.1.111.1.1

Let ff be a continuous function on a rectangle R={(x,y):a≤x≤b,c≤y≤d}.R = \{(x,y) : a \leq x \leq b, c \leq y \leq d\}\text{.} A double Riemann sum for ff over RR is created as follows.

  • Partition the interval [a,b][a, b] into mm subintervals of equal length Δx=b−am.\Delta x = \frac{b-a}{m}\text{.} Let x0,x_0\text{,} x1,x_1\text{,} …,\ldots\text{,} xmx_m be the endpoints of these subintervals, where a=x0<x1<x2<⋯<xm=b.a = x_0\lt x_1\lt x_2 \lt \cdots \lt x_m = b\text{.}
  • Partition the interval [c,d][c, d] into nn subintervals of equal length Δy=d−cn.\Delta y = \frac{d-c}{n}\text{.} Let y0,y_0\text{,} y1,y_1\text{,} …,\ldots\text{,} yny_n be the endpoints of these subintervals, where c=y0<y1<y2<⋯<yn=d.c = y_0\lt y_1\lt y_2 \lt \cdots \lt y_n = d\text{.}
  • These two partitions create a partition of the rectangle RR into mnmn subrectangles RijR_{ij} with opposite vertices (xi−1,yj−1)(x_{i-1},y_{j-1}) and (xi,yj)(x_i, y_j) for ii between 11 and mm and jj between 11 and n.n\text{.} These rectangles all have equal area ΔA=Δx⋅Δy.\Delta A = \Delta x \cdot \Delta y\text{.}
  • Choose a point (xij∗,yij∗)(x_{ij}^*, y_{ij}^*) in each rectangle Rij.R_{ij}\text{.} Then, a double Riemann sum for ff over RR is given by
    ∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA. \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A.

If f(x,y)≥0f(x,y) \geq 0 on the rectangle R,R\text{,} we may ask to find the volume of the solid bounded above by ff over R,R\text{,} as illustrated on the left of Figure 11.1.311.1.3. This volume is approximated by a Riemann sum, which sums the volumes of the rectangular boxes shown on the right of Figure 11.1.311.1.3.

fig_11_1_volume.svg fig_11_1_riemann_3_2.svg

Figure 11.1.311.1.3. The volume under a graph approximated by a Riemann Sum.

As we let the number of subrectangles increase without bound (in other words, as both mm and nn in a double Riemann sum go to infinity), as illustrated in Figure 11.1.411.1.4, the sum of the volumes of the rectangular boxes approaches the volume of the solid bounded above by ff over R.R\text{.} The value of this limit, provided it exists, is the double integral.

fig_11_1_riemann_4_4.svg fig_11_1_riemann_8_8.svg

Figure 11.1.411.1.4. Finding better approximations by using smaller subrectangles.
Definition 11.1.211.1.2

Let RR be a rectangular region in the xyxy-plane and ff a continuous function over R.R\text{.} With terms defined as in a double Riemann sum, the double integral of ff over RR is

∬Rf(x,y) dA=lim⁡m,n→∞∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA. \iint_R f(x,y) \, dA = \lim_{m,n \to \infty} \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A.

Some textbooks use the notation ∫Rf(x,y) dA\int_R f(x,y) \, dA for a double integral. You will see this in some of the WeBWorK problems.

Interpretation of Double Riemann Sums and Double integrals.

At the moment, there are two ways we can interpret the value of the double integral.

  • Suppose that f(x,y)f(x,y) assumes both positive and negatives values on the rectangle R,R\text{,} as shown on the left of Figure 11.1.511.1.5. When constructing a Riemann sum, for each ii and j,j\text{,} the product f(xij∗,yij∗)⋅ΔAf(x_{ij}^*, y_{ij}^*) \cdot \Delta A can be interpreted as a “signed” volume of a box with base area ΔA\Delta A and “signed” height f(xij∗,yij∗).f(x_{ij}^*, y_{ij}^*)\text{.} Since ff can have negative values, this “height” could be negative. The sum
    ∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A

    can then be interpreted as a sum of “signed” volumes of boxes, with a negative sign attached to those boxes whose heights are below the xyxy-plane.

fig_11_1_signed_graph.svg fig_11_1_signed_volume.svg

  • Figure 11.1.511.1.5. The integral measures signed volume.
    We can then realize the double integral ∬Rf(x,y) dA\iint_R f(x,y) \, dA as a difference in volumes: ∬Rf(x,y) dA\iint_R f(x,y) \, dA tells us the volume of the solids the graph of ff bounds above the xyxy-plane over the rectangle RR minus the volume of the solids the graph of ff bounds below the xyxy-plane under the rectangle R.R\text{.} This is shown on the right of Figure 11.1.511.1.5.
  • The average of the finitely many mnmn values f(xij∗,yij∗)f(x_{ij}^*, y_{ij}^*) that we take in a double Riemann sum is given by
    Avgmn=1mn∑j=1n∑i=1mf(xij∗,yij∗). \mbox{Avg} _{mn} = \frac{1}{mn} \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*).

    If we take the limit as mm and nn go to infinity, we obtain what we define as the average value of ff over the region R,R\text{,} which is connected to the value of the double integral. First, to view  Avg mn\text{ Avg } _{mn} as a double Riemann sum, note that

    Δx=b−am      and      Δy=d−cn. \Delta x = \frac{b-a}{m} \ \ \ \ \ \text{ and } \ \ \ \ \ \Delta y = \frac{d-c}{n}.

    Thus,

    1mn=Δx⋅Δy(b−a)(d−c)=ΔAA(R), \frac{1}{mn} = \frac{\Delta x \cdot \Delta y}{(b-a)(d-c)} = \frac{\Delta A}{A(R)},

    where A(R)A(R) denotes the area of the rectangle R.R\text{.} Then, the average value of the function ff over R,R\text{,} fAVG⁡(R),f_{\operatorname{AVG}(R)}\text{,} is given by

    fAVG⁡(R)=lim⁡m,n→∞1mn∑j=1n∑i=1mf(xij∗,yij∗)=lim⁡m,n→∞1A(R)∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA=1A(R)∬Rf(x,y) dA.\begin{align*} f_{\operatorname{AVG}(R)} & = \lim_{m,n \to \infty} \frac{1}{mn} \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*)\\[4pt] & = \lim_{m,n \to \infty} \frac{1}{A(R)} \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A\\[4pt] & = \frac{1}{A(R)} \iint_R f(x,y) \, dA. \end{align*}

    Therefore, the double integral of ff over RR divided by the area of RR gives us the average value of the function ff on R.R\text{.} Finally, if f(x,y)≥0f(x, y) \geq 0 on R,R\text{,} we can interpret this average value of ff on RR as the height of the box with base RR that has the same volume as the volume of the surface defined by ff over R.R\text{.}

Activity 11.1.311.1.3

Let f(x,y)=x+2yf(x,y) = x+2y and let R=[0,2]×[1,3].R = [0,2] \times [1,3]\text{.}

  1. Draw a picture of R.R\text{.} Partition [0,2][0,2] into 2 subintervals of equal length and the interval [1,3][1,3] into two subintervals of equal length. Draw these partitions on your picture of RR and label the resulting subrectangles using the labeling scheme we established in the definition of a double Riemann sum.
  2. For each ii and j,j\text{,} let (xij∗,yij∗)(x_{ij}^*, y_{ij}^*) be the midpoint of the rectangle Rij.R_{ij}\text{.} Identify the coordinates of each (xij∗,yij∗).(x_{ij}^*, y_{ij}^*)\text{.} Draw these points on your picture of R.R\text{.}
  3. Calculate the Riemann sum
    ∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A

    using the partitions we have described. If we let (xij∗,yij∗)(x_{ij}^*, y_{ij}^*) be the midpoint of the rectangle RijR_{ij} for each ii and j,j\text{,} then the resulting Riemann sum is called a midpoint sum.

  4. Give two interpretations for the meaning of the sum you just calculated.
Activity 11.1.411.1.4

Let f(x,y)=4−y2f(x,y) = \sqrt{4-y^2} on the rectangular domain R=[1,7]×[−2,2].R = [1,7] \times [-2,2]\text{.} Partition [1,7][1,7] into 3 equal length subintervals and [−2,2][-2,2] into 2 equal length subintervals. A table of values of ff at some points in RR is given in Table 11.1.111.1.1, and a graph of ff with the indicated partitions is shown in Figure 11.1.611.1.6.

Table 11.1.111.1.1. Table of values of f(x,y)=4−y2.f(x,y) = \sqrt{4-y^2}\text{.}
  −2-2 −1-1 00 11 22
11 00 3\sqrt{3} 22 3\sqrt{3} 00
22 00 3\sqrt{3} 22 3\sqrt{3} 00
33 00 3\sqrt{3} 22 3\sqrt{3} 00
44 00 3\sqrt{3} 22 3\sqrt{3} 00
55 00 3\sqrt{3} 22 3\sqrt{3} 00
66 00 3\sqrt{3} 22 3\sqrt{3} 00
77 00 3\sqrt{3} 22 3\sqrt{3} 00
fig_11_1_cylinder.svg
Figure 11.1.611.1.6. Graph of f(x,y)=4−y2f(x,y) = \sqrt{4-y^2} on R.R\text{.}
  1. Sketch the region RR in the plane using the values in Table 11.1.111.1.1 as the partitions.
  2. Calculate the double Riemann sum using the given partition of RR and the values of ff in the upper right corner of each subrectangle.
  3. Use geometry to calculate the exact value of ∬Rf(x,y) dA\iint_R f(x,y) \, dA and compare it to your approximation. Describe one way we could obtain a better approximation using the given data.

We conclude this section with a list of properties of double integrals. Since similar properties are satisfied by single-variable integrals and the arguments for double integrals are essentially the same, we omit their justification.

Properties of Double Integrals

Let ff and gg be continuous functions on a rectangle R={(x,y):a≤x≤b,c≤y≤d},R = \{(x,y) : a \leq x \leq b, c \leq y \leq d\}\text{,} and let kk be a constant. Then

  1. ∬R(f(x,y)+g(x,y)) dA=∬Rf(x,y) dA+∬Rg(x,y) dA.\iint_R (f(x,y) + g(x,y)) \, dA = \iint_R f(x,y) \, dA + \iint_R g(x,y) \, dA\text{.}
  2. ∬Rkf(x,y) dA=k∬Rf(x,y) dA.\iint_R kf(x,y) \, dA = k \iint_R f(x,y) \, dA\text{.}
  3. If f(x,y)≥g(x,y)f(x,y) \geq g(x,y) on R,R\text{,} then ∬Rf(x,y) dA≥∬Rg(x,y) dA.\iint_R f(x,y) \, dA \geq \iint_R g(x,y) \, dA\text{.}

Summary

  • Let ff be a continuous function on a rectangle R={(x,y):a≤x≤b,c≤y≤d}.R = \{(x,y) : a \leq x \leq b, c \leq y \leq d\}\text{.} The double Riemann sum for ff over RR is created as follows.
    • Partition the interval [a,b][a, b] into mm subintervals of equal length Δx=b−am.\Delta x = \frac{b-a}{m}\text{.} Let x0,x_0\text{,} x1,x_1\text{,} …,\ldots\text{,} xmx_m be the endpoints of these subintervals, where a=x0<x1<x2<⋯<xm=b.a = x_0\lt x_1\lt x_2 \lt \cdots \lt x_m = b\text{.}
    • Partition the interval [c,d][c, d] into nn subintervals of equal length Δy=d−cn.\Delta y = \frac{d-c}{n}\text{.} Let y0,y_0\text{,} y1,y_1\text{,} …,\ldots\text{,} yny_n be the endpoints of these subintervals, where c=y0<y1<y2<⋯<yn=d.c = y_0\lt y_1\lt y_2 \lt \cdots \lt y_n = d\text{.}
    • These two partitions create a partition of the rectangle RR into mnmn subrectangles RijR_{ij} with opposite vertices (xi−1,yj−1)(x_{i-1},y_{j-1}) and (xi,yj)(x_i, y_j) for ii between 11 and mm and jj between 11 and n.n\text{.} These rectangles all have equal area ΔA=Δx⋅Δy.\Delta A = \Delta x \cdot \Delta y\text{.}
    • Choose a point (xij∗,yij∗)(x_{ij}^*, y_{ij}^*) in each rectangle Rij.R_{ij}\text{.} Then a double Riemann sum for ff over RR is given by
      ∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA. \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A.
  • With terms defined as in the Double Riemann Sum, the double integral of ff over RR is
    ∬Rf(x,y) dA=lim⁡m,n→∞∑j=1n∑i=1mf(xij∗,yij∗)⋅ΔA. \iint_R f(x,y) \, dA = \lim_{m,n \to \infty} \sum_{j=1}^n \sum_{i=1}^m f(x_{ij}^*, y_{ij}^*) \cdot \Delta A.
  • Two interpretations of the double integral ∬Rf(x,y) dA\iint_R f(x,y) \, dA are:
    • The volume of the solids the graph of ff bounds above the xyxy-plane over the rectangle RR minus the volume of the solids the graph of ff bounds below the xyxy-plane under the rectangle R;R\text{;}
    • Dividing the double integral of ff over RR by the area of RR gives us the average value of the function ff on R.R\text{.} If f(x,y)≥0f(x, y) \geq 0 on R,R\text{,} we can interpret this average value of ff on RR as the height of the box with base RR that has the same volume as the volume of the surface defined by ff over R.R\text{.}

Credit and licence

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