Boise State Math 275 (Ultman)
Worksheet 3.1: Introduction to Double Integrals
Prerequisites
In order to learn the new skills and ideas presented in this worksheet, you must:
� Be able to integrate functions of a single variable.
Goals
In this worksheet, you will:
� Evaluate double integrals on rectangular and general domains.
� Change the order of integration when appropriate to evaluate double integrals.
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 1
Warm-Up
Consider the diagram below:
-5 -4 -3 -2 -1 0 1 2 3 4 5
-3
-2
-1
1
2
3
y = x - 12
y = x + 1
D
A. Sketch a grid in the region D using lines parallel to the x-axis and lines parallel to the
y -axis.
B. Choose one of the rectangles in your grid. Use “dx” to lable the length of a side parallel
to the x-axis, and use “dy” to label the length of the side parallel to the y -axis. Then, if
dA is the area of the rectangle:
dA =
C. What is the equation of the curve that forms the bottom boundary of the region D? What
is the equation of the curve that forms the top boundary of the region D above?
D. Find the the x- and y -coordinates of the two points of intersection of the curves bounding
D.
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 2
Integrating Over Rectangular Domains
1. (a) Sketch the rectangle 0 ≤ x ≤ 5, 0 ≤ y ≤ 3, and label the curves that make up the
boundary. This will be the region of integration D for the next few integrals. Pay
attention to the limits of integration of the integral, and the boundary curves of the
region.
-2 -1 0 1 2 3 4 5 6
-1
1
2
3
4
(b) On your sketch, draw a grid in the region D (the rectangle) using lines parallel to the
coordinate axes. Pick one of the small rectangles in your grid, and label a side parallel to
the x-axis as “dx”, and a side parallel to the y -axis as “dy”. What is the area dA of the
small rectangle, in terms of dx and dy? dA is called the area element.
dA =
(c) Evaluate the following integral using the order of integration dA = dx dy . First, evaluate
the “inside” integral with respect to x , treating y as a constant. Then evaluate what’s
left with respect to y :
¨D
4xy dA =
ˆ 30
ˆ 50
4xy dx dy =
ˆ 30
(ˆ 50
4xy dx
)dy
(d) Now evaluate the same integral using the order of integration dA = dy dx (evaluate
“inside” integral with respect to y treating x constant, then evaluate what’s left with
respect to x) and compare the result with your answer from #1. (They should be the
same.) ¨D
4xy dA =
ˆ 50
ˆ 30
4xy dy dx =
ˆ 50
(ˆ 30
4xy dy
)dx
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 3
Double Integrals Over General Regions
2. Suppose we want to integrate f (x, y) = 2x over the triangular domain D pictured below, using
the order of integration dA = dy dx .
(a) First, find the limits of integration with respect to y (since it is the “inside” integral).
Think about moving through the domain D vertically, from bottom to top.
Draw an arrow parallel to the y -axis, starting below the region D and ending above
the region D:
-1 0 1 2 3 4 5
-1
1
2
3
y = - x/2 + 2
D
y = 2
x = 4
• The lower limit of integration with respect to y is the equation of the line where the
arrow enters at the bottom of D.
• The upper limit of integration with respect to y is the equation of the line where
the arrow leaves at the top of D.
Lower limit with respect to y (arrow enters D) is: y =
Upper limit with respect to y (arrow leaves D) is: y =
continued
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 4
(b) Now, find the limits with respect to x . Since this is the “outside” integral, these limits
must be constant!
To find the limits with respect to x , project the region D onto the x-axis, and find the
x-values covered by the region:
-1 0 1 2 3 4 5
-1
1
2
3
y = - x/2 + 2
D
y = 2
x = 4
• The lower limit of integration with respect to x is the smallest x-value covered by
the region D.
• The upper limit of integration with respect to x is the largest x-value covered by
the region D.
Lower limit with respect to x (smallest x-value) is: x =
Upper limit with respect to x (largest x-value) is: x =
(c) Using the limits of integration found in parts (a) and (b), set up and evaluate the integral:
¨D
2x dA =
ˆ x=x=
ˆ y=y=
2x dy dx
=
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 5
(d) You can also use the other order of integration, dA = dx dy . The “inside” integral is
evaluated with respect to x , so draw the arrow parallel to the x-axis, from left to right.
To find the limits with respect to y (which will be constant), project the region D onto
the y -axis.
-1 0 1 2 3 4 5
-1
1
2
3
y = - x/2 + 2
D
y = 2
x = 4
• The lower limit of integration with respect to x is the equation of the line where the
arrow enters at the left of D. You will need to solve for x!
• The upper limit of integration with respect to x is the equation of the line where
the arrow leaves at the right of D.
Lower limit with respect to x (arrow enters D) is: x =
Upper limit with respect to x (arrow leaves D) is: x =
• The lower limit of integration with respect to y is the smallest y -value covered by
the region D.
• The upper limit of integration with respect to y is the largest y -value covered by
the region D.
Lower limit with respect to y (smallest y -value) is: y =
Upper limit with respect to y (largest y -value) is: y =
continued
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 6
Using the limits of integration found in part (d), set up and evaluate the double integral.
¨D
2x dA =
ˆ y=y=
ˆ x=x=
2x dx dy
=
(e) You should get the same answer for the two integrals in parts (c) and (d) — you are
integrating the same function (f (x, y) = 2x) over the same region (the triangle D). Was
it easier to integrate using the order of integration dA = dy dx from part (c), or the
order dA = dx dy from part (d)? Which one? Wny?
(f) Conclusions:
i. The outer limits of integration will always be constant.
Always True Sometimes True Never True
ii. To change the order of integration, just “swap” the limits from the first integral to
get the limits of the second integral.
Always True Sometimes True Never True
iii. The value of the integral is the same, regardless of which order of integration you use.
Always True Sometimes True Never True
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 7
Double Integrals and Area
3. (a) Consider the integral:
¨D
dA =
¨D
1 dA
=
ˆ 4−1
ˆ 20
1 dy dx
=
ˆ 4−1y∣∣∣20dx
=
ˆ 4−1
2− 0 dx
= 2
ˆ 4−1dx
= 10
i. Using the limits of integration, draw the rectangular domain D of this integral.
-2 -1 0 1 2 3 4 5 6
-1
1
2
3
4
ii. What does the integral in (a) represent with respect to the domain you drew in (b)?
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 8
(b) Consider the integral:
¨D
dA =
¨D
1 dA
=
ˆ 10
ˆ y0
1 dx dy
=
ˆ 10
x∣∣∣y0dy
=
ˆ 10
y − 0 dy
= y 2/2∣∣∣10
= 1/2
i. Using the limits of integration, draw the triangular domain D of this integral.
-2 -1 0 1 2 3 4 5 6
-1
1
2
3
4
ii. What does the integral in (a) represent with respect to the domain you drew in (b)?
(c) In general, what is represented by the integral:
¨D
dA =
¨D
1 dA?
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 9
Changing Order of Integration
4. Sometimes, it may be useful (or necessary!) to reverse the order of integration for a given
integral. Consider the integral:
¨D
ey2
dA =
ˆ x=2x=0
ˆ y=1y=x/2
ey2
dy dx.
(a) Sketch the domain of integration D using the limits of integration from this integral.
(b) Using your sketch from part (a), find the limits for the order of integration dA = dx dy .
Use these limits to set up and evaluate the integral with the order of integration reveresed.
(c) What do you gain by reversing the order of integration in this case?
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 10
Regions Requiring More Than One Double Integral
5. Go back to the region D pictured in the Warm-up.
(a) Set up the double integral˜Df (x, y) dA, using the order of integration dA = dy dx .
(You don’t know the function f (x, y), so you cannot evalute.)
(b) Now, set up the double integral˜Df (x, y) dA, using the order of integration dA = dx dy .
How many double integrals do you need for this order of integration? Why?
Boise State Math 275 (Ultman) Worksheet 3.1: Introduction to Double Integrals 11
Summary
Double integral of f (x, y) over a planar region D:
¨D
f (x, y) dA
◦ dA is called the area element. It measures the area of an infinitesimal rectangle in the
plane. In Cartesian coordinates, dA = dx dy = dy dx .
◦ D is the region of integration in the xy -plane. The boundary curves of D determine the
limits of integration.
◦ If f (x, y) is continuous, and the boundary curves of D are continuous, the double integral
can be evaluated as two single integrals.
• The “inside” integral is evaluated with respect to the indicated variable using tech-
niques from Calc I/II, while holding the other variable constant (compare to partial
derivatives!).
• The limits of integration of the “outside” integral are always constant.
Using the order of integration dA = dy dx, with g1(x) ≤ y ≤ g2(x) and a ≤ x ≤ b :
¨D
f (x, y) dA =
ˆ ba
ˆ g2(x)g1(x)
f (x, y) dy dx =
ˆ ba
(ˆ g2(x)g1(x)
f (x, y) dy
)dx
Using the order of integration dA = dx dy, with h1(y) ≤ x ≤ h2(y) and c ≤ y ≤ d :
¨D
f (x, y) dA =
ˆ dc
ˆ h2(y)h1(y)
f (x, y) dx dy =
ˆ dc
(ˆ h2(y)h1(y)
f (x, y) dx
)dy
◦ If f (x, y) is continuous, and the boundary curves of D are continuous, the order of inte-
gration does not matter:
¨D
f (x, y) dA =
ˆ ba
ˆ g2(x)g1(x)
f (x, y) dy dx =
ˆ dc
ˆ h2(y)h1(y)
f (x, y) dx dy
