Double Integrals over Rectangular Regions

Introduction to Double Integrals

We've handled limits and derivatives in three dimensions; what about integrals? Remember what we know about integrals in two dimensions: the integral of f(x)f(x) from aa to bb gives the area under that function between those limits.

Integral as area under a curve

When we go from two dimensions to three, area becomes volume, and instead of just having limits on xx, we'll also have limits on yy, forming a region RR in the xyxy plane.

Double integral as volume under a surface

To find the volume of this body, cut it into thin slices in one direction, as shown below.

Slicing the volume like a loaf of bread

The slice

Then the volume is given by the integral V=∫A(x) dxV=∫A(x)\ dx where the area is A(x)=∫f(x,y) dy.A(x)=\int f(x,y)\ dy.

Therefore, the volume is V=∬f(x,y) dy dx.V = \iint f(x,y)\ dy\ dx.

Fubini's Theorem

Let ff be continuous on the rectangular region R={(x,y) : a≤x≤b,c≤y≤d}.R={(x,y)\ :\ a \leq x \leq b, c \leq y \leq d}. Then ∬Rf(x,y) dA=∫cd∫abf(x,y) dx dy=∫ab∫cdf(x,y) dy dx.\iint_R f(x,y)\ dA = \int_c^d \int_a^b f(x,y)\ dx\ dy = \int_a^b \int_c^d f(x,y)\ dy\ dx.

In other words, the order of integration over a rectangular region doesn't matter.

How do we actually evaluate one of these double integrals? The answer is similar to what we found with partial derivatives; when integrating with respect to one variable, treat the other variable as a constant.

Evaluate the following integral. ∫13∫011+4xy dx dy\int_1^3 \int_0^1 1+4xy\ dx\ dy

Solution

Start with the inner integral and work outwards. Thus, integrate the function with respect to xx first, treating yy as a constant:

∫13∫011+4xy dx dy=∫13[x+2x2y]01 dy=∫13[(1+2y)−(0+0)] dy=∫131+2y dy=y+y2∣13=(3+9)−(1+1)=10\begin{aligned} \int_1^3 \int_0^1 &1+4xy\ dx\ dy\ &= \int_1^3 \bigg[ x+2x^2y \bigg]_0^1\ dy\ &= \int_1^3 \bigg[ (1+2y)-(0+0) \bigg]\ dy\ &= \int_1^3 1+2y\ dy\ &= y+y^2\bigg|_1^3\ &= (3+9)-(1+1) = \ans{10} \end{aligned}

Alternate Solution

Note that if we switch the order of integration, we get the same result (Fubini's Theorem):

∫01∫131+4xy dy dx=∫01[y+2xy2]13 dx=∫012+16x dx=2x+8x2∣01=10\begin{aligned} \int_0^1 \int_1^3 &1+4xy\ dy\ dx\ &= \int_0^1 \bigg[ y+2xy^2 \bigg]_1^3\ dx\ &= \int_0^1 2+16x\ dx\ &= 2x+8x^2\bigg|_0^1\ &= \ans{10} \end{aligned}

Find the volume of the solid bounded by the surface f(x,y)=4+9x2y2f(x,y)=4+9x^2y^2 over the region R={(x,y) : −1≤x≤1,0≤y≤2}.R={(x,y)\ :\ -1 \leq x \leq 1, 0 \leq y \leq 2}.

Solution

The volume of this solid is the double integral of the function over the given rectangular region:

V=∫−11∫024+9x2y2 dy dx=∫−11[4y+3x2y3]02 dx=∫−118+24x2 dx=8x+8x3∣−11=32\begin{aligned} V &= \int_{-1}^1 \int_0^2 4+9x^2y^2\ dy\ dx &= \int_{-1}^1 \bigg[ 4y+3x^2y^3 \bigg]0^2\ dx\ &= \int{-1}^1 8+24x^2\ dx\ &= 8x+8x^3 \bigg|_{-1}^1 = \ans{32} \end{aligned}

Evaluate the following integral, where R={(x,y) : 0≤x≤1,0≤y≤ln⁡2}.R={(x,y)\ :\ 0 \leq x \leq 1, 0 \leq y \leq \ln 2}. ∬Ryexy dA\iint_R ye^{xy}\ dA

Solution

Take a second to think before doing the integral. Since this is a rectangular region, we can integrate in either order, so we need to decide which order to use. If we integrate with respect to yy first, we'll need to use integration by parts, which isn't impossible, but more difficult than what we get if we integrate with respect to xx first.

∫0ln⁡2∫01yexy dx dy=∫0ln⁡2[exy]01 dy=∫0ln⁡2ey−1 dy=ey−y∣0ln⁡2=2−ln⁡2−1=1−ln⁡2\begin{aligned} \int_0^{\ln 2} \int_0^1 &ye^{xy}\ dx\ dy\ &= \int_0^{\ln 2} \bigg[ e^{xy} \bigg]_0^1\ dy\ &= \int_0^{\ln 2} e^y-1\ dy\ &= e^y-y\bigg|_0^{\ln 2}\ &= 2-\ln 2-1 = \ans{1-\ln 2} \end{aligned}

  1. Integrate f(x,y)=x2+xyf(x,y)=x^2+xy over the region R={(x,y) : 1≤x≤2,−1≤y≤1}.R={(x,y)\ :\ 1 \leq x \leq 2, -1 \leq y \leq 1}.

    2. 143\dfrac{14}{3}

  2. Integrate f(x,y)=4x3cos⁡yf(x,y)=4x^3\cos y over the region R={(x,y) : 1≤x≤2,0≤y≤π/2}.R={(x,y)\ :\ 1 \leq x \leq 2, 0 \leq y \leq \pi/2}.

    3. 1515

  3. Integrate f(x,y)=y1−x2f(x,y)=\dfrac{y}{\sqrt{1-x^2}} over the region R={(x,y) : 1/2≤x≤3/2,1≤y≤2}R={(x,y)\ :\ 1/2 \leq x \leq \sqrt{3}/2, 1 \leq y \leq 2}.

    4. π4\dfrac{\pi}{4}

  4. Integrate f(x,y)=xyf(x,y)=\sqrt{\dfrac{x}{y}} over the region R={(x,y) : 0≤x≤1,1≤y≤4}.R={(x,y)\ :\ 0 \leq x \leq 1, 1 \leq y \leq 4}.

    5. 43\dfrac{4}{3}

  5. Integrate f(x,y)=xysin⁡x2f(x,y)=xy \sin x^2 over the region R={(x,y) : 0≤x≤π/2,0≤y≤1}.R={(x,y)\ :\ 0 \leq x \leq \sqrt{\pi/2}, 0 \leq y \leq 1}.

    6. 14\dfrac{1}{4}

  6. Integrate f(x,y)=ex+2yf(x,y)=e^{x+2y} over the region R={(x,y) : 0≤x≤ln⁡2,1≤y≤ln⁡3}.R={(x,y)\ :\ 0 \leq x \leq \ln 2, 1 \leq y \leq \ln 3}.

    7. 12(9−e2)\dfrac{1}{2}(9-e^2)

Average Value

Recall from Calculus I that the average value of a function over an interval from aa to bb is f‾=1b−a∫abf(x) dx.\overline{f} = \dfrac{1}{b-a} \int_a^b f(x)\ dx.

If we generalize this to three dimensions, the length b−ab−a becomes the area of the region RR, and the integral becomes a double integral: f‾=1area of R∬Rf(x,y) dA\ans{\overline{f} = \dfrac{1}{\textrm{area of } R} \iint_R f(x,y)\ dA}

Find the average value of f(x,y)=sin⁡xsin⁡yf(x,y)=\sin x \sin y over the region R={(x,y) : 0≤x≤π,0≤y≤π}.R={(x,y)\ :\ 0 \leq x \leq \pi, 0 \leq y \leq \pi}.

Solution

The area of this region is π2π^2, so the average value is f‾=1π2∫0π∫0πsin⁡x sin⁡y dx dy=1π2∫0π[−cos⁡x sin⁡y]0π dy=1π2∫0π2sin⁡y dy=1π2[−2cos⁡y]0π=4π2\begin{aligned} \overline{f} &= \dfrac{1}{\pi^2} \int_0^\pi \int_0^\pi \sin x\ \sin y\ dx\ dy\ &= \dfrac{1}{\pi^2} \int_0^\pi \bigg[ -\cos x\ \sin y \bigg]_0^\pi\ dy\ &= \dfrac{1}{\pi^2} \int_0^\pi 2\sin y\ dy\ &= \dfrac{1}{\pi^2} \bigg[-2\cos y \bigg]_0^\pi\ &= \ans{\dfrac{4}{\pi^2}} \end{aligned}

  1. Find the average value of f(x,y)=4−x−yf(x,y)=4−x−y over the region R={(x,y) : 0≤x≤2,0≤y≤2}.R={(x,y)\ :\ 0 \leq x \leq 2, 0 \leq y \leq 2}.

    2. 22

  2. Find the average squared distance between the points of R={(x,y) : −2≤x≤2,0≤y≤2}R={(x,y)\ :\ -2 \leq x \leq 2, 0 \leq y \leq 2} and the origin.

    3. 83\dfrac{8}{3}