Unit 5, chapter 20

Accumulate over two coordinates

How a double integral over a rectangular region adds thin columns to give the volume under a surface, and how an iterated integral computes it through inner and outer integrals by holding one variable fixed.

  • Chapter 20
  • Double integrals and the order of integration

A roof over a square floor of side 11 follows the plane z=x+yz = x + y, so it has height 00 above one corner and height 22 above the opposite corner. How much space lies between the floor and the roof? A single integral adds strips along one axis, but every point of this floor needs two coordinates, and the height changes in both directions. This lesson builds the double integral from thin columns, evaluates it as an iterated integral one variable at a time, and finds that the volume is exactly 11.

Key terms

Rectangular region

Definition For real numbers a<ba < b and c<dc < d, the closed rectangle R=[a,b]×[c,d]R = [a, b] \times [c, d] is the set of points (x,y)(x, y) with a≤x≤ba \le x \le b and c≤y≤dc \le y \le d. Its sides are parallel to the axes and its area is (b−a)(d−c)(b - a)(d - c). On a rectangular region every bound is a constant. The range of yy is the same interval [c,d][c, d] for every xx, and the range of xx is [a,b][a, b] for every yy.

In plain words A rectangle with sides along the axis directions is the simplest base for a double integral. Slicing it in either direction gives segments with the same endpoints every time, so no bound depends on the other variable.

Example The unit square [0,1]×[0,1][0, 1] \times [0, 1] has area 11. Cutting each side into nn equal pieces splits it into n2n^2 small squares of side 1n\tfrac1n and area 1n2\tfrac{1}{n^2}. By contrast, the triangle of points with 0≤x≤y≤10 \le x \le y \le 1 is not a rectangular region, because at a given xx the variable yy runs only from xx to 11. Regions of that kind are the subject of the next lesson.

Double integral

Definition Let ff be a bounded function on R=[a,b]×[c,d]R = [a, b] \times [c, d]. Points a=x0<x1<⋯<xm=ba = x_0 < x_1 < \dots < x_m = b and c=y0<y1<⋯<yn=dc = y_0 < y_1 < \dots < y_n = d cut RR into the cells Rij=[xi−1,xi]×[yj−1,yj]R_{ij} = [x_{i-1}, x_i] \times [y_{j-1}, y_j], with areas ΔAij=Δxi Δyj\Delta A_{ij} = \Delta x_i \, \Delta y_j. Pick a sample point (xij∗,yij∗)(x_{ij}^*, y_{ij}^*) in each cell and form the double Riemann sum

∑i=1m∑j=1nf(xij∗,yij∗) ΔAij.\sum_{i=1}^{m} \sum_{j=1}^{n} f(x_{ij}^*, y_{ij}^*)\,\Delta A_{ij}.

Suppose these sums approach one number II as the longest cell diagonal shrinks to 00, whatever sample points are chosen. Then ff is integrable on RR, and II is its double integral,

∬Rf(x,y) dA=I.\iint_R f(x, y)\,dA = I.

Every function that is continuous on RR is integrable there. The symbol dAdA stands for the area of a small cell of the base.

In plain words Tile the base, stand a thin column on each tile as tall as ff at the chosen point, and add up the products of height and tile area. The double integral is the number these totals settle on as the tiles shrink in both directions.

Example Take f(x,y)=x+yf(x, y) = x + y on the unit square, cut into n2n^2 equal squares of area 1n2\tfrac{1}{n^2}. The function grows in both variables, so on each cell its smallest value is at the lower left corner and its largest value is at the upper right corner. With n=2n = 2, the four lower left corners give heights that add to 22, namely 00, 12\tfrac12, 12\tfrac12 and 11, and the four upper right corners give 11, 32\tfrac32, 32\tfrac32 and 22, which add to 66. So the lowest and highest sums on this grid are

L2=14⋅2=12,U2=14⋅6=32.\begin{aligned} L_2 &= \tfrac14 \cdot 2 = \tfrac12, \\[0.4em] U_2 &= \tfrac14 \cdot 6 = \tfrac32. \end{aligned}

For general nn the same count gives Ln=1−1nL_n = 1 - \tfrac1n and Un=1+1nU_n = 1 + \tfrac1n, and every Riemann sum on that grid lies between them. Both bounds approach 11, and ff is continuous, hence integrable, so

∬[0,1]2(x+y) dA=1.\iint_{[0, 1]^2} (x + y)\,dA = 1.

Volume under a surface

Definition Let ff be integrable on RR with f(x,y)≥0f(x, y) \ge 0 at every point of RR. The solid of points (x,y,z)(x, y, z) with (x,y)(x, y) in RR and 0≤z≤f(x,y)0 \le z \le f(x, y) has volume

V=∬Rf(x,y) dA.V = \iint_R f(x, y)\,dA.

If ff takes negative values, the integral is a signed volume. The space between the graph and the base counts as positive where the graph lies above the xyxy plane and as negative where it lies below. The volume between the graph and the base is then the integral of ∣f∣\lvert f \rvert.

In plain words A column with base area ΔA\Delta A and height ff has volume f ΔAf \, \Delta A, so adding columns adds volume. Reading the integral as a volume needs a height that never dips below the base.

Example The roof z=x+yz = x + y over the unit square is continuous and nonnegative, so the double integral 11 found above is the volume under it, in cubic units. Geometry confirms the value. The box [0,1]×[0,1]×[0,2][0, 1] \times [0, 1] \times [0, 2] has volume 22. The point reflection through its center sends (x,y,z)(x, y, z) to (1−x,1−y,2−z)(1 - x, 1 - y, 2 - z) and maps the box onto itself. A point satisfies z≤x+yz \le x + y exactly when its image satisfies 2−z≥(1−x)+(1−y)2 - z \ge (1 - x) + (1 - y), so the reflection swaps the part of the box below the roof with the part above it. The two parts are congruent and share only the roof, which has no volume, so each part has volume 11.

Iterated integral

Definition For a function ff on R=[a,b]×[c,d]R = [a, b] \times [c, d], the iterated integral in the order dy dxdy\,dx is

∫ab∫cdf(x,y) dy dx=∫ab[∫cdf(x,y) dy]dx.\begin{aligned} &\int_a^b \int_c^d f(x, y)\,dy\,dx \\[0.4em] &= \int_a^b \left[ \int_c^d f(x, y)\,dy \right] dx. \end{aligned}

The bracket is the inner integral, taken first in yy from cc to dd. Its value is then integrated by the outer integral, in xx from aa to bb. The differentials are read from the inside out, and each pair of limits belongs to the differential in the matching position, the innermost limits to the innermost differential. The iterated integral is defined when the inner integral exists for every xx in [a,b][a, b] and its value is an integrable function of xx. The order dx dydx\,dy is defined in the same way, with the roles of xx and yy exchanged. If ff is continuous on RR, both iterated integrals exist and both equal ∬Rf dA\iint_R f\,dA. This is Fubini’s theorem for a rectangle.

In plain words Add along one direction first, then add the results along the other. The inner integral sweeps out one slice of the solid, and the outer integral moves that slice across the base.

Example The volume under the roof z=x+yz = x + y is

∫01∫01(x+y) dy dx,\int_0^1 \int_0^1 (x + y)\,dy\,dx,

and the integrand is continuous on the unit square, so this iterated integral equals the double integral. The inner integral, with xx held fixed, is

∫01(x+y) dy=[xy+y22]y=0y=1=x+12.\begin{aligned} &\int_0^1 (x + y)\,dy \\[0.4em] &\quad = \left[ xy + \frac{y^2}{2} \right]_{y=0}^{y=1} \\[0.4em] &\quad = x + \frac12. \end{aligned}

The outer integral adds these values across the square,

∫01(x+12)dx=[x22+x2]x=0x=1=1.\begin{aligned} &\int_0^1 \left( x + \frac12 \right) dx \\[0.4em] &\quad = \left[ \frac{x^2}{2} + \frac{x}{2} \right]_{x=0}^{x=1} = 1. \end{aligned}

In the other order the inner integral is ∫01(x+y) dx=y+12\int_0^1 (x + y)\,dx = y + \tfrac12, and its integral from 00 to 11 is again 11.

Holding one variable fixed

Definition In the inner integral ∫cdf(x,y) dy\int_c^d f(x, y)\,dy, the outer variable xx is held fixed. It is treated as a constant while yy runs from cc to dd, so a term without yy integrates like a constant, and the evaluation substitutes y=dy = d and y=cy = c only. The result

A(x)=∫cdf(x,y) dyA(x) = \int_c^d f(x, y)\,dy

is a function of xx alone. When f≥0f \ge 0, the number A(x)A(x) is the area of the cross section of the solid by the plane perpendicular to the xx axis at that xx. The outer integral ∫abA(x) dx\int_a^b A(x)\,dx then adds these areas, as in the slicing method for volumes.

In plain words Freeze xx, and the problem becomes an ordinary integral in one variable. Its answer still depends on which xx was frozen, so it is the area of one slice and not yet the volume.

Example For the roof z=x+yz = x + y, freeze xx. Over the segment 0≤y≤10 \le y \le 1 the height rises in a straight line from xx to x+1x + 1, so the slice is a trapezoid of width 11 with parallel sides xx and x+1x + 1. Its area is 12(x+x+1)=x+12\tfrac12 (x + x + 1) = x + \tfrac12, which is exactly the inner integral. At x=0x = 0 the slice is a triangle of area 12\tfrac12, and at x=1x = 1 its area is 32\tfrac32. A common slip is to integrate the term xx as if xx were the variable and write x22\tfrac{x^2}{2}. Since xx is a constant here, its integral in yy is xyxy, which contributes x⋅(1−0)=xx \cdot (1 - 0) = x, the area of the rectangular part of the slice.

Check yourself

  1. Find the volume of the solid that lies above the rectangle [0,1]×[0,2][0, 1] \times [0, 2] and below the plane z=4−x−yz = 4 - x - y.
Show the answer

Every point of the rectangle has x+y≤3x + y \le 3, so the height is at least 11 and the integral is a volume. Holding xx fixed,

∫02(4−x−y) dy=[4y−xy−y22]y=0y=2=6−2x.\begin{aligned} &\int_0^2 (4 - x - y)\,dy \\[0.4em] &\quad = \left[ 4y - xy - \frac{y^2}{2} \right]_{y=0}^{y=2} \\[0.4em] &\quad = 6 - 2x. \end{aligned}

Then ∫01(6−2x) dx=6−1=5\int_0^1 (6 - 2x)\,dx = 6 - 1 = 5. As a check, the height is linear, so its average over the rectangle is its value at the center (12,1)(\tfrac12, 1), which is 52\tfrac52, and 52\tfrac52 times the area 22 is 55.

  1. Evaluate ∬R(x2+y) dA\iint_R (x^2 + y)\,dA over R=[0,1]×[0,2]R = [0, 1] \times [0, 2] in both orders. A classmate writes the order dx dydx\,dy as ∫01∫02(x2+y) dx dy\int_0^1 \int_0^2 (x^2 + y)\,dx\,dy, keeping the limits where they were. What does that expression compute?
Show the answer

In the order dy dxdy\,dx the inner integral is ∫02(x2+y) dy=2x2+2\int_0^2 (x^2 + y)\,dy = 2x^2 + 2, and ∫01(2x2+2) dx=23+2=83\int_0^1 (2x^2 + 2)\,dx = \tfrac23 + 2 = \tfrac83. In the order dx dydx\,dy the inner integral is ∫01(x2+y) dx=13+y\int_0^1 (x^2 + y)\,dx = \tfrac13 + y, and ∫02(13+y) dy=23+2=83\int_0^2 (\tfrac13 + y)\,dy = \tfrac23 + 2 = \tfrac83. The two orders agree, as they must for a continuous integrand on a rectangle.

The classmate’s expression pairs the inner limits 00 and 22 with dxdx, so xx runs over [0,2][0, 2] and yy over [0,1][0, 1]. That is a different rectangle, and the value is different too. The inner integral is ∫02(x2+y) dx=83+2y\int_0^2 (x^2 + y)\,dx = \tfrac83 + 2y, and

∫01(83+2y)dy=113.\int_0^1 \left( \frac83 + 2y \right) dy = \frac{11}{3}.

When the order changes, the limits move with their differentials.

  1. Evaluate ∫02∫01(x−1) dy dx\int_0^2 \int_0^1 (x - 1)\,dy\,dx. Is the result the volume of the region between the plane z=x−1z = x - 1 and the rectangle [0,2]×[0,1][0, 2] \times [0, 1]?
Show the answer

The inner integral is ∫01(x−1) dy=x−1\int_0^1 (x - 1)\,dy = x - 1, and ∫02(x−1) dx=0\int_0^2 (x - 1)\,dx = 0. The result is not that volume. The plane lies below the base where x<1x < 1 and above it where x>1x > 1, so the integral is a signed volume. The part over 0≤x≤10 \le x \le 1 contributes −12-\tfrac12 and the part over 1≤x≤21 \le x \le 2 contributes 12\tfrac12, and they cancel. The volume between the plane and the base is the integral of ∣x−1∣\lvert x - 1 \rvert, which is 12+12=1\tfrac12 + \tfrac12 = 1. Reading a double integral as a volume needs a nonnegative height.

How this lesson connects